This measure theory series is beneficial to humanity. Thank you for making it. After an exhaustive search, this is the best explanation for Lebesgue integral.
This series is so helpful. I’m trying to understand probability theory rigorously. Your video and some of my own reading just help me realize that random variables are nothing but measurable functions. This has transformed how I look at probabilities in general. Before I always thought there was some randomness involved. Now Using the concept of measure, I feel like the likelihood of events can be categorized by measure. This is a static global perspective, and it just makes everything easier to think about and work with. Very exciting stuff to learn!
Probability theory is literally just measure theory constrained to measure spaces (X, A, μ) such that μ(X) = 1. In other words, every probability space is isomorphic to a measure space of finite measure.
For those who would like to see the connection between the Riemann integral and the measure-theoretic integrals, I am going to go back to some definitions, and analyze the fundamental mathematical object underlying both definitions. Consider a function f : [a, b] -> R. A tag on the interval [a, b] is an n-tuple t : {m : 0 =< m =< n} -> [a, b], where if i < j, then t(i) < t(j). If such a tag t is defined, then the sum of f[t(i)]·w(i) is well-defined. Here, w is an n-tuple of real numbers, this time not constrained to the interval [a, b], and this is called a weight. This is the fundamental object of study. This sum is called a w-weighted sum of f over the tag t. Now, consider the set of closed subintervals of [a, b], denoted c([a, b]), and consider a map p : range(t) -> c([a, b]), with the consideration that the intersection of any two closed intervals in c([a, b]) is either the common endpoint of both intervals, or the empty set, with the consideration that the union of the range of p is [a, b], and with the consideration that t(i) is an element of the closed interval p[t(i)]. This turns p into an interval-partition with a tag t, and the endpoints of these intervals are denoted x(i). If we define w(i) := x(i + 1) - x(i), then this special case of the sum of f[t(i)]·w(i) is called a Riemann sum of f. Riemann sums of f are what we use to define the Riemann integral of f. When considering the interval-partition p, it may be more general to consider w(i) = μ(p[t(i)]), instead of w(i) := x(i + 1) - x(i). The idea here is that p[t(i)] is equal to a closed interval, whose value depends on the value of t(i), and the endpoints of this interval are x(i) and x(i + 1). μ can be interpreted to be a "size function" at the intuitive level, and so the "standard" size of an interval p[t(i)], with endpoints x(i + 1) and x(i), is x(i + 1) - x(i). In this generalization, one can allow for flexibility as to what μ is, as to not be restricted to the standard size of a closed interval. So the sum that you obtain from this is different from a Riemann sum, and is instead equal to f[t(i)]·μ(p[t(i)]), which you may call a μ-sum. The generalization can now have for f[t(i)] replaced by a lower bound of the set sp(f, p) := {f[t(i)] : x(i) =< t(i) =< x(i + 1)}. This lower bound can simply be called c(i), so the weighted sum is the sum of c(i)·μ(p[t(i)]). Now, this is really starting to resemble symbolically the kind of weigted sum that the measure-theoretic integrals are defined by, if μ is interpreted as being a measure, or measure-like. Finally, some generalizations can be made here: rather than defining p to assign a closed subinterval of [a, b] to every point of the tag t, p can simply assign a measurable set S(i) to every t(i), so long as the union of S(i) is still equal to [a, b], and so long as the pairwise intersection of S(i) is a null set. Furthermore, the domain of f allowed, rather than being a closed interval [a, b], can be generalized to simply be a set X, with (X, A) being a measurable space, and f being a measurable function. It is the supremum of the set of every such sum in this generalization that is equal to the μ-integral. So what is the Riemann integral with respect to this μ-theory? It is an integral with respect to μ, where μ is specifically the Lebesgue measure, restricted to closed intervals, and with c(i) := f[t(i)], and with the added restriction that it only is the integral of f if the infimum of the aforementioned set, where c(i) is instead an upper bound of sp(f, p), rather than a lower bound. The gauge integral is also defined using these weighted sums of functions over tags, together with restrictions as well.
Actually, the connection between these different integrals is simpler, more obvious, and more intuitive if we start with the Darboux integral, rather than the Riemann integral. Firstly, we go back to the weighted sums of f[t(i)]·w(i). Each w(i) can immediately be interpreted as being the output of a function of the partition intervals of [a, b]: that is, a function of [x(i), x(i + 1)]. So w(i) = μ([x(i), x(i + 1)]). Now, we have the sums of f[t(i)]·μ([x(i), x(i + 1)]). If μ([x(i), x(i + 1)]) = Δx(i), then we are back to the Riemann sums f[t(i)]·Δx(i), but rather considering these Riemann sums, we can go to the upper Darboux sums, and the lower Darboux sums, giving us sums of sup({c(i) : c(i) = f[t(i)] & t(i) is in [x(i), x(i + 1)]})·Δx(i), and inf({c(i) : c(i) = f[t(i)] & t(i) is in [x(i), x(i + 1)]})·Δx(i) respectively. The idea is now to generalize these Darboux sums, by going back to looking at Δx(i), and again considering μ([x(i), x(i + 1)]), where Δx results from the special case when μ is the Lebesgue measure. When applied to lower Darboux sums specifically, you sum with respect to this measure μ, and then you take the supremum of every such sum, obtaining something analogous to the lower Darboux integral, but with respect to μ. This lower μ-integral is precisely the definition of the Lebesgue integral of f with respect to μ, as presented in the video, though this definition is restricted only to measurable functions f : [a, b] -> R with the Borel σ-algebra on both spaces, which can be made more general by allowing a domain X and an arbitrary partition (X(0), ..., X(n)), where the sums are now sup({c(i) : c(i) = f[t(i)] & t(i) is an element of X(i)})·μ[X(i)] and inf({c(i) : c(i) = f[t(i)] & t(i) is an element of X(i)})·μ[X(i)]. This definition also restricts μ to being a finite measure. The approach in the video is retained if this definition is made to apply to f+ and f- individually, rather than f itself, and if 0·♾ = 0. So in summary, the lower Darboux integral is exactly equal to the Lebesgue integral with respect to the Lebesgue measure, where the integrable functions are those whose nonnegative parts are Borel-measurable, specifically.
I am going to follow a few lectures about Lebesgue integrals. A friend of mine who is good at maths warned me it will be very difficult. These videos give me a very clear idea and intuitions about what I will be learning. It gives me a framework everything will fit in hopefully. Thank you so much!
I have an exam of Real Analysis I for thesis qualification, and you just give me a brief understanding of measure. Wish me luck cuz I really did not get familiar with proofs yet, and already no much time left. Thank you, your videos really helpful.
Thank you so much. You're a life saver. My professor just sent me notes, no lecture, and I had a hard time understanding her notes, until I saw your videos. Thank you.
These videos are great, usually when I'm watching math videos on youtube they are quite old. I was shocked when I saw how recently these have come out. Looking forward to more!
Such a good video at understanding the reasoning behind how we are approximating the integral. I feel like many books/videos don't do a good job at providing the reasoning behind taking the supremum. Thanks for this!
These are the best videos on the measure theory one can make. It would be great if you could make more videos on other areas of mathematics like TOPOLOGY,REAL ANALYSIS
I'd like to pile onto that by saying that if you went into measure-theoretic probability, that would actually be really neat. By that, I mean if you talked about, say, the normal distribution and that sort of thing and cast it all the way back down into measure theory and then built it up to the computational statistics we do with it.
@@ryanhutchins2634 Distributions is more a topic of statistics and probability than probability theory, but nonetheless, it would be nice if this channel could cover that.
I met this topic in Real Analysis during my MS yet I can't remember anything. Now that I am taking my PhD I am starting to appreciate it. I am majoring Statistics but Real Analysis is a corequisite for Probability Theory. Please make lectures on Theory of Probabilities. Thank you.
Some are tough to believe but we need to believe and learn further...I imagine always how do I proceed further when one thing is not yet clear. But its a nice feeling of the subject!
Prove the excision property for the outer measure m∗ , that is “If A is a measurable set and B ⊇ A, then prove m∗ (B \ A) = m∗ (B) − m∗ (A).” (Note that B \ A is the set minus {x ∈ B : x /∈ A}.
Thx a lot you help me a lot to understand how it all waorks, I ve watched all of your videos of measure theory and now I am the one who teach my friends what measure theroy is :D
Wow! Thanks! I watched already several videos about Lebesque measure and read scripts about it. It was always a definition for which I tried to learn by heart. I wonder how easily, you could visualise it, and now I, finally, understood the reasons for defying it like it is!
People watching this video should keep in mind that simple functions are in general not considered the same as step functions. They are a superset of step functions. (4:45)
This video was really good. I needed to learn about the Lebesgue integral and all the book resources seem really unmotivated or intuitive. Your video helped me understand!!! Thank you!!!
Hi, I am not sure the quizz for this lesson is consistent. According to question 2, one of the properties of a simple function is f >= 0. Essentially, question 2 restricts the simple functions to those in the set S^+ without making this explicit. But then question 4 asks if a linear combination of indicator functions that can take negative values is a simple function, and expects the answer "yes".
Will you explain how Borel Sigma algebra on R ist different from the power set of R and how we are still able to measure everything without the power set? I think this would also help me to better understand topology.
i have a question regarding the notation of the lebesgue integrals. namely, when i was introduced to them during my real analysis course i was first told that the notation Integral(X; fdμ), this dμ in particular, is just the notation that was adopted from the riemann integrals. not to be understood as a differential of some sort of this measure. moreover, measure has been very clearly introduced as a funtion on the sets, not set elements, such as numbers. which brings me to my question. how should i understand the notation Integral(X; f(x)dμ(x)) -- i have also seen Integral(X; f(x)μ(dx)) being used here btw -- ? Is dμ a function on set elements now? what is going on? what is even worse, once the lebesgue integral with respect to lebesgue measure is introduced it is often written as Integral(X; fdx), which is indistinguishable from the riemann integral one except maybe that lebesgue integrals tend to operate on sets whereas riemann itegrals on lower and upper bounds. and another question. i have recently learnt about differential forms and integral with respect to these differential forms. it is quite beautiful actually and gives differentials a solid ground. but, how does this relate to the lebesgue integrals. it must be all beautifully connected somehow, right?
A very good question. The notation makes indeed a lot of sense but let me better explain this a suitable video! Regarding one part of your question: Do you really see it as a problem that you can distinguish the Riemann integral and the Lebesgue integral in the notation?
I think you answered your own question. The notation in question should be understood as a generalization of the notation for Riemann integrals, and nothing else.
You are an amazing lecturer, thank you! It is not clear to me the following: taking the supremum of simple function is done aiming to take a finer integral on the function, but, considering your graphical representation, how can the seupremum change anything from not taking it?
20:26 Can we say that \int f d\mu is the limit "as the intervals on the codomain approach zero in length" of "the integral of the corresponding simple functions"? That way the previous picture would be a full explanation rather than just an intuition.
Since we are assuming the measurable maps to be non-negative, so the Borel sigma algebra is defined only on the non-negative reals in this case right? Would it be a problem if codomain were taken to be the entire real line? inverse of the R would be X, but inverse of something smaller would still be X, but that wouldn't cause any problem right? Thanks for the videos, I'd be glad if you can confirm if my points are right.
The Borel sigma algebra is defined on the whole real line but then we just consider non-negative functions for the definition. This makes everything easier because we don't have expressions like ∞ - ∞.
@@sunilrampuria9339 An alternative approach is to consider arbitrary measurable functions f : X -> R rather than functions f : X -> [0, +♾), but then you choose to work with finite measures, or you choose partitions of the domain for which every element of the partition has finite measure.
Thank you for great videos! I've previously only seen the integral defined to take real values, not extended real; is this common practice? In e.g. Friedman (Foundations of Modern Analysis) a simple function is integrable if \mu(E_i)
Thank you. Indeed, it is very common to start with the extended reals and later only to consider finite integrals. At beginning, the extended reals are just easier to work with. That's all :)
The reason why you must consider finite integrals in this case is relatively simple: because the definition needs to stay equivalent to the definition arrived at when the approach, rather than restricting to _nonnegative_ measurable functions, instead restricts to finite measures on arbitrary measurable functions. In the latter approach, the integral is obviously finite, so in the alrernate approach, the integral should be finite as well. Finiteness allows for linearity properties, and possible arithmetic with the integral operators, which is part of what is desired from a rigorous integral definition. My approach for intuitively motivating measure-theoretic integration starts by trying to generalize from the lower Darboux integral, and so it avoids making direct appeals to measure theory in the beginning, and it avoids having to deal with the ideal, infinite points of the affinely extended real line. From there, one can reason about how the generalization, if wants to obey certain desired axioms, has to restrict the measures we integrate with respect with to being finite, motivating a kind of finitist measure theory. Then the problem of arbitrarily choosing integrability to be defined by finite integrals or the problem of arbitrarily choosing the problem-prone convention of 0·♾ = 0, does not occur with my approach. The disadvantage with my approach is that it results in a more "limited" measure theory, since measures that assign +♾ to sets are not allowed, but this is a sacrifice I am willing to make as an educator, if it makes measure-theoretic integrals easier to understand, and the entire theory itself easier to understand, and less arbitrary.
Thank you this video is really great. Sadly I am only in AP Calc AB and can only understand half of what you are saying. How does the Lebesgue integral apply to integrate functions such as x^2 on an interval (that aren't exactly part of measure theory fiasco I guess)?
x^2 is a polynomial, not a function. What you want is a function f : X -> Y, f(x) = x^2, where f is a measurable function, and that measurability is given by an appropriate choice of σ-algebras of X and Y. You need to specify what X and Y are, in order for the choices to be well-defined, and you also must define what x^2 means for an element x of X, before you can proceed.
Prove the excision property for the outer measure m∗ , that is “If A is a measurable set and B ⊇ A, then prove m∗ (B \ A) = m∗ (B) − m∗ (A).” (Note that B \ A is the set minus {x ∈ B : x /∈ A}.are you can solve it
This ends with a defiition reminiscent of lower Darboux sums. In fact choosing the rectangles below the curve seemed rather arbitrary compared to e.g. choosing the rectangles above. Could we have defined the Lebesgue itegral as the infimum of the rectangles above te curve instead, similar to the upper Darboux sums? Are the two equivalent?
Yeah, these definitions are equivalent. For Lebesgue integration there is no need for distinguishing upper and lower sums. (EDIT: Apparently, I didn't get your question, see answers below)
@@vassillenchizhov290 Okay, maybe I misunderstand your question. Our idea here is that we use positive functions and approximate them from below. If you approximated them from above, you would to have to avoid infinities. However, it can not happen that we have a finite "upper sum" and a finite "lower sum" that are not the same. That is what I mean with that we don't have to distinguishing upper and lower sums for the Lebesgue integral. We just choose the lower sums together with positive functions for the definition.
@@brightsideofmaths Thank you very much. If I understood this correctly the upper Darboux sums would then correspond to you choosing not the lower bar to be $c_i$ at 17:55, but rather the higher bar. And then we would have to take the infimum. As noted one should be more careful however with infinities in that case. TH-cam deleted my previous comment for some reason - I assume because of the link to math stackexchange.
Shouldn't c_i > 0 instead of c_i >= 0. Consider the zero function, inorder to show that the integral on the real line is zero, we would need 0*infinity=0 if we choose the latter option.
You can require that, if you want to, and I would argue that it is objectively more convenient for all purposes to do so, but it is not technically necessary, as long as care with the notation is taken.
can the set X contain any mathematical objects, or just real numbers? doing some research and if it were only the latter it would really simplify my work.
Hello, is it not enough to assume that the pre-image of any set in the powerset of the codomain is an element of the sigma-algebra of the domain in order to define the Lebesgue integral of f ? (Do we need the measurability of the codomain ?) btw : I love the way you teach things !
@@brightsideofmaths :) ! Yeah my question was weirdly put, I m sorry. What's on my mind is could you define the concept of a measurable map without defining a sigma-algebra on the codomain ? I don't grasp why we need this assumption in the further definitions and theorems.
11:16 I do not understand where you got that the measurable function "only takes finitely many values". I was also not sure whether those values you were referring to belonged to the domain or the codomain. I am really enjoying these videos! They are helping me get out of a rut and I'm having great fun watching them.
Those values obviously refer to the ones in the codomain, and this is true by the definition of what the codomain of a function is. Also, you are confusing "measurable functions" with "simple functions".
@@brightsideofmaths maybe I am using too much visualization for my argument but if some sets A1 and A2 intersect in the domain wouldn't that create some overlappng areaand be counted twice? I mean I would think that the area would the sum of these 2 guys without the overlapping area( inclusion-exclusion principle)?
@@brightsideofmaths aaaaahh, fair enough. Thanks a lot for answering my question even sfter 6 months of making the video. You usually don't see creators answering questions like these, even educational creators like you
I don't find trivial to see that I(f) is well-defined for a S+ function. Usually the integral is defined using the canonical representation, but not here. So IMO we must prove that all representations lead to the same integral value. Any hint to prove that ?
Maybe you can use your knowledge of the canonical representation to transform every other representation to that one. That you can compare the integrals :)
@@brightsideofmaths yes thank you that was what I started trying but I did not find it so straightforward. Anyway I'm surprised you said in the course that it was "well defined" as if this was trivial. Defining using the canonical representation would have been easier for me...
I didn't say "trivial". "Well-defined" is a common mathematical term and important for the definition here. I try to keep the video compact so I skip technical details. Using the canonical representation does not change the fact that you need to show the same thing in the end.
Great video! Not sure I understand why the integral is dMu... I'm used to dx or dy, i.e. the mini-steps are w.r.t. a certain variable. And here they are w.r.t. to the measure?
One doubt - in the definition we write integral f(x) dMu(x) , but the point 'x' is coming from space X of a measurable function but the measure is defined only on sigma algebra of that space X right. What if x is from X but does not belong to sigma algebra
The integral you are familiar with is the Darboux integral, or the Riemann integral, which is equivalent to the former. For the Darboux integral, you actually are technically integrating with respect to the Lebesgue measure, but there are also a number of restrictions, such as having the domain be of the function be a closed interval of R, and restricting the supremum of sums to be only over closed interval partitions of the domain, rather than arbitrary measurable set partitions. Another restriction is that the corresponding upper Darboux integral has to be equal to the lower Darboux integral. Measure-theoretic integration is just a generalization of lower Darboux integration, so there is no extension of the upper Darboux integral to consider when dealing with μ-integrals. However, the reason you think there is no measure involved in the Darboux integration is because our notation for the differential does it with respect to the variable of integration, rather than with respect to the Lebesgue measure. The notation is different.
1. This is a very nice set of lectures, in a subject where most of the current TH-cam offerings are either poorly explained, overly complex, or simply downright tedious. Thanks. 2. I'd be interested to know what system/software you're using to write these notes, as I'm looking for something similar myself.
@@brightsideofmaths I've never tried Xournal but I will now take a look. Thanks. BTW, will you be covering the Caratheodory criterion for measurable sets at any point? I'd be interested in seeing something about that. Danke im Voraus!
@@scollyer.tuition I also find Carathéodory's criterion very interesting. However, it is very special and I have to do some other videos before it makes to talk about this. I put it on my list :) Thank you!
@@brightsideofmaths I would like to point out that the characteristic functions DO NOT form a basis of S, but instead a SPANNING SET of S (because the characteristic functions are not in general, linearly independent). Just a small mistake worth pointing out here. Great video as always. I've already studied measure theory, and these material familiar to me, but with a cursory watch, I think your videos are clear and great.
I love how you fast forward the writing to save time. Great!
This measure theory series is beneficial to humanity. Thank you for making it. After an exhaustive search, this is the best explanation for Lebesgue integral.
You're very welcome! And thanks for you support :)
This series is so helpful. I’m trying to understand probability theory rigorously. Your video and some of my own reading just help me realize that random variables are nothing but measurable functions. This has transformed how I look at probabilities in general. Before I always thought there was some randomness involved. Now Using the concept of measure, I feel like the likelihood of events can be categorized by measure. This is a static global perspective, and it just makes everything easier to think about and work with. Very exciting stuff to learn!
Probability theory is literally just measure theory constrained to measure spaces (X, A, μ) such that μ(X) = 1. In other words, every probability space is isomorphic to a measure space of finite measure.
For those who would like to see the connection between the Riemann integral and the measure-theoretic integrals, I am going to go back to some definitions, and analyze the fundamental mathematical object underlying both definitions.
Consider a function f : [a, b] -> R. A tag on the interval [a, b] is an n-tuple t : {m : 0 =< m =< n} -> [a, b], where if i < j, then t(i) < t(j). If such a tag t is defined, then the sum of f[t(i)]·w(i) is well-defined. Here, w is an n-tuple of real numbers, this time not constrained to the interval [a, b], and this is called a weight. This is the fundamental object of study. This sum is called a w-weighted sum of f over the tag t.
Now, consider the set of closed subintervals of [a, b], denoted c([a, b]), and consider a map p : range(t) -> c([a, b]), with the consideration that the intersection of any two closed intervals in c([a, b]) is either the common endpoint of both intervals, or the empty set, with the consideration that the union of the range of p is [a, b], and with the consideration that t(i) is an element of the closed interval p[t(i)]. This turns p into an interval-partition with a tag t, and the endpoints of these intervals are denoted x(i). If we define w(i) := x(i + 1) - x(i), then this special case of the sum of f[t(i)]·w(i) is called a Riemann sum of f. Riemann sums of f are what we use to define the Riemann integral of f.
When considering the interval-partition p, it may be more general to consider w(i) = μ(p[t(i)]), instead of w(i) := x(i + 1) - x(i). The idea here is that p[t(i)] is equal to a closed interval, whose value depends on the value of t(i), and the endpoints of this interval are x(i) and x(i + 1). μ can be interpreted to be a "size function" at the intuitive level, and so the "standard" size of an interval p[t(i)], with endpoints x(i + 1) and x(i), is x(i + 1) - x(i). In this generalization, one can allow for flexibility as to what μ is, as to not be restricted to the standard size of a closed interval. So the sum that you obtain from this is different from a Riemann sum, and is instead equal to f[t(i)]·μ(p[t(i)]), which you may call a μ-sum. The generalization can now have for f[t(i)] replaced by a lower bound of the set sp(f, p) := {f[t(i)] : x(i) =< t(i) =< x(i + 1)}. This lower bound can simply be called c(i), so the weighted sum is the sum of c(i)·μ(p[t(i)]). Now, this is really starting to resemble symbolically the kind of weigted sum that the measure-theoretic integrals are defined by, if μ is interpreted as being a measure, or measure-like. Finally, some generalizations can be made here: rather than defining p to assign a closed subinterval of [a, b] to every point of the tag t, p can simply assign a measurable set S(i) to every t(i), so long as the union of S(i) is still equal to [a, b], and so long as the pairwise intersection of S(i) is a null set. Furthermore, the domain of f allowed, rather than being a closed interval [a, b], can be generalized to simply be a set X, with (X, A) being a measurable space, and f being a measurable function. It is the supremum of the set of every such sum in this generalization that is equal to the μ-integral.
So what is the Riemann integral with respect to this μ-theory? It is an integral with respect to μ, where μ is specifically the Lebesgue measure, restricted to closed intervals, and with c(i) := f[t(i)], and with the added restriction that it only is the integral of f if the infimum of the aforementioned set, where c(i) is instead an upper bound of sp(f, p), rather than a lower bound. The gauge integral is also defined using these weighted sums of functions over tags, together with restrictions as well.
Actually, the connection between these different integrals is simpler, more obvious, and more intuitive if we start with the Darboux integral, rather than the Riemann integral. Firstly, we go back to the weighted sums of f[t(i)]·w(i). Each w(i) can immediately be interpreted as being the output of a function of the partition intervals of [a, b]: that is, a function of [x(i), x(i + 1)]. So w(i) = μ([x(i), x(i + 1)]). Now, we have the sums of f[t(i)]·μ([x(i), x(i + 1)]). If μ([x(i), x(i + 1)]) = Δx(i), then we are back to the Riemann sums f[t(i)]·Δx(i), but rather considering these Riemann sums, we can go to the upper Darboux sums, and the lower Darboux sums, giving us sums of sup({c(i) : c(i) = f[t(i)] & t(i) is in [x(i), x(i + 1)]})·Δx(i), and inf({c(i) : c(i) = f[t(i)] & t(i) is in [x(i), x(i + 1)]})·Δx(i) respectively.
The idea is now to generalize these Darboux sums, by going back to looking at Δx(i), and again considering μ([x(i), x(i + 1)]), where Δx results from the special case when μ is the Lebesgue measure. When applied to lower Darboux sums specifically, you sum with respect to this measure μ, and then you take the supremum of every such sum, obtaining something analogous to the lower Darboux integral, but with respect to μ. This lower μ-integral is precisely the definition of the Lebesgue integral of f with respect to μ, as presented in the video, though this definition is restricted only to measurable functions f : [a, b] -> R with the Borel σ-algebra on both spaces, which can be made more general by allowing a domain X and an arbitrary partition (X(0), ..., X(n)), where the sums are now sup({c(i) : c(i) = f[t(i)] & t(i) is an element of X(i)})·μ[X(i)] and inf({c(i) : c(i) = f[t(i)] & t(i) is an element of X(i)})·μ[X(i)]. This definition also restricts μ to being a finite measure. The approach in the video is retained if this definition is made to apply to f+ and f- individually, rather than f itself, and if 0·♾ = 0.
So in summary, the lower Darboux integral is exactly equal to the Lebesgue integral with respect to the Lebesgue measure, where the integrable functions are those whose nonnegative parts are Borel-measurable, specifically.
I am going to follow a few lectures about Lebesgue integrals. A friend of mine who is good at maths warned me it will be very difficult. These videos give me a very clear idea and intuitions about what I will be learning. It gives me a framework everything will fit in hopefully. Thank you so much!
Lebesgue? More like "These videos are the best!" Thanks again for making and sharing such a high-quality resource.
I have an exam of Real Analysis I for thesis qualification, and you just give me a brief understanding of measure. Wish me luck cuz I really did not get familiar with proofs yet, and already no much time left.
Thank you, your videos really helpful.
Thank you so much. You're a life saver. My professor just sent me notes, no lecture, and I had a hard time understanding her notes, until I saw your videos. Thank you.
How did the rest of your class go?
These videos are great, usually when I'm watching math videos on youtube they are quite old. I was shocked when I saw how recently these have come out. Looking forward to more!
Thank you very much! The next videos are in preparation :)
Such a good video at understanding the reasoning behind how we are approximating the integral. I feel like many books/videos don't do a good job at providing the reasoning behind taking the supremum. Thanks for this!
Nice! Thanks a lot! You can check out my website in the description if you need more :)
These are the best videos on the measure theory one can make.
It would be great if you could make more videos on other areas of mathematics like TOPOLOGY,REAL ANALYSIS
I'd like to pile onto that by saying that if you went into measure-theoretic probability, that would actually be really neat. By that, I mean if you talked about, say, the normal distribution and that sort of thing and cast it all the way back down into measure theory and then built it up to the computational statistics we do with it.
@@ryanhutchins2634 Distributions is more a topic of statistics and probability than probability theory, but nonetheless, it would be nice if this channel could cover that.
This lecture I like the most so far. I see now why they use positive step functions. I knew that they use positive functions but I didn’t know why.
I met this topic in Real Analysis during my MS yet I can't remember anything. Now that I am taking my PhD I am starting to appreciate it. I am majoring Statistics but Real Analysis is a corequisite for Probability Theory. Please make lectures on Theory of Probabilities. Thank you.
That's awesome! What are you researching, and how has that been going?
Stop being lazy and do your own study
아주 차분하고, 독자 입장에서 잘 설계된 고급 수학 강의를 들으니 참 좋습니다. 배우고 갑니다. 감사합니다.
Amazing, I think I finally understood what a Lebesgue Integral is!
Some are tough to believe but we need to believe and learn further...I imagine always how do I proceed further when one thing is not yet clear. But its a nice feeling of the subject!
Thank you very much for best explanation in this series.
Thanks a lot sir
you are my life savior
Prove the excision property for the outer measure m∗
, that is
“If A is a measurable set and B ⊇ A, then prove
m∗
(B \ A) = m∗
(B) − m∗
(A).”
(Note that B \ A is the set minus {x ∈ B : x /∈ A}.
Thank you a lot! Helped me catch up with my unit. My stats professor read the slides instead of explaining the concept :).
How did the rest of your class go?
wow that was very cool how lebesgue integral is basically like riemann integration tipped on it's side (plus step functions)
Thx a lot you help me a lot to understand how it all waorks, I ve watched all of your videos of measure theory and now I am the one who teach my friends what measure theroy is :D
Wow! Thanks! I watched already several videos about Lebesque measure and read scripts about it. It was always a definition for which I tried to learn by heart. I wonder how easily, you could visualise it, and now I, finally, understood the reasons for defying it like it is!
Thank you very much :)
Thank you for these videos, they're the only thing keeping me afloat in my measure theory course
How did the rest of your course go?
love you bro
that's a tremendous work, thank you
Can you post PDFs of what you write in the videos in the links below the video? Maybe public Google Docs of something.
That will be extremely helpful.
Fantastic videos. Thank you so much!
People watching this video should keep in mind that simple functions are in general not considered the same as step functions. They are a superset of step functions. (4:45)
There are a lot of definitions around and I didn't want to go into the details here: some authors use step functions in a broader sense than others.
This video was really good. I needed to learn about the Lebesgue integral and all the book resources seem really unmotivated or intuitive. Your video helped me understand!!! Thank you!!!
Thank you. I am very happy that you like it and that it is helpful :)
marvelous, thank you!
Thank you too! :)
Gerçekten anladım. Emeğiniz için teşekkürler ♡
It would be great if you could make more videos on optimization.
It would be great if you stop being laxy and just do your own study
Hi, I am not sure the quizz for this lesson is consistent. According to question 2, one of the properties of a simple function is f >= 0. Essentially, question 2 restricts the simple functions to those in the set S^+ without making this explicit. But then question 4 asks if a linear combination of indicator functions that can take negative values is a simple function, and expects the answer "yes".
I will check that! Thanks :)
It's fixed! Thanks for pointing out the mistake :)
@@brightsideofmaths , thank you!
Will you explain how Borel Sigma algebra on R ist different from the power set of R and how we are still able to measure everything without the power set? I think this would also help me to better understand topology.
I have a video about non-measurable sets.
Look up the Vitali set construction!
For the simple function at 7:11, do Ai's need to be disjoint? Thanks
No, they don't.
Thank you, you are the best
Thank you, these are great and really growing my understanding
Can I donate a specific amount instead of subscribing? I want to support but also be somewhat aware of my spending. LMK, Thanks man
i have a question regarding the notation of the lebesgue integrals. namely, when i was introduced to them during my real analysis course i was first told that the notation Integral(X; fdμ), this dμ in particular, is just the notation that was adopted from the riemann integrals. not to be understood as a differential of some sort of this measure. moreover, measure has been very clearly introduced as a funtion on the sets, not set elements, such as numbers. which brings me to my question. how should i understand the notation Integral(X; f(x)dμ(x)) -- i have also seen Integral(X; f(x)μ(dx)) being used here btw -- ? Is dμ a function on set elements now? what is going on? what is even worse, once the lebesgue integral with respect to lebesgue measure is introduced it is often written as Integral(X; fdx), which is indistinguishable from the riemann integral one except maybe that lebesgue integrals tend to operate on sets whereas riemann itegrals on lower and upper bounds. and another question. i have recently learnt about differential forms and integral with respect to these differential forms. it is quite beautiful actually and gives differentials a solid ground. but, how does this relate to the lebesgue integrals. it must be all beautifully connected somehow, right?
A very good question. The notation makes indeed a lot of sense but let me better explain this a suitable video!
Regarding one part of your question: Do you really see it as a problem that you can distinguish the Riemann integral and the Lebesgue integral in the notation?
I think you answered your own question. The notation in question should be understood as a generalization of the notation for Riemann integrals, and nothing else.
Terrific!
You are an amazing lecturer, thank you!
It is not clear to me the following: taking the supremum of simple function is done aiming to take a finer integral on the function, but, considering your graphical representation, how can the seupremum change anything from not taking it?
Thank you! It's the supremum of the whole set. So the supremum of the integrals :)
@@brightsideofmaths Thank you very much, it makes more sense
20:26 Can we say that \int f d\mu is the limit "as the intervals on the codomain approach zero in length" of "the integral of the corresponding simple functions"?
That way the previous picture would be a full explanation rather than just an intuition.
That would not be accurate, since we are working with suprema of sets, not with sequences.
Since we are assuming the measurable maps to be non-negative, so the Borel sigma algebra is defined only on the non-negative reals in this case right? Would it be a problem if codomain were taken to be the entire real line? inverse of the R would be X, but inverse of something smaller would still be X, but that wouldn't cause any problem right?
Thanks for the videos, I'd be glad if you can confirm if my points are right.
The Borel sigma algebra is defined on the whole real line but then we just consider non-negative functions for the definition. This makes everything easier because we don't have expressions like ∞ - ∞.
@@brightsideofmaths Thank you for the reply. I got it.
@@sunilrampuria9339 An alternative approach is to consider arbitrary measurable functions f : X -> R rather than functions f : X -> [0, +♾), but then you choose to work with finite measures, or you choose partitions of the domain for which every element of the partition has finite measure.
Thank you for great videos! I've previously only seen the integral defined to take real values, not extended real; is this common practice? In e.g. Friedman (Foundations of Modern Analysis) a simple function is integrable if \mu(E_i)
Thank you. Indeed, it is very common to start with the extended reals and later only to consider finite integrals. At beginning, the extended reals are just easier to work with. That's all :)
The reason why you must consider finite integrals in this case is relatively simple: because the definition needs to stay equivalent to the definition arrived at when the approach, rather than restricting to _nonnegative_ measurable functions, instead restricts to finite measures on arbitrary measurable functions. In the latter approach, the integral is obviously finite, so in the alrernate approach, the integral should be finite as well. Finiteness allows for linearity properties, and possible arithmetic with the integral operators, which is part of what is desired from a rigorous integral definition.
My approach for intuitively motivating measure-theoretic integration starts by trying to generalize from the lower Darboux integral, and so it avoids making direct appeals to measure theory in the beginning, and it avoids having to deal with the ideal, infinite points of the affinely extended real line. From there, one can reason about how the generalization, if wants to obey certain desired axioms, has to restrict the measures we integrate with respect with to being finite, motivating a kind of finitist measure theory. Then the problem of arbitrarily choosing integrability to be defined by finite integrals or the problem of arbitrarily choosing the problem-prone convention of 0·♾ = 0, does not occur with my approach. The disadvantage with my approach is that it results in a more "limited" measure theory, since measures that assign +♾ to sets are not allowed, but this is a sacrifice I am willing to make as an educator, if it makes measure-theoretic integrals easier to understand, and the entire theory itself easier to understand, and less arbitrary.
Thank you this video is really great. Sadly I am only in AP Calc AB and can only understand half of what you are saying. How does the Lebesgue integral apply to integrate functions such as x^2 on an interval (that aren't exactly part of measure theory fiasco I guess)?
x^2 is a polynomial, not a function. What you want is a function f : X -> Y, f(x) = x^2, where f is a measurable function, and that measurability is given by an appropriate choice of σ-algebras of X and Y. You need to specify what X and Y are, in order for the choices to be well-defined, and you also must define what x^2 means for an element x of X, before you can proceed.
I think it's awesome that you're already checking this stuff out while you're in calc BC! How have your studies been going?
Thank you for your helpful videos. I still have a hard time following. I enjoyed Lecture 1, but soon got lost with the concepts. Any solutions?
Maybe, my other courses like Real Analysis can help you to get a foundation. Afterwards you can come back here :)
Prove the excision property for the outer measure m∗
, that is
“If A is a measurable set and B ⊇ A, then prove
m∗
(B \ A) = m∗
(B) − m∗
(A).”
(Note that B \ A is the set minus {x ∈ B : x /∈ A}.are you can solve it
awesome
This ends with a defiition reminiscent of lower Darboux sums. In fact choosing the rectangles below the curve seemed rather arbitrary compared to e.g. choosing the rectangles above. Could we have defined the Lebesgue itegral as the infimum of the rectangles above te curve instead, similar to the upper Darboux sums? Are the two equivalent?
Yeah, these definitions are equivalent. For Lebesgue integration there is no need for distinguishing upper and lower sums.
(EDIT: Apparently, I didn't get your question, see answers below)
@@vassillenchizhov290 Okay, maybe I misunderstand your question. Our idea here is that we use positive functions and approximate them from below. If you approximated them from above, you would to have to avoid infinities. However, it can not happen that we have a finite "upper sum" and a finite "lower sum" that are not the same. That is what I mean with that we don't have to distinguishing upper and lower sums for the Lebesgue integral. We just choose the lower sums together with positive functions for the definition.
@@brightsideofmaths Thank you very much. If I understood this correctly the upper Darboux sums would then correspond to you choosing not the lower bar to be $c_i$ at 17:55, but rather the higher bar. And then we would have to take the infimum. As noted one should be more careful however with infinities in that case.
TH-cam deleted my previous comment for some reason - I assume because of the link to math stackexchange.
Is this definition equivalent to the Riemann integration if I have a continuous function from R to R?
Yes, for continuous functions on intervals [a,b], it's the same.
Thanks a lot for all of your efforts
Dont you need f to be bounded in order to partition the y-axis in n parts?
Nope. That would be an issue if the partitions were required to be the same size, but they're not.
Shouldn't c_i > 0 instead of c_i >= 0. Consider the zero function, inorder to show that the integral on the real line is zero, we would need 0*infinity=0 if we choose the latter option.
Yeah, we set 0*infinity = 0 in these cases, here.
You can require that, if you want to, and I would argue that it is objectively more convenient for all purposes to do so, but it is not technically necessary, as long as care with the notation is taken.
can the set X contain any mathematical objects, or just real numbers? doing some research and if it were only the latter it would really simplify my work.
X can be any set.
Hello, is it not enough to assume that the pre-image of any set in the powerset of the codomain is an element of the sigma-algebra of the domain in order to define the Lebesgue integral of f ? (Do we need the measurability of the codomain ?) btw : I love the way you teach things !
Thank you very much! Maybe I don't understand your question. A measurable set is exactly an element of the sigma algebra :)
@@brightsideofmaths :) ! Yeah my question was weirdly put, I m sorry. What's on my mind is could you define the concept of a measurable map without defining a sigma-algebra on the codomain ? I don't grasp why we need this assumption in the further definitions and theorems.
It is important to note these sets A1,...,An should be disjoint
very good video tho!
Thank you. The sets don't need to be disjoint :)
you do right? If some sets are overlapping and you take the sum over all measures you take some double
@@florissomers1596 An that is not a problem because the value of the function gets also added.
They need not be disjoint, they simply need to have the measure of their intersections be 0.
Hi, I have a question at 14:11 that: why the integral of f is on d(mu) but not dx. Btw, the series is very useful for me, thank you.
Thanks a lot! This is simply the notation we choose for mentioning the corresponding measure.
@@brightsideofmaths oh, i got it. Thank you.
11:16 I do not understand where you got that the measurable function "only takes finitely many values". I was also not sure whether those values you were referring to belonged to the domain or the codomain. I am really enjoying these videos! They are helping me get out of a rut and I'm having great fun watching them.
Those values obviously refer to the ones in the codomain, and this is true by the definition of what the codomain of a function is. Also, you are confusing "measurable functions" with "simple functions".
Would the integral of a simpe function still mean the area unde the functions even when the measurable sets A1, ..., An overlapp each other?
Yes!
@@brightsideofmaths maybe I am using too much visualization for my argument but if some sets A1 and A2 intersect in the domain wouldn't that create some overlappng areaand be counted twice? I mean I would think that the area would the sum of these 2 guys without the overlapping area( inclusion-exclusion principle)?
@@quantumgaming9180 But wouldn't you also add the y-values, making the rectangle taller?
@@brightsideofmaths aaaaahh, fair enough. Thanks a lot for answering my question even sfter 6 months of making the video. You usually don't see creators answering questions like these, even educational creators like you
@@quantumgaming9180 It's kind of the deal. People support me (on Steady, Patreon, and so on) and I answer their questions :)
Wow!!!!! just wow!!!!!!
I don't find trivial to see that I(f) is well-defined for a S+ function. Usually the integral is defined using the canonical representation, but not here. So IMO we must prove that all representations lead to the same integral value. Any hint to prove that ?
Maybe you can use your knowledge of the canonical representation to transform every other representation to that one. That you can compare the integrals :)
@@brightsideofmaths yes thank you that was what I started trying but I did not find it so straightforward. Anyway I'm surprised you said in the course that it was "well defined" as if this was trivial. Defining using the canonical representation would have been easier for me...
I didn't say "trivial". "Well-defined" is a common mathematical term and important for the definition here. I try to keep the video compact so I skip technical details. Using the canonical representation does not change the fact that you need to show the same thing in the end.
But I am sorry for the confusion. I could have been more clear :)
@@brightsideofmaths no problem
Great!
Great video! Not sure I understand why the integral is dMu... I'm used to dx or dy, i.e. the mini-steps are w.r.t. a certain variable. And here they are w.r.t. to the measure?
Yeah, they are with respect to a measure because your "mini-steps" now have to be measured first :)
One doubt - in the definition we write integral f(x) dMu(x) , but the point 'x' is coming from space X of a measurable function but the measure is defined only on sigma algebra of that space X right. What if x is from X but does not belong to sigma algebra
@@premkumar-so3ff We actually do not write f(x)·dμ(x), we just write f·dμ. The former is just abuse of notation, there is nothing to interpret there.
The integral you are familiar with is the Darboux integral, or the Riemann integral, which is equivalent to the former. For the Darboux integral, you actually are technically integrating with respect to the Lebesgue measure, but there are also a number of restrictions, such as having the domain be of the function be a closed interval of R, and restricting the supremum of sums to be only over closed interval partitions of the domain, rather than arbitrary measurable set partitions. Another restriction is that the corresponding upper Darboux integral has to be equal to the lower Darboux integral. Measure-theoretic integration is just a generalization of lower Darboux integration, so there is no extension of the upper Darboux integral to consider when dealing with μ-integrals.
However, the reason you think there is no measure involved in the Darboux integration is because our notation for the differential does it with respect to the variable of integration, rather than with respect to the Lebesgue measure. The notation is different.
1. This is a very nice set of lectures, in a subject where most of the current TH-cam offerings are either poorly explained, overly complex, or simply downright tedious. Thanks.
2. I'd be interested to know what system/software you're using to write these notes, as I'm looking for something similar myself.
1. Thank you very much :)
2. I just use Xournal for writing and capture the screen.
@@brightsideofmaths I've never tried Xournal but I will now take a look. Thanks.
BTW, will you be covering the Caratheodory criterion for measurable sets at any point? I'd be interested in seeing something about that. Danke im Voraus!
@@scollyer.tuition I also find Carathéodory's criterion very interesting. However, it is very special and I have to do some other videos before it makes to talk about this. I put it on my list :) Thank you!
8:50 problem with the simple function
How do I prove that I(f) is well defined? Any hint? :)
If I(h) is well-defined, then I(f) is as well because it's just a supremum, which always exists.
So do the charasteristic functions form a basis of S?
Yes, the characteristic functions corresponding to measurable sets form a basis of S.
@@brightsideofmaths well thanks for answering that fast
@@brightsideofmaths I would like to point out that the characteristic functions DO NOT form a basis of S, but instead a SPANNING SET of S (because the characteristic functions are not in general, linearly independent). Just a small mistake worth pointing out here.
Great video as always. I've already studied measure theory, and these material familiar to me, but with a cursory watch, I think your videos are clear and great.
it is only a laying theory bcs the the sum h of seris ci.xAi doesn't have analytical expression.
What do you mean?
Who came from Classroom of the Elite?
Who is Cote?
@@brightsideofmaths Cote is a Novel in which they talked about Lebesgue integrals in one scene so I figured someone who read Cote would look it up
@@L.Lawliet.3301 Nice! Thanks for the hint :)
@@L.Lawliet.3301 you should write it as classroom of the elite. It is easy to know then.
@@vizzyb8400 you‘re probably right xD