Nice one!

@vcubingx

4 жыл бұрын

Thanks Grant!!

@Exachad

4 жыл бұрын

Acknowledged by the man, the myth, the legend.

@NovaWarrior77

4 жыл бұрын

He's here everybody! Shhhh! Get into your positions!

@friedkeenan

4 жыл бұрын

It warms my heart that Grant watched this video from a small youtuber

@BleachWizz

4 жыл бұрын

YES YES YES REALLY NICE ONE. I was thinking "but can't instead I just f^-1(x) and solve for the same integral?" But when "f(x) = constant for x in a range" it's literally impossible to do that. And with this and the way you can apply it to probability where you can't just invert the function is brilliant!

As the King of Calculus, Riemann, faced the Dirichlet function, Riemann asked the function, “Are you Riemann integrable because you are continuous? Or are you continuous because you are Riemann integrable?” (note: this iff statement is not actually true) Riemann began to open his domain *”Malevolent Shrine”* and *cleaved* the area under the curve into slices approaching infinity. However, the Dirichlet function is continuous nowhere, and by *contradiction* the function simply stated “Stand proud. You're strong but Nah, I’d win.” Bernhard Riemann’s domain crumbled. In his dying moments, Riemann uttered the phrase “With this treasure, I summon…” Because in sets of measure zero always bet on *Lebesgue.* The proof was trivially concluded, and those who pioneered the techniques of calculus, the ones who formalised the integral, they would all bear witness to the one who is free. To the one who left it all behind, and his overwhelming *integrability!*

So cool, more 3b1b inspired youtubers.. I am slightly jealous of kids today who are getting this level of education for almost free.

@asherasher9249

2 жыл бұрын

The day may come when you can download knowledge onto your brain like a computer. I'm not sure how universally jealous people would be of that though lol

@agrajyadav2951

Жыл бұрын

one of the best things to happen to this world since 1940

@twt2718

8 ай бұрын

I had to figure EVERYTHING out on my own!! I had a textbook and professors that tenure and were pissed about the number of courses they had to “teach”.

@nonamedyet

8 ай бұрын

Yet many kids out there spend tons of hours watching TikTok dances

@joshuat6124

8 ай бұрын

Yes, learning resources are more prevalent and easily accessible, but so are distractions imo

Clciked and thought it was 3b1b, glad I made that mistake, great video!

Great video Vivek! =)

@vcubingx

4 жыл бұрын

Thanks Papa

@patryk_49

4 жыл бұрын

Now I am waiting for some cursed meme about factoring f(x) and integrating Lebesgue Integeral with respect to miu.

@yusufklc7821

4 жыл бұрын

Papa is here too omg

@Sciencedoneright

4 жыл бұрын

@@vcubingx is he your father?

@HKCREATESUNIQUE

3 жыл бұрын

@@Sciencedoneright his name is papa

At 7:19, E(x) should be the integral of x*P(x), not just P(x). The integral of P(x) is 1!

@Pabloparsil

2 жыл бұрын

Yes and he should have written f(x) or something else, P(x) suggests probability but it's the density what he is plotting and is used to compute expectations. Amazing that the author hasn't pinned your comment yet

@jarosawszyc8287

2 жыл бұрын

Exactly right!

@Mr_Hassell

2 жыл бұрын

Yes

@Tito-xg3wb

8 ай бұрын

@@Pabloparsil Exactly but it remains a confusion between x the variable of integration and X the random variable. I would have written E[X] = integration of x *f(x) dx (f the density) or integration of x dP(x) with P the measure of the law of X (don't know of to say it properly sorry)

This is excellent. Every time I've tried to understand lebesgue integration in the past it was presented as something complicated and hard to understand, but you've made it so simple. I'm sure there's more detail you didn't cover, since once video can only do so much, but you've actually given some good intuition what what a lebesgue integral actual IS, which is something I haven't been able to find in any of the many other explanations I've found. Thank you very much!

This is so cool for sure. Animated presentation is really good

One thing I'll give as an advice : don't stop. I looked back at myself from after 6 years and I'm ashamed of how better I was at 14/15/16/17. You're at an awesome level. Make sure it's not your peak! Awesome video :)

@alphazero339

8 ай бұрын

What field of maths did you learn at that age when you were younger so I can compare with myself. I feel like I'm learning too slowly to ever accomplish anything..

7:25 Isn't it supposed to be the integral of x p(x) dx

@stefanjenkins6196

4 жыл бұрын

For the expected value function yes. The integral in the video sums to 1

@mehmetm9392

4 жыл бұрын

I think it is P(X) = x*f(x), with f being the PDF.

@stefanjenkins6196

4 жыл бұрын

Mehmet M Mehmet M if that is the case that would be an unusual way of notating that. Usually the Probability = ∫ PDF(x)dx. In this case his P(x) is his PDF. The integral for E should be of E = ∫ x p(x) dx

@mehmetm9392

4 жыл бұрын

@@stefanjenkins6196 I know, that this was a typo. But nevertheless, I appreciate your nice explanation.

@uhu2323

3 жыл бұрын

This comment should be on the top pinned. I hate that the corrections always get buried under the fangirl comments. We get it... It's a great video. Can we get to scientific content distribution 2.0 already where this problem is sorted?

These animations just look so good :)

@vcubingx

4 жыл бұрын

Thank you so much! Your videos were incredibly useful when I was learning this topic.

@brightsideofmaths

4 жыл бұрын

@@vcubingx Thanks. Glad I could help :)

@mikevaldez7684

3 жыл бұрын

@@vcubingx weak sauce.....not enough examples & theory....

@pragyanpranay3681

8 ай бұрын

i follow both of the channels and man this feels very nice of everyone coming together and learning the topic so nicely. I really love the way you have presented lesbegue integral... I need to watch it more times for sure to get a little more understanding, but this is feels so nice! just switching how you construct rectangles to find area can do so much... would have never imagined that...

I think you haven't understood the Lebesgue integral at all. The animation with horizontal rectangles is quite flawed. That's not how it works!! Rectangles are vertical, what happens is that we split the range of the image into different sections. In each of them, we take an arbitrary point, and multiply it by the inverse function of the section. That may be one or more VERTICAL rectangles.

Good video! You have an amazing talent of effectively illustrating complex ideas in a simple manner. Keep up the good work!!

@vcubingx

4 жыл бұрын

Thank you very much!

The "limits of rectangles" explanation is what they teach you in high school calc, is how Newton/Leibniz thought of integration, and is correct to a good approximation, but it's not technically what a Riemann Integral is. Riemann Integrals (which are taught as a rigorous framework in Real Analysis class) are actually more powerful, and are able to integrate any function which is continuous "almost everywhere". ----------- Take Tomae's Function, which is extremely similar to Dirichlet's Function that you showed: f(x) = 0 if x is irrational, 1/n if x = m/n (rational) This function has infinite discontinuities. However, it's discontinuous at every rational number and continuous at every irrational number(!!). Since there are only countably-many discontinuities, it's continuous "almost everywhere" and is therefore Riemann Integrable! ----------- In fact, it was discovered in the 50's that a small tweak to the definition of Riemann Integrals makes them strictly stronger than Lebesgue Integrals - that is, every function which can be integrated using Lebesgue's Method can also be integrated using these new "Generalized Riemann Integrals", but the converse is not true!

@ThisIsANameBruh

4 жыл бұрын

damn, i shouldn't have taken up engineering.

@xCorvus7x

4 жыл бұрын

This one, right? en.wikipedia.org/wiki/Henstock-Kurzweil_integral

@BlueRaja

4 жыл бұрын

@@xCorvus7x Yes, that's the Generalized Riemann Integral.

@angelmendez-rivera351

3 жыл бұрын

@@BlueRaja *Riemann integrals are actually more powerful, and are able to integrate any function which is continuous "almost everywhere".* This is incorrect. There is only one Riemann integral, and the Riemann integral cannot integrate the indicator function of the rational numbers embedded in the real numbers, even though this function is bounded and continuous almost everywhere. In fact, this is one notable point of weakness of the Riemann integral that led to the development of the Lebesgue integral in the first place. Also, the claim that the Riemann integral is more powerful is just false. A well-known theorem of real analysis is that a function f : I -> R is Darboux integrable if and only if it is Riemann integrable. *Yes, that's the generalized Riemann integral.* Calling it this is misleading and incorrect, as this is not the name of the integral. Yes, the gauge integral is a direct generalization of the Riemann integral, but it is not called the generalized Riemann integral. Also, one important detail you neglect is that the gauge integral is only a stronger integral when we limit ourselves to functions in R^I. The measure-theoretic framework for the Lebesgue integral is more robust in that one can perform integration on arbitrary measurable spaces (X, Σ), rather than just (R, B(R)), which is the space in which the gauge integral is defined on. A more in-depth exploration of the subject reveals that the direction in which one generalizes from Riemann integration to measure-theoretic integration is a different one than the direction in which one generalizes from Riemann integration to gauge integration. It is possible to produce a tweak like the one from Riemann integration to gauge integration whilst simultaneously applying the Lebesgue framework of generalization of spaces. This shows that while the gauge integral is better behaved, it is not a broadening of scope in the same way Lebesgue's framework is.

@BlueRaja

3 жыл бұрын

@@angelmendez-rivera351 The "Generalized Riemann Integral" is the name we learned for it in Real Analysis class, and also the name my Real Analysis textbook uses ("Introduction to Real Analysis", Bartle+Sherbert, 1999), and is one of the names listed on the Wikipedia page. So I feel pretty confident in saying I'm not incorrect in calling it that. That same textbook also has this: Theorem 7.3.12 "Lebesgue's Integrability Criterion": A bounded function f : [a,b] → ℝ is Riemann Integrable if and only if it is continuous almost everywhere on [a,b] So I feel pretty confident about that claim, too. Your mistake is that "the indicator function of the rational numbers embedded in the reals", aka. Dirichlet's Function, is continuous nowhere (see en.wikipedia.org/wiki/Dirichlet_function#Topological_properties for a proof) --- When I said "Riemann Integrals are more powerful", I meant "more powerful than the limits of rectangles explanation would imply". Not "more powerful than Lebesgue integrals"

3:14 this is a bit too quick. While the probability for this one number is 0, we also have an infinite amount of other rational numbers it can be. So the total probability is an infinite sum of things. To determine that that is 0 you need some more arguments.

@andrewkarsten5268

8 ай бұрын

Came to same this

This channel is gonna explode. Thanks for the explanation.

The idea that somehow the Lebesgue integral is "done horizontally" is misleading. The Lebesgue integral is defined _exactly_ like the Riemann integral, except that for the Lebesgue integral the simple functions are linear combinations of characteristic functions of _measurable_ sets rather than just intervals. Then, once you've defined it, you prove that the integral is the area of the subgraph, you prove Fubini's theorem, and the "Cavalieri principle": all this allows you to rewrite the integral "horizontally" so to speak. No need to directly define the Lebesgue integral "horizontally".

This channel has so nice quality! Congrats! :)

@vcubingx

4 жыл бұрын

Thank you so much!

Superb!!!!! This video just inspired me to extend this result into a deep area of number theory!

So much better than my Lebesgue Integration Course!!!

Mate I have an lesbegue integration exam in a few hours and was nearly having a panic attack till this video came along, thanks so much! Subscribed!

7:15 shouldn’t it be the integral of P(x)*x?

This channel deserves much more subscribers than it has now

Please do more of these vidoes! Good explanation. Would love some intuative measure theory / functional analysis.

Your English has gotten very good!

@vcubingx

4 жыл бұрын

Thanks, I'm very happy with my improvement too

Vivek my man ! Ur killing it with those animations !!! 🙌

Where textbook lag; you proceed with your animation and make sense what the hell mathematics underneath which is very appealing to visualize absurd abstract idea from where concept are taken. Nice work bro; mathematics channel like you are life saver for viewer.

Overall a good video, though I have a few comments. The figure featuring horizontal slabs on your thumbnail does not actually describe the Lebesgue integral. That figure indicates that we sum slabs of a fixed height and varying widths to approximate an integral. This is not accurate, as you describe yourself in your video. What happens instead is that we decompose the image of the function into a partition of intervals. The low end points of these intervals will be our weights for approximating an integral. We then collect the preimages of those intervals, which may be disconnected sets. However, these preimages will be measurable (by, e.g., continuity of the function. More generally by measurability of the function.) We then multiply the weights by measure of the preimages. In the coin counting analogy, the function values (which I'm calling weights) are the "number of coins of each type" and the measures of the preimages are the "values labeled on each coin." Also, I don't believe that the step function example is a good explanation as to why the Riemann integral is not preferable. There isn't any concern of mismatching upper & lower approximating sums in this case. Showing a visual of a more clear refinement demonstrates this. The problem with partitioning domains in high dimensional space is a good reason, but it isn't due to some conceptual complexity with domain decomposition. It is more that proofs on the properties of the multivariate Riemann integral require rather laborious notational baggage from multi-indexing high dimensional boxes combined with a time spent concerning the reader with the definition of measure zero sets while not actually trying to teach measure theory. Best reason to care about Lebesgue is going to be the use limiting properties, e.g. dominated convergence theorem.

1:01 Not to ruin what is an otherwise perfect video but continuity has a typo

@vcubingx

4 жыл бұрын

Good catch! I'll see what I can do about it.

@josebonilla5493

3 жыл бұрын

Yep. THere's an extra i the way he spelled it.

Ayanokoji learned this as a 5 year old

@localsimp4579

4 ай бұрын

Reason why I came here bro Ayanokoji is just HIM.

Can't wait to see how your content grows! Thanks for the video.

@vcubingx

4 жыл бұрын

Thanks Aaron!

Hey, nice video! Clear presentation 👌 If I may, can you skip less steps? Since it's a new concept for the viewer, it would help if you go slower even if you are using an argument you already went over earlier. It would help if you repeat the logic of the argument again while using it, just for clarity and connectedness. I just discovered your channel and Im excited to look at the rest of the videos! Keep goingg!! 👌👍

@vcubingx

4 жыл бұрын

Thanks for the feedback! I'll definitely keep this in mind when I make my next video!

Is the Graph at 6:50 correct? I'm not very deep into the mechanics, but as far as I understood the y_i * mu(A_i) is rectangle below the measurable Function similar to Rieman, but instead of equal x-length for the dx intervals the x-length is determined by dmu intervals on the y-axis. Please correct me if I am wrong.

Real life: It's easy in theory, but not so easy in practice. Mathematics: It's easy conceptually, but not so easy in theory.

I really like the analogy to expected value. It's much more helpful than the coin analogy in Lebesgue's quote. The coin thing is too vague, but expected value is much more specific and so is a much more satisfying analogy. I also really appreciate that you gave a specific real world example of when Lebesgue integration would be useful. Personally, I particularly like the example with circuits as I'm an Electrical Engineering major.

@vcubingx

4 жыл бұрын

Glad it was helpful! Thanks for watching!

@Lucky10279

4 жыл бұрын

@@vcubingx Thanks for making the video. 3b1b linked to it on his patreon and said it had similar animations to his videos. When I saw it was on Lebesgue integration I was excited, as every other explanation I'd found on the topic was too technical. A video similar in style to 3b1b that explained it sounded right up my alley and it was. I also watched your video on fractional calculus which was pretty interesting too.

I don’t like your 0.123123... It could be 0.12312345645612312345645600000...Rational but not matching your decimal string in the nth place. Still, you’re right of course that the integral is one, and if you switch to f(x)=1 if x is rational and 0 if x is irrational, the integral is 0. The rationals form an everywhere dense set of measure 0 on the interval.

If you take an increasing sequence of Riemann integrable functions, the limit might not be Riemann integrable. And even if you happen to know the limit is Riemann integrable, it is still very tedious to prove that the limit of the integrals is the integral of the limit (a result, which seems obvious). With Lebesgue integrand this is almost trivial to prove. Historically this was very important. Another main motivation historically was Fubini's Theorem. There the Riemann integral is again in trouble with "integrability issues" . With Lebesgue integral Fubini is a breeze really.

You are the 3Blue1Brown.Jr great video.... !!!

@plasma7851

4 жыл бұрын

😂 I thought it was him when I clicked on the video

Are you planning to take real/complex analysis in high school? You're still in your junior year right? I think you'll have plenty of time! Btw, did 3b1b's probability series inspire you to do this video on Lebesgue integrals or was it just your own curiosity? Your videos are awesome, keep up the great work!

@vcubingx

4 жыл бұрын

I don't think I will. The college I take my math courses at right now doesn't have an analysis course.

i keep seeing this vid's thumbnail and thinking it says "lesbian integrals". hats off for finding a way to stick out in my mind. youve earned my view

Aren''t the integration components are still vertically distributed, because,it is still y*measure(or length) across x-axis? as per your diagram , it looks more like xdy instead of ydx, which is still a reimann integral with change in axis ?

@angelmendez-rivera351

3 жыл бұрын

Yes, you are correct. The claim that the Lebesgue integral can be intuitively visualized as being a vertical integral is just incorrect, and it leads to a false intuition of what the integral is. A much better intuitive explanation of the integral would remind us of the fact that there is no intrinsic reason to partition the interval of integration into closed intervals that serve as the base of rectangles, and that rearranging the points in those intervals without changing the value of the function at those points, intuitively, should not change the integral. Being able to partition the interval into arbitrary sets, instead of only other intervals, means that we need to have a well-defined notion of what the "length of a set" is, one which generalizes the intuitive idea of "length of an interval" that we already have. Lebesgue integration, and measure theory, are precisely one way to solve this problem.

@pedroteran5885

2 жыл бұрын

The diagram is misleading in that it does not represent what is later described in the video. However, it is true that, for a non-negative real function f, there holds [Lebesgue integral of f against a finite measure m] = [Improper Riemann integral of m({f>=t})dt]. The diagram does represent the right-hand side. One advantage of the RHS is that you can replace m by a more general set function which is not a measure, in which case it is called a Choquet integral.

and it get paid out in the currency that somebody get some new knowledge, simply you get our satisfaction of knowing something, thank you, it is good, correct and sustainable.

Am I completely lost ? At 7:00 you say : _"consider trying to calculate the expected value of some probabilistic event if we're dealing with a probability distribution p of x like this normal distribution the expected value is just an integral from negative infinity to positive infinity of the distribution"_ But this is incorrect right ? Integrating the probability distribution (probability density for continuous variable) will simply give you 1 right ? This is one of Kolmogorov's probability axiom isn't it ? That the sum of the probabilities of all possible outcome in Ω be 1 ? In order to get the expected value, we need a weighted average, which means the integral of the value of the random variable, weighted by its probability.

I think the explanation will be even clearer if you mentioned more on what 'Lebesgue measure' is, at 5:07. Other than that great work!

@vcubingx

4 жыл бұрын

I agree! Watching it back I realized that I skimmed over the idea of the Lebesgue measure, but it's definitely an incredibly important topic.

@rzhang3927

4 жыл бұрын

​@@vcubingx Maybe you can make another video on that topic. I think Tao's book did a good job of explaining it. (Chp 1.1 to 1.3 I think, I am learning this myself now too) An animated version of that will be fantastic!

@vcubingx

4 жыл бұрын

@@rzhang3927 I'll look into it! Definitely a fantastic idea.

Great video! Lot of thanks for clarifying a frightening concept

THANK YOU SO MUCH FOR THIS! Now I can learn the difference between classic reimann integration and this technique.

@vcubingx

4 жыл бұрын

Glad it was helpful!

This video is worth watching.... I really liked the way you described lebesgue integral really learnt something interesting subscribed ❤

Very nice explanation, it has been awhile since I took a measure theory class so my memory of it has diminished, but your explanation perfectly encapsulates the main concept behind Lebesgue integration.

Nice , very nice . Please which App did you use for the graphic visualization of those area

1:02 I don't see how this makes any sense. Either it works or it doesn't. You don't have to "visualize" anything.

Sir ! Which software did you use to make these awasome videos.

This method is also the best way to integrate a normal distribution put into the 3rd dimension using multiplicative integrals but to do that it’s almost ALWAYS just 2 of them.

I don't really like this "horizontal integral" analogy. That seems to me like Riemann but with a change of variables. You can consider a "partition" of the function's range with a similar intuition as in a cumulative distribution function, but the rectangles that are formed should not be horizontal, but still vertical. Lebesgue is about taking the inside vertical rectangles and the outside ones (taking the sup in the infinitesimal interval), showing their sum is essentially equal and then taking one of them as the integral, which turns out to be exactly as Riemann's (where the latter can be calculated). The fundamental change is in the fact that we do not care about which partition of an interval we take, but instead we consider all the simple functions (hence all possible values that f can reach, which to me is different from a partition of the range) and come to our result by density arguments. That's why it is called the "weak" integral, since proceeding by density we allow for functions that behave oddly in countable points and treat them as if they were normal, that's why Dirichlet's can be summed by Lebesgue. Of course this is an abstract construction in that you can't take all the simple functions when trying to calculate an integral. In fact numerical methods are based on Riemann's integral. Here is an article about why this horizontal visualization is actually misleading: en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics/Archive/2007/Jun#Mathematical_eyes_sought

@user-nf7on7zj1o

2 ай бұрын

I totally agree with you. The animation about integration using horizontal strips is very misleading, if not down right wrong.

Dude, you are the god of animations. Lebesgue might be cool, but your animations are cooler. Maybe, you animate Itô the next time. :)

I didn't get why the probability of .123123123.... being rational will be 0. From what I understand, the probability of choosing 0.123123.. from [0,1] is 0. Can you please explain this elaborately or give a link to where it is explained?

@vcubingx

4 жыл бұрын

Another kind viewer gave a really good explanation to this, here's his comment if you can't find it: " This is true. I don't think he was trying to give a formal explanation though, so much as just trying to give a little bit of intuition. A slightly more formal version would invoke the concept of "countability" and of different sizes of infinity. I'm sure you can find a more extensive explanation online; however, in brief: we say that two sets have the "same size" or "same cardinality" if they can be put into a one-to-one correspondence (that is, there's a function from one set to the other set which is one-to-one and onto, and thus can be "inverted" or "undone" -- in fancy math language, this is called a "bijection"). This is indeed a reasonable definition: when you count something (say, a collection of sheep) you assign each sheep to a natural number in a strictly increasing way ("sheep number 1, sheep number 2, etc."), such that each sheep gets a number and no sheep is given multiple numbers (no double counting happens). You then see what subset of the natural numbers you're in correspondence with (if you've put the sheep in correspondence with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, you say that there are 10 sheep; if you've put the sheep in correspondence with the set {1, 2, 3, ..., N}, you say that there are N sheep). The advantage of this definition, however, is that it allows you to compare the size of infinite sets. For instance, and slightly paradoxically, using this definition there are just as many positive even integers as there are positive integers: we can see this by the function from natural numbers to even natural number given by f(n) = 2n. In fact, this property is characteristic of infinite sets -- one definition of an infinite set is a set which can be put into one-to-one correspondence with a proper subset of itself. Using this definition, one can show that there are just as many rational numbers as there are positive integers, but that there are more real numbers than positive integers. To see this, let's introduce what's knows as the notion of a "height" of a rational number. The "height" of a rational number p/q (written in "simplest," or "reduced" form, with p and q sharing no common factors) is simply the larger of |p| or |q|, where the vertical bars denote absolute value, and can be thought of as sort of a measure of the "complexity" of a given rational number (so 1/2 is simpler than 47/355). Clearly, there's a finite number of rational numbers at any given height h, since there's finitely many integers between -h and h. Now, imagine listing the rational numbers in order of increasing heights (so, we write out all of the rational numbers with height 1 in some order, then all of the rational numbers with height 2 in some order, then all of the rational numbers with height 3 in some order, and so an and so on). Now that you've listed them, you've got a one-to-one correspondence: assign "1" to the first rational number on that list, "2" to the second, and so on and so on. Since every rational number has some height, every rational number appears somewhere on this list, and by construction appears only once, so this correspondence is indeed one-to-one. Therefor, we say that the "cardinality" of the rational numbers is the same as that of the positive integers, or that the two have the same "size." However, the same is not true of the real numbers. To see this, we use a "proof by contradiction." Imagine that the set of just the real numbers between 0 and 1 had the same size as the natural numbers. We can then imagine listing them out one after another in an infinite list, as we did with the rational numbers. We can also imagine writing out each real number as its infinite decimal expansion. Now, construct another number as follows: in the 1st decimal place of our new number, we write "7" if the 1st decimal place of the 1st number was NOT a 7, and if it was, we write "1". In the second decimal place of our new number, we write "7" if the 2nd decimal place of the 2nd number was NOT a 7, and if it was, we write "1." More generally, in the nth decimal place we write "7" if the nth decimal place of the nth number was NOT a 7, and if it was, we write "1." Now I claim this number is nowhere on our list: if it was somewhere on our list, then we could consider the number associated with the place it could be found on the list (that is, it would be at the mth place on this list for some value of m). But we know that the number we created disagrees with the mth number on our list in the mth decimal place--this yields a contradiction, since the number we created is clearly a real number, yet we had assumed that we'd listed all real numbers, and we've now created a real number which we failed to list! Since the set of real numbers includes those real numbers between 0 and 1, it certainly cannot be put into one-to-one correspondence with the positive integers, and since the set of real numbers includes all of the positive integers, the set of real numbers can be said to be the "larger" set. Now, we get to what vcubingx was saying -- because there are far far more real numbers than rational numbers, there must be far far more irrational numbers than rational numbers (since the set of reals is just the set of rationals combined with the set of irrational numbers). Thus, it is reasonable to say as he said in the video, that there are infinitely more real numbers than rational numbers. Of course, connecting this "cardinality" argument to something specific about "measure" requires more technical machinery than I've described here (in fact, more technical machinery than I yet have), but this long argument describes some of the intuition. Given its length, you can see why vcubingx didn't really want to get into all of this in the video. "

@subaruyagami2327

4 жыл бұрын

@@vcubingx ohh, I get the intuition behind it. Thanks! Great video, love it.

Thanks, and it's so easy & simple!

You make such amazing videos 🙌

3:19 but there should be infinitely many possible rational numbers like 0.12311111... and 0.1231212121212...

@omarino99

4 жыл бұрын

If it’s rational at some point it it has to repeat itself though, 123 was just an example. You can always say that at some point it has to stop and repeat itself otherwise it wouldn’t be rational.

@nathanielkingsbury6355

4 жыл бұрын

This is true. I don't think he was trying to give a formal explanation though, so much as just trying to give a little bit of intuition. A slightly more formal version would invoke the concept of "countability" and of different sizes of infinity. I'm sure you can find a more extensive explanation online; however, in brief: we say that two sets have the "same size" or "same cardinality" if they can be put into a one-to-one correspondence (that is, there's a function from one set to the other set which is one-to-one and onto, and thus can be "inverted" or "undone" -- in fancy math language, this is called a "bijection"). This is indeed a reasonable definition: when you count something (say, a collection of sheep) you assign each sheep to a natural number in a strictly increasing way ("sheep number 1, sheep number 2, etc."), such that each sheep gets a number and no sheep is given multiple numbers (no double counting happens). You then see what subset of the natural numbers you're in correspondence with (if you've put the sheep in correspondence with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, you say that there are 10 sheep; if you've put the sheep in correspondence with the set {1, 2, 3, ..., N}, you say that there are N sheep). The advantage of this definition, however, is that it allows you to compare the size of infinite sets. For instance, and slightly paradoxically, using this definition there are just as many positive even integers as there are positive integers: we can see this by the function from natural numbers to even natural number given by f(n) = 2n. In fact, this property is characteristic of infinite sets -- one definition of an infinite set is a set which can be put into one-to-one correspondence with a proper subset of itself. Using this definition, one can show that there are just as many rational numbers as there are positive integers, but that there are more real numbers than positive integers. To see this, let's introduce what's knows as the notion of a "height" of a rational number. The "height" of a rational number p/q (written in "simplest," or "reduced" form, with p and q sharing no common factors) is simply the larger of |p| or |q|, where the vertical bars denote absolute value, and can be thought of as sort of a measure of the "complexity" of a given rational number (so 1/2 is simpler than 47/355). Clearly, there's a finite number of rational numbers at any given height h, since there's finitely many integers between -h and h. Now, imagine listing the rational numbers in order of increasing heights (so, we write out all of the rational numbers with height 1 in some order, then all of the rational numbers with height 2 in some order, then all of the rational numbers with height 3 in some order, and so an and so on). Now that you've listed them, you've got a one-to-one correspondence: assign "1" to the first rational number on that list, "2" to the second, and so on and so on. Since every rational number has some height, every rational number appears somewhere on this list, and by construction appears only once, so this correspondence is indeed one-to-one. Therefor, we say that the "cardinality" of the rational numbers is the same as that of the positive integers, or that the two have the same "size." However, the same is not true of the real numbers. To see this, we use a "proof by contradiction." Imagine that the set of just the real numbers between 0 and 1 had the same size as the natural numbers. We can then imagine listing them out one after another in an infinite list, as we did with the rational numbers. We can also imagine writing out each real number as its infinite decimal expansion. Now, construct another number as follows: in the 1st decimal place of our new number, we write "7" if the 1st decimal place of the 1st number was NOT a 7, and if it was, we write "1". In the second decimal place of our new number, we write "7" if the 2nd decimal place of the 2nd number was NOT a 7, and if it was, we write "1." More generally, in the nth decimal place we write "7" if the nth decimal place of the nth number was NOT a 7, and if it was, we write "1." Now I claim this number is nowhere on our list: if it was somewhere on our list, then we could consider the number associated with the place it could be found on the list (that is, it would be at the mth place on this list for some value of m). But we know that the number we created disagrees with the mth number on our list in the mth decimal place--this yields a contradiction, since the number we created is clearly a real number, yet we had assumed that we'd listed all real numbers, and we've now created a real number which we failed to list! Since the set of real numbers includes those real numbers between 0 and 1, it certainly cannot be put into one-to-one correspondence with the positive integers, and since the set of real numbers includes all of the positive integers, the set of real numbers can be said to be the "larger" set. Now, we get to what vcubingx was saying -- because there are far far more real numbers than rational numbers, there must be far far more irrational numbers than rational numbers (since the set of reals is just the set of rationals combined with the set of irrational numbers). Thus, it is reasonable to say as he said in the video, that there are infinitely more real numbers than rational numbers. Of course, connecting this "cardinality" argument to something specific about "measure" requires more technical machinery than I've described here (in fact, more technical machinery than I yet have), but this long argument describes some of the intuition. Given its length, you can see why vcubingx didn't really want to get into all of this in the video.

@kcsj9000

4 жыл бұрын

@@nathanielkingsbury6355 diagonal argument?

@nathanielkingsbury6355

4 жыл бұрын

@@kcsj9000 Yeah, precisely. Wasn't sure precisely how much background you had, so I wanted to do the argument out rather than simply use the analytic magic wand of saying "by Cantor's diagonal argument..."

@tomkerruish2982

4 жыл бұрын

@@nathanielkingsbury6355 From my long, rambling post above, you can show that any countable set has measure 0. Essentially, take epsilon paint, and cover the first point with half of it, the second with a quarter, the third with an eighth, etc. You paint the entire set with epsilon paint, which can be as small as you like; this is what is meant by having Lebesgue measure 0.

To Visualize 4D object, think blue cube, split in half, and the second half is color coded sides that represent each cube value in a color.

@ccshumshum8104

8 ай бұрын

Doesn't make sense

@davimedina5165

5 ай бұрын

Faulty reasoning, truth is we have no conceivable idea of what the 4th dimension would be, just like a 2D being can't conceive what the 3th dimension is, purely our assumptions that are probably wrong

The sources are awesome, I couldn't understand wikipedia but with "(Riemann) Integration Sucks!!!" from Peyam i could understand a lot more. Good video

Beautiful explanation! Thanks so much for this video :)

@vcubingx

4 жыл бұрын

Glad you enjoyed it!

You're talented my dude, keep it up.

Yes, it is the range of the function which is partitioned when dealing with Lebesgue integration. But the picture you draw in 3:40 is not what a simple function approximating the Lebesgue integral would look like and is slightly misleading.

I still don't get how it follows that lebesgue integration is just adding up strips horizontally just from your example of the lebesgue integration of the dirichlet function alone

@NateROCKS112

4 жыл бұрын

The Lebesgue measure is the most important concept. It essentially generalizes the concept of length. So you're iterating over the range and taking the width of the corresponding domain. The Dirichlet function isn't continuous, but that's where the Lebesgue measure comes in.

@angelmendez-rivera351

3 жыл бұрын

@@NateROCKS112 Yes, but you missed the point of the objection. The video, and importantly, the thumbnail and title of the video, present the integral as being a vertical way of doing integration, as opposed to Riemann integration, which is horizontal, but this is just not the case at all.

can you make a step by step solution to some Lebesgue Integral?

According to Riemann integration, f(x)=1 if x is rational, f(x)=0 if x is irrational, then the integral of f over [0,1] does not exist. But according to Lebesgue integration, the integral of f over [0,1] exists. Why? Can you explain clearly for me?

If the only difference between the Riemann and Lebesgue integral was dividing up along the x-axis vs y-axis, then the Jordan measure (which uses boxes and doesn't care about the x or y axis) should be able to find the area under the graph of any Lebesgue-integrable function, right? The fact that this is not possible I think means there is some other deeper difference between the two integrals.

What blows my mind is that the integral of the Dirichlet function is equal to 0, while the integral over the _negated_ Dirichlet function is equal to 1. The argument is the same, and I understand it in both cases, but still, it blows my mind.

great explanation! very cool

@vcubingx

4 жыл бұрын

hackerman

@gregoriousmaths266

4 жыл бұрын

lol

Damn this video was hot!!!!! Pure flame xD i never understood lebesque integral since I'm an engineer and you explained it in such a perfect way 👌🏻 hope you become an university profesor one day :)

At 7:38, the verbal emphasis on $5 is the same as that on $10, so (listening, not watching closely) I got confused in the subsequent math and had to back up to see what I'd missed. Not a complaint, just a note that inflection is relied upon by at least one viewer :). Thanks for the video!

Nice explanation, will looking forward for other technique from you. 😀

Great video mate

@vcubingx

4 жыл бұрын

Thanks!

7:05 There is an x missing under the integral? Otherwise the integral is just 1.

Thank you for this video! What about visualization of Ito Integral in contrast to Riemann/Lebesgue Integration please?

@vcubingx

2 жыл бұрын

Glad you liked it! That's a tough topic to cover, so probably not anytime soon

I didn't understand anything but the visuals are amazing. Great job!

Your videos are great. Just an idea: Maybe it'd be nice to change a little bit more the font used in the videos to give them its own visual identity, right now they look too similar to 3b1b. Anyway, that's not a problem! Just something I think would be nice to the channel!

@jorgemarcelo4708

4 жыл бұрын

If its not possible to change the font then maybe it'd be a cool idea to choose a color pattern for the channel. Black background with blues graphs is the one 3b1b uses so maybe we could choose something else ! Again, just an idea I think would be beneficial to the channel.. love your content!

@vcubingx

4 жыл бұрын

@@jorgemarcelo4708 Thanks for your idea, it's definitely something I'm gonna start doing. I should try using the same font which I use in my thumbnails, but I don't have any ideas for the background color. What do you think?

@jorgemarcelo4708

4 жыл бұрын

@@vcubingx In the website [colorhunt.co] it's possible to find lists of colors that go well together. It is even possible to search with tags like "pastel" colors. Maybe if you find a color pattern that you like the darkest color could be used in the background and the lighter ones in graphs.

@jorgemarcelo4708

4 жыл бұрын

I suggest typing "Winter" in the search bar, it'll give you "serious-looking" color palettes

@vcubingx

4 жыл бұрын

@@jorgemarcelo4708 Thanks! I'm definitely gonna try these out. One of my friends also suggested [coolors.co] to generate some pallets, so I'll give both of these a try.

The original idea of Lebesgue integral is the measure of the area under the graph which is later discovered to be coincided with sup of integrals of simple functions

I learned the Darboux integral, not the Riemann integral. While they integrate precisely the same class of functions, they're conceptually different. With the Darboux integral, we consider the integrals of step functions less than or equal to the given function (the lower integral), and analogously for the upper integral. (Step functions are easy-peasy to integrate.) If the sup of the lower integrals equals the inf of the upper integrals, then that's the integral of the original function. (Tl;dr we approximate the function from above and below, using the method of exhaustion.) This generalizes to the Lebesgue integral; rather than approximate a function from below by a finite linear combination of characteristic functions of intervals, we substitute measurable sets for intervals. (Of course, now we need to define Lebesgue measure and Lebesgue-measurable sets.) To show that the rational numbers have measure 0, suppose we have epsilon paint, for some epsilon greater than 0. Since the rationals are denumerable, we order them r_1, r_2, r_3,... We use half our paint to cover r_1, one-fourth to cover r_2, one-eighth to cover r_3, etc. At the end, we've covered all the rationals with our paint, showing that they take up at most epsilon room. However, we can make epsilon as small as we like, so they take up less than any positive amount of room, and thus actually take up 0 room. (This vague concept of "taking up room" is what's meant by Lebesgue measure.)

@tracyh5751

4 жыл бұрын

I also learned the Darboux integral instead of the Riemann integral in my undergrad analysis class.

@tomkerruish2982

4 жыл бұрын

@@tracyh5751 Were you using Apostol?

@subaruyagami2327

4 жыл бұрын

Riemann integral and Darboux integral are equivalent, so in a way you learned Riemann integral too!

@tomkerruish2982

4 жыл бұрын

@@subaruyagami2327 True, but I find them different conceptually. Specifically, it's very "hand-wavy" (imo) to claim that every Riemann sum of a function will converge to the same limit as the mesh size goes to 0. Of course, this can be made rigorous, but it seems to me that the Darboux integral is easier to establish, in that it is often simple to find partitions of an interval such that the associated upper and lower Darboux sums are within a given epsilon of the sought-after result, whereas showing the same result for every Riemann sum below a certain mesh size appears daunting. Again, this is based on my experiences and opinions, although I note that the Wikipedia entry on the Darboux integral states, "The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral." (By no means the best-regarded source of information, but it's rather late for me and I'm headed to bed.)

@tracyh5751

4 жыл бұрын

@@tomkerruish2982 No, instructor's notes.

This was amazing to watch 😊

Cool video. Btw, in statistics and probability it's more convenient sometimes to use stieltjes integral rather than lebesgue's. It's a bit simplier to calculate

@ramanunnikrishnan7354

3 жыл бұрын

thanks that was useful

After looking at a couple of your videos I wondered, do you use 3b1b's manim python library for your animations?

@pragyanpranay3681

8 ай бұрын

I'm very sure he does... although you would have discovered it by now...

whoa ! this is so cool :O which book/site are you referring here?

thankyou, that was a very clear explanation

Wow. Until I heard his name said aloud, my brain automatically respelled Lebesgue's name as Lebesque even while reading it!

Heyo ! Thanks for the vid, it's great :) what's your outro music btw ?

Everybody gangster until the integral suddenly turns lebqsgue

I remember this when we studied Cantor sets

Aaaahhh yes. I remember these

@joshuacoleman6122

8 ай бұрын

It's me 3 years later!

very lovelly explanation

Excellent video lecture.

Very clearly explained! Congrats!

@vcubingx

4 жыл бұрын

Thanks a lot!

Very nicely explained!!!

This video is great. But you teased im the beginning with 3 and 4 dimensional volume problems. So I was expecting a follow up on why Lebesgue integration is more natural or intuitive in that setting. Do you plan on a follow up?

@pedroteran5885

2 жыл бұрын

If not, you may want to look for some video on Lebesgue measure in R^n.

This is the best video 👍👍...Please sir give us more video about this topic and give us more example...

i read that as lesbian integral

@vcubingx

8 ай бұрын

that's crazy

Lebesgue integrals confused me a bit before, but it's kinda more clear now I guess.

Easy to understand! Thank you!

2:40 How is the probability of getting 1 equal to 1/10 . There are 3 numbers it can be (1, 2, 3) and therefore it must be 1/3? And it was set up so as the decimal represents repeating 1,2,3 decimal, so why are we guessing the value of the next digit, when we know its 1, 2, 3?

Shouldn't the expectation be the integral of the probability distribution multiplied by x?