Riemann vs Lebesgue Integral
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- Опубліковано 22 лют 2018
- In this video, I show how to calculate the integral of x^3 from 0 to 1 but using the Lebesgue integral instead of the Riemann integral. My hope is to show you that they indeed produce the same answer, and that in fact Riemann integrable functions are also Lebesgue integrable. Enjoy!
I'm a physics Ph.D student and I must say you are a gift doctor. We need more people to enthusiastically teach advanced math in such a clear and concise way with a smile. You are doing an outstanding service, keep the excellent work.
Thanks so much!!!! ☺️
Important caveat: The Lebesgue integral is only an extension of the normal Riemann integral, not of the improper Riemann integral. The Dirichlet integral, for example, (Int(0..inf, (sin x)/x dx)) is improperly Riemann integrable, but not Lebesgue integrable. At least, that's what Wikipedia claims.
Correct!
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Hehe. I love your channel.
You explain everything with huge smile and a lot of passion i really enjoyed watching this !!
Riemann integral: partition [a,b] where a=x0
Greetings from Colombia doc.
Thanks so much for the explanations
Good job man!You have a good personality and an interesting presentation way.
Thank you. You just made me more educated ...
This is such a great video! Thank you!
What a matchup. Truly the Logan Paul vs KSI of its day.
Hello Professor,
I write from Italy while i am studying for Analysis 3 at Physics in Milan and i was wondering:
Is it possible to define the outer measure for any set?
If yes, given that Vitali's set is not Lebesgue-measurable, what is its exterior measure?
I learnt that with the Lebesgue integral the area under the curve is sliced horizontally.
Or did I miss something?
Same for me. Now I'm lost.
Ye, that's the basic definition and the most important intuition for Lebesgue integrals. The idea of slicing horizontally is useful because you no longer need any notion of "intervals" or even "order" on the x axis, so you can actually replace it with any set that allows us to measure the "area"/"size" of some of it's subsets, which is an enormous generalization compared to (naive) Riemann integration.
A surprising thing you get with this definition is good behavior of integration of functional sequences and their limits (dominated convergence and some other less famous theorems). As it turns out, you can give an equivalent definition for the Lebesgue integral as the supremum of integrals of all "simple" functions that are bounded by the function that we want to integrate. This definition is a bit similar to the one with slicing the area under the curve horizontally, as whenever you slice that area in some way you basically get a simple function that approximates the integral. And one can show, that for well-behaved (measurable) functions it does not matter how we approximate the function, so it suffices to pick only those approximations that are bounded by our function, therefore allowing us to replace some vague limiting process with arbitrary partitions of the y axis by a simple and elegant supremum.
If you want to get a better understanding of the underlying theory I'd suggest reading a book on measure theory, functional analysis or (advanced) probability theory as those are the areas where Lebesgue integration is used the most, at least according to my experience.
Yeah, I'm not any less confused...
nice explanation, interesting presentation,
Can you integrate e^(x^2) with Lebesque Integral ?
So I understand the Lebesgue integral is a true extension when the Riemann integral does not work, but for "simple" functions like polynomials the resulting funtioncs will be same same. What about trig, exp, ln functions and their combinations? Do we need the chain rule, U-substition and the like as well? Thank you very much! And continue your great work!
Karsten Meinders For differentiable functions, the fundamental theorem of calculus and the chain rule etc. still apply, so Lebesgue integrating those functions is exactly the same way as Riemann integrating! In one of the comments below I explain what happens for example with cos.
Shouldn't the width written at 5'10" be 1/N, instead of i/N?
Yeah, it’s a small typo, but fortunately the rest of the proof is correct!
We need more push on Analyse II :P Can you show us Taylor's Theorem?
Can you do some max min values of two or more variables with graphs or something like will be nice to see.
Taylor’s Theorem would be nice :) There’s already a video on the max min values, it’s called “The True Second Derivative Test”
The idea is to show us some solving limit example with O notation. It's really interesting.
respect
I wonder.... What is the actual length of the line between f(a) and f(b)? Like for f(x)=sqrt(1-x) the length from -1 to 1 is π, so what about x^3 and others?
It’s the integral from a to b of sqrt(1+ (f’(x))^2) dx
This looks like a special case, since we know the formula for the sum of the cubes of the integers from 1 to N, but what about the global case, does it exist for Lebesgue integral, like *F(b) - F(a)* for Riemann integral? How to integrate cos(x) with Lebesgue, for example?
Thanks for the video btw :)
Same process!
Here fn = sum from 1 to n of cos(x_i) 1_(xi-1,xi),
So the integral of cos from a to b is lim
n of fn = lim n Delta x sum cos(xi)
This limit of sums is equal to sin(b) - sin(a) by the fundamental theorem of calculus, and therefore the Lebesgue integral of cos(x) from a to b is sin(b) - sin(a)
Thank you very much 😍😍😍❤❤
When Fn converges to f. Is it pointwise or uniform?
Pointwise
Excellent
I can't figure out how you find the indicator function of the f(x)
Think of it by working backwards:
You have the term f(xi) Delta x = f(xi) (b-a)/n = f(xi) times (xi - xi-1)
Now this term is the area of a rectangle of width xi - xi-1 and height f(xi)
On the other hand, consider the function that has value f(xi) on the interval (xi-1,xi) and zero everywhere else. The integral of this function is precisely equal to f(xi) times (xi - xi-1).
But in general, a function which has value c on an interval A and 0 everywhere else is c times a function which has value 1 on A and 0 everywhere else, but this is just c times the indicator function of A.
So our function here is precisely f(xi) times the indicator function of (xi-1,xi)
Now i think i got it! Thank you Master!
I thought Riemann was vertical rectangles and Lebesgue was horizontal rectangles (based off pictures on Wikipedia article).
What’s an example of neither integrable?
1/x from 1 to infinity, or the indicator function of a non-measurable set. But as you see it’s hard for a function not to be Lebesgue integrable!
Dr. Peyam's Show
Dr. Peyam's Show did you mean 1/x, [0,1]? Because when i looked up the example that is the interval given.
Or it doesn’t matter because [0,1] is symmetrical to [1, inf)?
By the integral convergence test the integral of 1/x dx from 1 to infinity converges iff the sum from a to infinity of 1/x converges, set a=1 and:
1+1/2+1/3+...=1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+...>1+1/2+1/2+1/2+....=1+sum from 1 to infinity of 1/2, which diverges(Oresme's proof), hence the integral from 1 to infinity of 1/x dx also diverges
Both are not integrable, actually :)
well, as 1/|0|->infinity we have the integral f^+ du->infinity and because 1/x>0 for x>0 the integral f^- du is undefined hence also from 0 to 1 doesn't exists, but symmetry is not the right reasoning
the symmetry is over the line x=y, so if you say that the integral from 0 to 1 of 1/x - x dx=integral from 1 to infinity of 1/x dx+the integral from 0 to 1 of x dx by symmetry and then prove that the integral from 0 to 1 of x exists you can argue that they both not integrable by symmetry, but it is a lot more annoying
Nice! now do it with function f:R2 to R2 !
how can I prove that the sum of the fns converge to x^3?
Look at a random point x in [0,1] and show that |f(x) - fn(x)| goes to 0
I LIKE 5:45 see the middle triangle smiling 😃 😊 with close eyes ( i-1)/N
Thanks sir
excuse my stupid question but when does the riemann integral doesn't work?
ua-cam.com/video/jBl3nsg0rhg/v-deo.html
Love your German jokes, weil ich deutscher bin 😂
So the bottom line is that Lebesgue integral is useful whenever the function is not Riemann integrable. For this Riemann integrable function x^3 the proof presented here seems to me to be kind of circular reasoning. You start with Riemann, go over to Lebesgue and back to Riemann.
Yep
Lol best cartoon ever
How come no one on the internet can explain the Lebesgue integral in simple terms. ffs
Have you checked out my Lebesgue integral video?
@@drpeyam I have. As someone who is not studying math though (I study food science), but still wants to learn the concept because it may be useful for me, as math is the language of all sciences; I have consulted the internet. Now to clarify I am confident in my ability that I would be able figure it out if I went and read the entire Wikipedia page or chapter in a calculus book, but that would require too big of a time investment as I don't yet know when this knowledge will come in handy and have more important things to do in my studies.
If you want to make content that appeals to a wider range of fields, I would recommend working on simplification of your explanation. Take as inspiration the 3Blue1Brown channel. Analyze his work and you will see that besides fancy math words he uses trivially simple analogies and graphics.
If you wanted to explain for instance the Lebesgue integration to a person like me effectively, you would at the start of the video have to: sum it up in
You essentially ask yourself
"How long is the interval on which f has a value of a?"
And you do that "for all real a"
Since that isnt possible tho, you first restrict yourself to simple functions, functions which only take on a finite number of values and then define your function as a limit of simple functions
Examples for simple functions:
g(x) = 1 for all x
Or f(x) = 0 for x=0
The integrals over these examples would diverge (i hope thats obvious)
But if you take f(x) = 2 for x in [0,1] u [2, pi), then the integral of f would just be
2 * [ (1-0) + (pi - 2) ] = 2(pi-1)
These may be quite simple cases, but they are the foundation since now we could define fn(x) = k for x in [k, k+1) and 0
@@why_though BRO, if u wanted to know the concept of Leb. , plz don't see this type of video which must lead u to go the wrong way.
Makes no sense the Lebesgue integral part.
:(
Why does he talk like that? Does he do that on purpose to make the video funny?
You look like you are new to Dr. Peyam's channel, instead of saying dumb things like that go get yourself a life!!
🤣😂🤣😂
Can you integrate e^(x^2) with Lebesque Integral ?