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Lebesgue integral

by woundedtiger4
Tags: integral, lebesgue
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woundedtiger4
#1
May7-12, 06:04 PM
P: 176
can anyone please show me that how to apply Lebesgue integral by showing some example like we do Riemann integral in calculus class? for example:
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Congruent
#2
May7-12, 07:47 PM
P: 9
It is a standard fact that if f is a real valued function defined on a closed interval which is integrable in the Riemann sense, then the function is measurable and integrable (in the Lebesgue sense) and the two integrals are equal. So if what you're asking for is an example of finding an anti-derivative of functions as is done in calculus, then the procedure is to note the above and integrate.

However, the power and utility of the Lebesgue integral is the ease with which more complicated functions can be integrated (the integral can easily be broken up over various sets which have specific properties and has powerful convergence theorems; there also exist functions which may fail to be integrable in the Riemann sense; see the characteristic function of the rationale over any closed interval). But these don't have general calculus style approaches. These sorts of integration arguments require more care than applying a set of procedural formulas.
woundedtiger4
#3
May7-12, 09:33 PM
P: 176
Quote Quote by Congruent View Post
It is a standard fact that if f is a real valued function defined on a closed interval which is integrable in the Riemann sense, then the function is measurable and integrable (in the Lebesgue sense) and the two integrals are equal. So if what you're asking for is an example of finding an anti-derivative of functions as is done in calculus, then the procedure is to note the above and integrate.

However, the power and utility of the Lebesgue integral is the ease with which more complicated functions can be integrated (the integral can easily be broken up over various sets which have specific properties and has powerful convergence theorems; there also exist functions which may fail to be integrable in the Riemann sense; see the characteristic function of the rationale over any closed interval). But these don't have general calculus style approaches. These sorts of integration arguments require more care than applying a set of procedural formulas.
thanks a tonne

Bacle2
#4
Aug15-12, 04:37 AM
Sci Advisor
P: 1,169
Lebesgue integral

Sorry for the necropost; I ran into this in a search. Feel free to Ignore if you don't need
it anymore:

Just like the case with a Riemann integral, where you do Riemann sums, in the Lebesgue case, you do Lebesgue sums, where you partition the range instead of the domain, and then the value of the Lebesgue integral is the sum of these values:

If your range likes in [y1,y2], then you partition your range, select some points yi, i.e., you select a point in each interval of the range ---remember you select points in your domain when you do Riemann sums--then your sums are of the form:

Ʃy=1n yi*m(A)

where A is the preimage set under f, so that you sum a value selected in your range. There are , of course, conditions for this integral to exist/converge; some of these conditions can be described with simple functions, but I think it is important to note that, just like with Riemann, the value of the sum should approach a limit as the number of partitions increases.

Is that what you were asking?

times the measure of the associated preimage.
woundedtiger4
#5
Aug15-12, 11:34 AM
P: 176
Quote Quote by Bacle2 View Post
Sorry for the necropost; I ran into this in a search. Feel free to Ignore if you don't need
it anymore:

Just like the case with a Riemann integral, where you do Riemann sums, in the Lebesgue case, you do Lebesgue sums, where you partition the range instead of the domain, and then the value of the Lebesgue integral is the sum of these values:

If your range likes in [y1,y2], then you partition your range, select some points yi, i.e., you select a point in each interval of the range ---remember you select points in your domain when you do Riemann sums--then your sums are of the form:

Ʃy=1n yi*m(A)

where A is the preimage set under f, so that you sum a value selected in your range. There are , of course, conditions for this integral to exist/converge; some of these conditions can be described with simple functions, but I think it is important to note that, just like with Riemann, the value of the sum should approach a limit as the number of partitions increases.

Is that what you were asking?

times the measure of the associated preimage.
thanks a tonne for helping me to understand it, though I have managed to clarify this topic from different texts but your answer is truly great.
once again thanks a tonne.

Edited note: what is m here? is it some function? if it takes A (which is domain) then what does map into ? range? can you please guide me in the context of probability if it's possible. Thanks
pwsnafu
#6
Aug15-12, 05:47 PM
Sci Advisor
P: 820
Quote Quote by woundedtiger4 View Post
Edited note: what is m here? is it some function? if it takes A (which is domain) then what does map into ? range? can you please guide me in the context of probability if it's possible. Thanks
The Riemann integral assumes you partition up the domain and approximate with rectangles.
The Lebesgue integral assumes you partition up the range and approximate.
I use word assume because its not obvious you can do this and and even achieve a unique answer.

In Riemann, you take an interval [a,b] and split it into subintervals [x0, x1], [x1, x2], ...
You then define a way to "measure the length" of these subintervals ##m([\alpha,\beta]) = \beta-\alpha##. This gives you the width of each rectangle.

In Lebesgue, we split up the range, then look at the sets of form ##f^{-1}(E) = \{ x \in X | f(x) \in E\}##. This is the pullback, all the x that f sends to E. However, in general this set will look nasty even if E looks nice. So we define a "measure", a function whose input is subsets of the domain of f and outputs a number (the "length" of that set). That is m.

The theory is called measure theory and if you are doing probability you will need to know this.
Bacle2
#7
Aug15-12, 06:45 PM
Sci Advisor
P: 1,169
In a very basic and informal way, this is part of how it is laid out:

You have a measure triple (X, m, Sigma). X is a set/space, m is a measure (see below), and the sigma algebra is a collection of subsets of X.

You have a measure m, which is a function defined on a certain collection of subsets that is countably subadditive (measure of the union of sets is ≤ sum of the measures), non-negative and monotone (A Subset B, then m(A) < m(B)) .

You normalize your space, so that m(X)=1 . Then the probability of an event is its measure. So that, for example, the probability of hitting a rational with a dart thrown
at random is the measure of the subset of rationals in X . In the Lebesgue measure , the
measure of all rationals is 0 .

I need to go, but let me know if there is something else that interests you.
woundedtiger4
#8
Aug16-12, 10:04 AM
P: 176
Quote Quote by Bacle2 View Post
In a very basic and informal way, this is part of how it is laid out:

You have a measure triple (X, m, Sigma). X is a set/space, m is a measure (see below), and the sigma algebra is a collection of subsets of X.

You have a measure m, which is a function defined on a certain collection of subsets that is countably subadditive (measure of the union of sets is ≤ sum of the measures), non-negative and monotone (A Subset B, then m(A) < m(B)) .

You normalize your space, so that m(X)=1 . Then the probability of an event is its measure. So that, for example, the probability of hitting a rational with a dart thrown
at random is the measure of the subset of rationals in X . In the Lebesgue measure , the
measure of all rationals is 0 .

I need to go, but let me know if there is something else that interests you.
so m in lebesgue gives the width of rectangle as it gives in reimann?

I am lil' confused that, is m=f^-1 (E) ?????
Bacle2
#9
Aug16-12, 11:56 AM
Sci Advisor
P: 1,169
Quote Quote by woundedtiger4 View Post
so m in lebesgue gives the width of rectangle as it gives in reimann?

I am lil' confused that, is m=f^-1 (E) ?????
m is a function that is intended to abstract/generalize the properties you would want

to find in the measure of a set. So, the Lebesgue measure of an interval is its length,

and, with some adjustments, the measure of a rectangle is the product of the lengths

of the intervals forming the sides of the rectangle.

f^-1(E) is just the set {x: f(x) is in E}. What you do is you break down your range

--like the way you break down your domain -- as {y1, y2,...,yn} , and --again,like

the case of the Riemann integral-- you select one point yi* in each interval [yi,yi+1)

and then you consider the measure of the preimage.The Lebesgue integral is then

the sum yi*m(Ei). There are rules for the (net ) convergence of the Lebesgue

integral: the Leb integral equals, 'L' if,given any e>0 , there must be some del.>0 with

||P||< del. implies the integral of any of these partitions is within less than e of L,

i.e. | Int f -L| <e when ||P||< del.

Along the lines of what Pwsnafu said, you define the Lebesgue measure (in R

here, since you want to do probability; you can define measures in general topological

space) by defining
declaring the length of an interval (a,b) to be b-a and then defining the measure of a

subset of R to be the infimum of the measures over all covers of a set.
woundedtiger4
#10
Aug16-12, 07:57 PM
P: 176
Quote Quote by Bacle2 View Post
m is a function that is intended to abstract/generalize the properties you would want

to find in the measure of a set. So, the Lebesgue measure of an interval is its length,

and, with some adjustments, the measure of a rectangle is the product of the lengths

of the intervals forming the sides of the rectangle.

f^-1(E) is just the set {x: f(x) is in E}. What you do is you break down your range

--like the way you break down your domain -- as {y1, y2,...,yn} , and --again,like

the case of the Riemann integral-- you select one point yi* in each interval [yi,yi+1)

and then you consider the measure of the preimage.The Lebesgue integral is then

the sum yi*m(Ei). There are rules for the (net ) convergence of the Lebesgue

integral: the Leb integral equals, 'L' if,given any e>0 , there must be some del.>0 with

||P||< del. implies the integral of any of these partitions is within less than e of L,

i.e. | Int f -L| <e when ||P||< del.

Along the lines of what Pwsnafu said, you define the Lebesgue measure (in R

here, since you want to do probability; you can define measures in general topological

space) by defining
declaring the length of an interval (a,b) to be b-a and then defining the measure of a

subset of R to be the infimum of the measures over all covers of a set.
Sir, thank you so much.
pwsnafu
#11
Aug16-12, 08:45 PM
Sci Advisor
P: 820
One thing you will be surprised with: it is possible to construct a definition of integral which is equivalent to the Lebesgue integral on ℝ, but nonetheless partitions the domain. It's called the McShane integral, but you'll want to read about the Henstock-Kurzweil integral first.
woundedtiger4
#12
Aug17-12, 05:47 PM
P: 176
Quote Quote by pwsnafu View Post
One thing you will be surprised with: it is possible to construct a definition of integral which is equivalent to the Lebesgue integral on ℝ, but nonetheless partitions the domain. It's called the McShane integral, but you'll want to read about the Henstock-Kurzweil integral first.
WOW, how many integrals are out in the world? I thought Riemann & Lebesgue are the only integrals :D Does any of the Henstock-Kurzweil & McShance have any use or role in probability?

Thanks a tonne for sharing this wonderful information with me.
pwsnafu
#13
Aug17-12, 09:14 PM
Sci Advisor
P: 820
[QUOTE=woundedtiger4;4038516]WOW, how many integrals are out in the world?[QUOTE]

There's an infinite number. The Riemann integral, McShane integral and Henstock-Kurzweil integral are all the same in the sense they are all Riemann sums. The trick is changing which partitions you allow. If you allow lots, you need to restrict f to guarantee the limit existing (Riemann). If you restrict the partitions you can weaken f (H-K). McShane (and hence Lebesgue) exists in between these two. You can imagine other selection criteria, and you'd get different integrals.


Does any of the Henstock-Kurzweil & McShance have any use or role in probability?
Nope. Of course you could try and study other integrals in prob, but you'd need to know the theory well enough and show these integrals have properties which are important in prob and yet don't exist in Lebesgue.

For a prob theory perspective you'll want:
  • measure spaces and Polish spaces
  • Lebesgue integral
  • Ito integral (definitely for stochastic stuff)

Things like H-K integrals are obscure in the literature. You'll know them if you do real analysis, but otherwise never hear of them. I'd be amazed if 3 other people on this forum know how to abstract the H-K integrals
woundedtiger4
#14
Aug18-12, 02:33 PM
P: 176
[QUOTE=pwsnafu;4038677][QUOTE=woundedtiger4;4038516]WOW, how many integrals are out in the world?

There's an infinite number. The Riemann integral, McShane integral and Henstock-Kurzweil integral are all the same in the sense they are all Riemann sums. The trick is changing which partitions you allow. If you allow lots, you need to restrict f to guarantee the limit existing (Riemann). If you restrict the partitions you can weaken f (H-K). McShane (and hence Lebesgue) exists in between these two. You can imagine other selection criteria, and you'd get different integrals.




Nope. Of course you could try and study other integrals in prob, but you'd need to know the theory well enough and show these integrals have properties which are important in prob and yet don't exist in Lebesgue.

For a prob theory perspective you'll want:
  • measure spaces and Polish spaces
  • Lebesgue integral
  • Ito integral (definitely for stochastic stuff)

Things like H-K integrals are obscure in the literature. You'll know them if you do real analysis, but otherwise never hear of them. I'd be amazed if 3 other people on this forum know how to abstract the H-K integrals
excellent answer.......... thank you Sir


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