Limits and Lebesgue Integration

In summary, the conversation discusses using Lebesgue's Dominated Convergence Theorem to prove that the integral over a set E is equal to the limit of the integrals over that set. It also provides examples to show that this is not always true if the limit of the integrals is infinity. The solution involves using Fatou's Lemma.
  • #1
iomtt6076
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Homework Statement


Let [tex](X,\Sigma,\mu)[/tex] be a measure space. Suppose that {fn} is a sequence of nonnegative measurable functions, {fn} converges to f pointwise, and [tex] \int_X f = \lim\int_X f_n < \infty[/tex]. Prove that [tex] \int_E f = \lim\int_E f_n[/tex] for all [tex]E\in\Sigma[/tex]. Show by example that this need not be true if [tex] \int_X f = \lim\int_X f_n = \infty[/tex].


Homework Equations




The Attempt at a Solution


Since [tex]\lim\int_X f_n < \infty[/tex], I think that the sequence [tex]\{\int_X f_n\}[/tex] is bounded, say by M. If I define g(x) to be the constant function g(x)=M, can I just apply Lebesgue's Dominated Convergence Theorem?
 
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  • #2
A constant function g won't work. Here is an example which shows why. All my examples will use ordinary Lebesgue measure, and X=some subset of the real line, in fact, X=some subset of [tex](0,\infty)[/tex]. Draw pictures of all these examples!

Example: [tex]f(x)=1/x^2[/tex] on [tex]X=(1,\infty)[/tex]. f_n doesn't matter too much, but let's say

[tex]f_n(x)=f\cdot\chi_{[1,n]}[/tex] where [tex]\chi_A[/tex] denotes the characteristic function of A (i.e., 1 on A and 0 elsewhere).

Then f is integrable, i.e. [tex]\int|f|<\infty[/tex] but the space [tex]X=(1,\infty)[/tex] doesn't have finite measure, so (nonzero) constant g won't be integrable.

-----

OK, so can you use DCT anyway with a different g? I know two main versions of DCT. Version 1 requires a single dominating function g that dominates all the f_n simultaneously. Version 2 only requires a sequence of dominating functions g_n.

Version 1 is certainly in your book and/or notes. Look to see if you have Version 2 available:

DCT 2: Suppose g_n is a sequence of integrable functions, lim g_n=g (a.e.), g is integrable, f_n is a sequence of measurable functions, [tex]|f_n|\le g_n[/tex], lim f_n=f (a.e.), and

[tex]\lim\int g_n=\int g[/tex]

Then [tex]\lim\int f_n=\int f[/tex]

Here is an example which shows that Version 1 (with a single g) is not going to work:

Example: [tex]X=(0,\infty)[/tex], [tex]f_n(x)=\frac1n\cdot\chi_{[n,n+1]}[/tex] and [tex]f\equiv0[/tex]. Note that [tex]\int f_n=1/n[/tex]

Any g that would dominate all f_n simultaneously would have [tex]\int g\ge\sum_{n=1}^{\infty} \frac1n=\infty[/tex]

-----

Version 2 will work. Do you have it available to use?
 
  • #3
(continued)

As for the example requested, the three main counterexamples to try in all these problems are (a) [tex]n\chi_{(0,1/n)}[/tex] (b) [tex]\chi_{(n,n+1)}[/tex] and (c) [tex]\frac1n\cdot\chi_{(0,n)}[/tex] (draw!)

These examples illustrate integral of the limit not being equal to the limit of the integrals. You can't use them as is for your counterexample on X, but maybe one of them would work for the counterexample on E. Well, (b) and (c) involve the entire domain X=(0,infty) or X=(1,infty), so it looks like (a) is a good candidate on, say, E=(0,1). Now extend the domain of the sequence (a) to all of X=(0,infty) in such a way that you make the limit of the integrals over X become infinite, and the integral over X of the limit is also infinite.

Try all this and ask more if you need to.
 
  • #4
All you need is Fatou's Lemma to solve this problem.
 

What is the concept of limits in calculus?

Limits in calculus refer to the value that a function approaches as the input approaches a particular value or point. It is used to describe the behavior of a function near a certain point, and it is an essential tool in understanding derivatives and integrals.

What is the difference between Riemann integration and Lebesgue integration?

Riemann integration is based on the concept of dividing a function into small intervals and finding the area under each interval. Lebesgue integration, on the other hand, is based on the concept of dividing a function into measurable sets and calculating the measure of those sets. Lebesgue integration is more general and can be used to integrate a wider range of functions than Riemann integration.

What is the fundamental theorem of Lebesgue integration?

The fundamental theorem of Lebesgue integration states that the Lebesgue integral of a function can be calculated by taking the limit of the Riemann sums of the function. In other words, the Riemann sums converge to the Lebesgue integral as the size of the intervals approaches zero. This theorem allows us to use the simpler Riemann integral to calculate Lebesgue integrals.

What is the significance of Lebesgue measure in integration?

Lebesgue measure is a measure of the size or extent of a set in n-dimensional space. It is used in Lebesgue integration to define the measure of a set of real numbers and to calculate the Lebesgue integral of a function. Lebesgue measure is also important in probability theory and has applications in various fields of mathematics.

What are the main applications of Lebesgue integration?

Lebesgue integration has various applications in mathematics, physics, economics, and other fields. It is used to solve problems in probability theory, differential equations, and analysis. It also has applications in signal processing, image processing, and data analysis. Lebesgue integration is a powerful tool for calculating integrals of complex functions and has revolutionized the field of integration.

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