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Limits and Lebesgue Integration

  • Thread starter iomtt6076
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  • #1
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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?
 

Answers and Replies

  • #2
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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.

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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]

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Version 2 will work. Do you have it available to use?
 
  • #3
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(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
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All you need is Fatou's Lemma to solve this problem.
 

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