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Suppose $f$ is a nonnegative $\mathcal{M}-\text{measurable}$ function and $\{E\}_{n=1}^\infty\subset\mathcal{M}$ with $E_1\supset E_2 \supset \cdot \cdot \cdot $. Further suppose $\int_\mathbb{R}f \ d \lambda<\infty.$ Prove that $$\large\int_{\cap_{n=1}^\infty E_n}f \ d\lambda=\lim_{n \rightarrow \infty}\int_{E_n}f \ d\lambda .$$ I can use the fact that if $\{f_n\}_{n=1}^\infty$ is a monotone nonincreasing sequence of nonnegative Lebesgue measurable functions, and $\int f_1 \ d\lambda<\infty $. Then $$ \large \int_E \lim_{n\rightarrow \infty}f_n \ d\lambda = \lim_{n\rightarrow \infty}\int f_n \ d\lambda.$$ I was thinking that I shuold use $f_n = f \chi_{E_n} $ then $f_n$ is nonincreasing. So we have $$ \large\int_{E_n}\lim_{n \rightarrow \infty}f_n \ d\lambda = \lim_{n\rightarrow \infty}\int_{E_n} f_n \ d\lambda .$$ I also know that $\cap_{k=1}^nE_k=E_n$ and if we take the limit of both sides we get $\cap_{n=1}^\infty E_n=\lim_{n\rightarrow \infty} E_n$. I am just not sure what I should do next.

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Using $f_n=f\chi_{E_n}$ is a good idea. After slightly rewriting, you have that

$\int \lim_{n\rightarrow \infty} f_n d\lambda=\lim_{n\rightarrow \infty} \int_{E_n} f d\lambda$

Note that you have the right hand side of the equality you want to show. What remains is to show that the integrand on the left hand side converges pointwise to the function you want (after rewriting $\int_{{\cap_{n=1}^\infty}E_n}fd\lambda$ as $\int f\chi_{{\cap_{n=1}^\infty}E_n}d\lambda$) ; that is, show that $\lim_{n\rightarrow \infty} f_n=f\chi_{{\cap_{n=1}^\infty}E_n}$.

Showing that $f\chi_{E_n}\rightarrow f\chi_{{\cap_{n=1}^\infty}E_n}$ pointwise is equivalent to showing that $\chi_{E_n}\rightarrow \chi_{{\cap_{n=1}^\infty}E_n}$ pointwise, which is easy to show using the assumptions on the $E_n$.

Let $\epsilon >0$ be given, and let $x\in (-\infty, \infty)$. We want to show that there exists a positive integer $N$ such than $|\chi_{{\cap_{n=1}^\infty}E_n}(x)-\chi_{E_n}(x)|<\epsilon$ whenever $n\geq N$. Either $x\in \cap_{n=1}^\infty E_n$ or $x\not \in \cap_{n=1}^\infty E_n$.

If $x\in \cap_{n=1}^\infty E_n$, then $\chi_{E_n}(x)=1$ for all $n$, and $\chi_{{\cap_{n=1}^\infty}E_n}(x)=1$ also. So let $N=1$. Then for all $n\geq N$, we have that $|\chi_{{\cap_{n=1}^\infty}E_n}(x)-\chi_{E_n}(x)|=0<\epsilon$ and we're done.

If $x\not \in \cap_{n=1}^\infty E_n$, then $x\not \in E_k$ for some integer $k$. Then, since $E_k\supseteq E_{k+1}\supseteq$..., $x\not \in$...

Can you finish it from here?

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  • $\begingroup$ but $f_n$ is nonincreasing so how can we use the monotone convergence theorem? $\endgroup$ – tmpys Mar 11 '14 at 6:21
  • $\begingroup$ I misspoke. I meant to cite the fact you mentioned that you are allowed to use. $\endgroup$ – user122916 Mar 11 '14 at 6:28
  • $\begingroup$ Ahh sorry I should have read on after and I would have realized that. $\endgroup$ – tmpys Mar 11 '14 at 6:33
  • $\begingroup$ If $x\not \in \cap_{n=1}^\infty E_n$, then $\chi_{E_n}(x)=0$ for all $n$, and $\chi_{{\cap_{n=1}^\infty}E_n}(x)=0$ and this is enough to show $\chi_{E_n}\rightarrow \chi_{{\cap_{n=1}^\infty}E_n}$ ? $\endgroup$ – tmpys Mar 11 '14 at 6:38
  • $\begingroup$ Be careful. $x \not \in \cap_{n=1}^\infty E_n$ means that there exists some $k$ such that $x\not \in E_k$... $\endgroup$ – user122916 Mar 11 '14 at 6:42

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