Question

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 \[\large\int_{\cap_{n=1}^\infty E_n}f \ d\lambda=\lim_{n \rightarrow \infty}\int_{E_n}f \ d\lambda .\]

Answer 1

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 \(\cup_{k=1}^nE_k=E_n\) and if we take the limit of both sides we get \(\cup_{n=1}^\infty E_n=\lim_{n\rightarrow \infty} E_n\). I am just not sure what I should do next.

Answer 2

@eliassaab

Answer 3

I'm sorry you have to prove this boring, obvious statement. Good luck.

Answer 4

@eliassaab you around?

Answer 5

Let
\[\Large
E=\cap_{n=1}^\infty E_n\\ \Large
f_E = f 1_E \\ \Large
f_n= f 1_{E_n}\\\Large
f_E(x) = \lim_{n\to \infty} f_n(x)d \lambda\\ \Large
\int_E f d \lambda =\int_E f_E d \lambda=\lim_{n\to \infty} \int_E f_n(x)d \lambda=\\
\Large
\lim_{n\to \infty} \int_{E_n} f(x)d \lambda
\]
We are done

Answer 6

If I have a sequence of measurable sets, and \(E=\cup_{n=1}^\infty E_n\) how can I show that \(\chi_{E_n} f\rightarrow\chi_Ef\) p.w. ?