$\int_{A_n} f(x) d\mu$ converge to $\int_{A } f(x) d\mu$?

Question

Suppose $f$ is integrable on sets $A_1 \supset A_2 \supset \ldots \supset A_n \supset \ldots$ and let $A = \cap_{n=1}^{\infty} A_n$.

Does $\int_{A_n} f(x) d\mu$ converge to $\int_{A } f(x) d\mu$?

I think this is false but I can't prove it, If you can give me some hints please, I'm really stuck in this problem.

Thanks for your time and help.

Answer 1

So, first do the following manipulation: $$ \int_{A_1}fd\mu - \int_{A_n}fd\mu = \int_{A_1\setminus A_n}fd\mu. $$ Now, note that from the given condition, $$ 0 \leq f\mathcal{X}_{A_1 \setminus A_n}\nearrow f\mathcal{X}_{A_1 \setminus A}. $$ Hence, from MCT we get, $$ \lim_{n\to\infty}\int_{A_1\setminus A_n}fd\mu = \int_{A_1\setminus A}f d\mu \implies \lim_{n\to\infty}\left(\int_{A_1}fd\mu - \int_{A_n}fd\mu\right)=\int_{A_1}fd\mu - \int_{A} fd\mu \\ \implies \lim_{n\to\infty}\int_{A_n}fd\mu = \int_A fd\mu $$ using the fact that $f\mathcal{X}_{A_1}\in L^1(\mu)$. Note that in the absence of this, we would not be able to apply MCT.

Remark

This proof somehow mimics the proof regarding continuity of measure, when the sets are diminishing. In that case, if one recalls, there was an additional requirement that measure of one set is finite, which allows us to take set differences and apply monotonicity argument that is valid other way.

Answer 2

Set $f_n := f \cdot 1_{A_n}$, (so you get $f(x)$ if $x \in A_n$ and zero otherwise). Then, since the $A_n$ are decreasing, $f_n \downarrow f_0 := f \cdot 1_A$ pointwise (a point being in $A$ only if it is in each $A_n$). Now you can apply Monotone Convergence (or Dominated Convergence using $f_1$) to get $\lim_n \int f_n = \int f_0$.