Let $(\Omega,S,\mu)$ be a measure space. Suppose $f$ is integrable on $A_1\supset A_2 \supset A_3\dots$, a decreasing sequence of measurable sets $\{A_n\}_n\subset S$ and denote $$A:=\bigcap_nA_n.$$ Prove or disprove that $$\lim_{n\to\infty}\int\limits_{A_n}fd\mu\to\int\limits_A fd\mu.$$
Intuitively that's not correct but I can't pick an counterexample. Can it be that the claim is correct?
It looks right to me. This can be shown via the Dominated Convergence Theorem.
Your sequence of functions is $\{ f\chi_{A_{n}} \}_{n = 1}^{\infty}$. Then we have $\int \limits_{\Omega} |f \chi_{A_{n}}| \,d\mu < \infty$ and for all $n$ (since $f$ is integrable on $A_{n}$), $|f \chi_{A_{n}}| \leq |f \chi_{A_{1}}|$. By the Dominated Convergence Theorem, it follows that $$\lim \limits_{n \to \infty} \int \limits_{A_{n}} f\,d\mu =\lim \limits_{n \to \infty} \int \limits_{\Omega} |f \chi_{A_{n}}| \,d\mu \to \int \limits_{\Omega} |f \chi_{A}| \,d\mu = \int \limits_{A} f \,d\mu$$