I found the following question in the book of kolomogorov fomin introductory real analysis and I don't know how to solve it. Does anyone have any ideas?
Suppose $f$ is integrable on sets $A_1,A_2,....A_n,...$ such that $A_1 \supset A_2 \supset....\supset A_n\supset ....,$
and let $A = \cap A_n$. Does $\int_{A_n}$$f(x)dm$ converge to $\int_{A}$$f(x)dm$?
What I tried so far is the following $\forall$n $\int_{A_n}$$f(x)dm$ -$\int_{A}$$f(x)dm$=$\int_{A_n-A}$$f(x)dm$
Moreover I have noticed
that $m(A_n-A)$ converges to 0 but I am not able to prove that this implies $\int_{A_n-A}$$f(x)dm$ converges to 0 but I cannot think of a counterexample for this statement.
Thanks in advance for any help
For the sake of completeness, I will post my answer in the comments.
To prove the desired statement, Consider $f_n = \chi_{A_n} \cdot f$, where $\chi_{A_n}$ is the characteristic function on $A_n$. Since $f$ is integrable and $\chi_{A_n}$ is bounded, $f_n$ is integrable on $A_1$. Also, $\vert f_n \vert \le \vert f \vert$. The sequence $f_n$ converges pointwise to $\chi_A \cdot f$, which is also integrable on $A_1$. Therefore, by the Lebesgue dominated convergence theorem,
$$ \int_{A_n} f = \int_{A_1} \chi_{A_n} \cdot f = \int_{A_1} f_n \to \int_{A_1} \chi_A \cdot f = \int_A f. $$
Moreover, note that $m(A_n-A)$ need not converge to zero. Consider $A_n = \mathbb{R}-[0,n]$. Then $A=(-\infty,0)$ and $A_n-A = (n,\infty)$ which has infinite measure for every $n$.