Show $\lim\limits_k \int_{A_k} f_k = \lim\limits_k\int_A f_k,\;$ given $f_k \in\mathcal{L}^1(\mathbb{R}^n),\; \lim\limits_k\lambda(A_k\Delta A) = 0$.

Question

Show $\lim\limits_k \int_{A_k} f_k = \lim\limits_k\int_A f_k,\;$ given $f_k \in\mathcal{L}^1(\mathbb{R}^n),\; \lim\limits_k\lambda(A_k\Delta A) = 0$. Here $\{A_k\}$ and $A$ are Lebesgue measurable. And if needed, $\lim\limits_k\int_A f_k < \infty.$

What I have so far: $$ \left|\int_{A_k}f_k - \int_Af_k\right| = \left|\int(\chi(A_k)-\chi(A))f_k\right| \le \int |(\chi(A_k)-\chi(A))f_k| = \int \chi(A_k\Delta A)|f_k| $$

I'm trying to show $\int \chi(A_k\Delta A)|f_k| \to 0$ but of no avail.

Please help! Thank you.

Answer 1

Hint: Here is a fact about integrability that you could use. Given $\epsilon >0$ there exist $\delta$ such that

$$ \int_A |f_{k}(x)|\lambda(dx) \;\;<\;\;\epsilon $$

whenever $\lambda(A) < \delta$.

There exists $N$ so large such that $\lambda (A_{k} \Delta A) < \delta$ for all $k > N$.