I'm trying to prove this convergence result. Could you please have a check on my attempt?
Let $(X, \mu)$ be a probability space and $f:X \to \mathbb R$ such that $f \in L_1 (\mu)$. Assume there is a sequence $(B_n)$ of measurable sets such that $\mu(B_n) \to 1$. Then $$ \int_{B_n} f \mathrm d \mu \to \int_{X} f \mathrm d \mu. $$
Proof: Because $(1_{B_n} f)_n$ does not necessarily converges to $f$ $\mu$-a.e., we are unable to apply dominated convergence theorem. It suffices to prove that $$ \int_{B^c_n} f \mathrm d \mu \to 0. $$
We need the following lemma whose proof is given at the end.
Lemma: For each $\varepsilon>0$, there is $\delta >0$ such that if $\mu (B) < \delta$ then $\int_{B} |f| \mathrm d \mu < \varepsilon$.
- Given $\varepsilon >0$, there is $\delta>0$ such that $\int_{B} |f| \mathrm d \mu < \varepsilon$ for all $\mu(B) <\delta$ by our Lemma.
- Given $\delta >0$, there is $N >0$ such that $\mu(B_n^c) < \delta$ for all $n>N$.
It follows that for each $\varepsilon>0$ there is $N>0$ such that $\int_{B^c_n} |f| \mathrm d \mu <\varepsilon$ for all $n>N$. Notice that $|\int_{B^c_n} f \mathrm d \mu| \le \int_{B^c_n} |f| \mathrm d \mu$. This completes the proof.
Proof of the lemma: We have $f \in L_1 (\mu)$ implies $|f| \in L_1 (\mu)$, so there is a simple function $0 \le g \le |f|$ such that $$ \int_X (|f| -g) \mathrm d \mu < \frac{\varepsilon}{2}. $$
We assume $g$ has a form $$ g = \sum_{i=1}^m a_i 1_{A_i} $$ where $a_i \ge 0$ for $i = 1, \ldots, m$ and $(A_i)_{i=1}^m$ is pairwise disjoint sequence of measurable sets. We have $$ \int_B g \mathrm d \mu = \sum_{i=1}^m a_i \mu (A_i \cap B) \le \mu(B) \sum_{i=1}^m a_i. $$
If $\mu(B) \le \dfrac{\varepsilon}{2 \sum_{i=1}^m a_i}$, then $$ \begin{align} \int_B |f| \mathrm d \mu &= \int_B (|f| - g) \mathrm d \mu +\int_B g \mathrm d \mu \\ &\le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align} $$
