Let $f:X \to \mathbb R_{\ge 0}$ be measurable and $(B_n)$ such that $\mu(B_n) \to 1$. Then $\int_{B_n} f \mathrm d \mu \to \int_{X} f \mathrm d \mu$ I'm trying to prove this convergence result. Could you please elaborate on how to finish the proof?

Let $(X, \mu)$ be a probability space and $f:X \to \mathbb R_{\ge 0}$ measurable. Assume there is a sequence $(B_n)$ of measurable sets such that $\mu(B_n) \to 1$. Then
$$
\int_{B_n} f \mathrm d \mu \to \int_{X} f \mathrm d \mu.
$$

My attempt:  Notice that $(1_{B_n} f)_n$ does not necessarily converges to $f$ $\mu$-a.e. Let $f_m := f \wedge m$. Then $f_m \in L_1 (\mu)$ for all $m$ and $f_m \nearrow f$. It follows from this result that
$$
\lim_n \int_{B_n} f_m \mathrm d \mu = \int_{X} f_m \mathrm d \mu \quad \forall m.
$$
The proof would be complete if we could show that
$$
\lim_m \lim_n \int_{B_n} f_m \mathrm d \mu = \lim_n \int_{B_n} f \mathrm d \mu.
$$
Could you elaborate on how to proceed?

Update: I already proved for the case $f \in L_1 (\mu)$ here. The remaining burden is on the case $\int_{X} f \mathrm d \mu = +\infty$ :v
 A: For all simple function $g$ such that $0 \le g \le f$, we have:
$$\int f \mathrm{d} \mu\ge \limsup_{n} \int_{B_n} f \mathrm{d} \mu \ge \liminf_{n} \int_{B_n} f \mathrm{d} \mu  \ge \liminf_{n} \int_{B_n} g \mathrm{d} \mu =\int g\mathrm{d}\mu,$$
Therefore,
$$\limsup_{n} \int_{B_n} f \mathrm{d} \mu = \liminf_{n} \int_{B_n} f \mathrm{d} \mu =\int f\mathrm{d}\mu.$$
Hence, the conclusion.
A: Step 1. Let $f:X\to[0,\infty)$ be measurable. If $\{a_n\}$ is a positive sequence with $a_n\to\infty$, then
$$\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu\to \int_X f\,d\mu.\tag{1}$$
This can be proved using Fatou: $\lim_{n\to\infty} f(x)1_{\{x\in X: f(x)\leq a_n\}}(x)=f(x)$ for all $x\in X$ (note that $f(x)<\infty$ for all $x$.)
Step 2. We may assume WLOG that $\mu(B_n)>0$ for all $n$. For any $\epsilon>0$, we consider $a_n=\frac{\epsilon}{\mu(B_n^c)}$. Since
\begin{align*}
\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu&=\int_{B_n\cap\{x\in X: f(x)\leq a_n\}}f\,d\mu+\int_{B_n^c\cap\{x\in X: f(x)\leq a_n\}}f\,d\mu\\
&\leq\int_{B_n}f\,d\mu+a_n\mu(B_n^c)=\int_{B_n}f\,d\mu+\epsilon,
\end{align*}
we have
$$\liminf_{n\to\infty}\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu\leq \liminf_{n\to\infty}\int_{B_n}f\,d\mu+\epsilon.$$
Since $a_n=\frac{\epsilon}{\mu(B_n^c)}\to\infty$, $(1)$ implies that for each $\epsilon>0$,
$$\liminf_{n\to\infty}\int_{\{x\in X: f(x)\leq a_n\}}f\,d\mu=\int_Xf\,d\mu.$$
Hence,
$$\int_Xf\,d\mu\leq \liminf_{n\to\infty}\int_{B_n}f\,d\mu+\epsilon,\qquad \forall \epsilon>0,$$
i.e.,
$$\int_Xf\,d\mu\leq \liminf_{n\to\infty}\int_{B_n}f\,d\mu\leq \limsup_{n\to\infty}\int_{B_n}f\,d\mu\leq \int_Xf\,d\mu.$$
