Interchanging limits for an integral Assume that $\mu$ and $(\mu_n)_{n \in \mathbb{N}}$ are finite measures on $(\Omega, \mathcal{F})$ such that for all $A \in \mathcal{F}$
$$
\lim_{n \rightarrow \infty} \mu_n (A) = \mu(A). \tag{1}
$$
Now let $f : \Omega \rightarrow \mathbb{R}$ be measurable and bounded.

Is it true that
$$
\lim_{n \rightarrow \infty} \int_{\Omega} f \, d\mu_n = \int_{\Omega} f \, d\mu \tag{2}
$$

Clearly, $(1)$ means that $(2)$ holds for all indicator functions $1_A$, $A\in \mathcal{F}$, and the linearity of the integral ensures that $(2)$ is true of all non-negative simple functions. If $f$ is $\mathcal{F}$-measurable and non-negative, then there exists a sequence of increasing simple functions $(f_m)_{m \in \mathbb{N}}$ converging to $f$. Assume that
$$
f_m = \sum_{i=1}^{k_m} a_{m, i} 1_{A_{m, i}}
$$
for some $k_m \in \mathbb{N}$. By the moonstone convergence theorem, we get
$$
\lim_{n \rightarrow \infty} \int_{\Omega} f \, d\mu_n = \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int_{\Omega} \sum_{i=1}^{k_m} a_{m, i} 1_{A_{m, i}} \, d\mu_n.
$$
Now, is it possible to show that the limits can be interchanged? If so, the conclusion is straightforward.
 A: Indeed the claim is true for any bounded measurable $f : \Omega \to \mathbb{R}$.

Proof. Let $s $ be any simple function such that $s \leq f$ everywhere. Then
\begin{align*}
\int s \, \mathrm{d}\mu
= \lim_{n\to\infty} \int s \, \mathrm{d}\mu_n
\leq \liminf_{n\to\infty} \int f \, \mathrm{d}\mu_n.
\end{align*}
Next, choose a sequence $(s_k)$ of simple functions such that $s_k \uparrow f$ everywhere. (It is clear that such a sequence exists.) Then $\int s_k \, \mathrm{d}\mu \uparrow \int f \, \mathrm{d}\mu$ by the monotone convergence theorem. So, we conclude
\begin{align*}
\int f \, \mathrm{d}\mu
= \lim_{k\to\infty} \int s_k \, \mathrm{d}\mu
\leq \liminf_{n\to\infty} \int f \, \mathrm{d}\mu_n. \tag{1}
\end{align*}
Now applying the above inequality to $-f$ instead, we obtain
$$ \limsup_{n\to\infty} \int f \, \mathrm{d}\mu_n \leq \int f \, \mathrm{d}\mu. \tag{2} $$
Combining $\text{(1)}$ and $\text{(2)}$, the desired conclusion follows.

Addendum. The proof that leads to $\text{(1)}$ also applies without modification to the case where $f$ is non-negative but not necessarily bounded. However, $\text{(2)}$ no longer need to hold for unbounded $f$. So, in this case, we cannot conclude that $\int f \, \mathrm{d}\mu_n$ tends to $\int f \, \mathrm{d}\mu$.

Counter-Example. Let $\mu_n$ be the Borel measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ defined by
$$ \mu_n(\mathrm{d}x) = n^{3/4} \mathbf{1}_{[n-\frac{1}{n}, n)}(x) \, \mathrm{d}x. $$
Since $\mu_n(\mathbb{R}) = n^{-1/4} \to 0$, we have $\mu_n(A) \to 0$ for any Borel subset $A$ of $\mathbb{R}$, and hence the condition holds with $\mu \equiv 0$. On the other hand, if we set
$$ f(x) = \frac{1}{\sqrt{1 + \lfloor x \rfloor - x}}, $$
then $\int_{\mathbb{R}} f \, \mathrm{d}\mu_n = 2n^{1/4} \to \infty$, whereas we clearly have $\int_{\mathbb{R}} f \, \mathrm{d}\mu = 0$.

