Monotone convergence theorem for a sequence of measures? How can you prove the following:

Let $(\mu_n)_{n\in \mathbb N}$ a sequence of measures on a measure space $(\Omega, \mathcal F)$, such that $\mu_n \uparrow \mu$ for a measure $\mu$, in the sense that $(\mu_n)$ is monotone ($=$non-decreasing) and $\mu_n \to \mu$ pointwise. Then $\int\limits f\,d\mu_n \uparrow \int\limits f\,d\mu$ for all non-negative and measurable functions $f$.

I seem to be able to verify it for simple functions. But the general case (using the plain monotone convergence theorem) involves switching limits of sequences and I'm not familiar with any conditions that allow this (specifically none that obviously apply here).

Motivation: If $\mu$ is $\sigma$-finite we can define $\mu_n(A):=\mu(A\cap F_n)$, where $F_n \uparrow \Omega$ with $\mu(F_n)<\infty$. Then $(\mu_n)$ is a sequence of finite measures with $\mu_n \uparrow \mu$. So it seems in some cases it may be easier to first prove a statement about finite measures and then extend it through the above theorem.
 A: Since $\mu_n \leqslant \mu$ for all $n$, we have
$$\int f\,d\mu_n \leqslant \int f\,d\mu$$
for all $n$, and therefore
$$\lim_{n\to\infty} \int f\,d\mu_n = \sup_n \int f\,d\mu_n \leqslant \int f\,d\mu.$$
Conversely, we have
$$\int f\,d\mu = \sup \Biggl\{ \int s\,d\mu : 0 \leqslant s \leqslant f,\; s\text{ simple}\Biggr\}.$$
So fix an arbitrary simple $s$ with $0 \leqslant s \leqslant f$. By the case for simple functions,
$$\int s\,d\mu = \lim_{n\to\infty} \int s\,d\mu_n \leqslant \lim_{n\to\infty} \int f\,d\mu_n.$$
Taking the supremum over all admissible $s$ gives
$$\int f\,d\mu \leqslant \lim_{n\to\infty} \int f\,d\mu_n,$$
so together with the first part, we have the monotone convergence theorem, supposing your argument for simple functions is correct. (It should be, that case is simple.)
A: As regards your question about "switching limits", there is a general principle that can be applied here: Consider  a doubly-indexed collection $\{a_{m,n}\}_{m,n\in\Bbb N}$ of real numbers. Then $\sup_{m,n} a_{m,n}=\sup_n\sup_m a_{m,n}=\sup_m\sup_n a_{m,n}$. If, in addition, $a_{m,n}$ is monotone increasing in each index (that is, $a_{m,n}\le a_{m+1,n}$ and $a_{m,n}\le a_{m,n+1}$ for each $m\in\Bbb N$ and each $n\in\Bbb N$), then the suprema can be reinterpreted as limits, and you obtain $\lim_{m,n} a_{m,n}=\lim_n\lim_m a_{m,n}=\lim_m\lim_n a_{m,n}$.
