If $\lambda_n\to\lambda$ setwisely, can we show $\int f\:{\rm d}\lambda_n\to\int f\:{\rm d}\lambda$? Let $(E,\mathcal E)$ be a measurable space and $\lambda,\lambda_n$ be measures on $(E,\mathcal E)$ for $n\in\mathbb N$ such that $(\lambda_n)_{n\in\mathbb N}$ is nondecreasing and $$\lambda_n\xrightarrow{n\to\infty}\lambda\tag1$$ ("setwisely").
If $f:E\to[0,\infty]$ is $\mathcal E$-measurable, are we able to show that $\int f\:{\rm d}\lambda_n\to\int f\:{\rm d}\lambda$?
Clearly, we can find $\mathcal E$-measurable $f_k:E\to[0,\infty]$ with $|f_k(E)|\in\mathbb N$ for $k\in\mathbb N$ such that $(f_k)_{k\in\mathbb N}$ is nondecreasing and $$f_k\xrightarrow{k\to\infty}f\tag2.$$
By $(1)$, $$\lambda_ng_k\xrightarrow{n\to\infty}\lambda g_k\;\;\;\text{for all }k\in\mathbb N\tag3.$$ And by the monotone convergence theorem, $$\int g_k\:{\rm d}\lambda_n\xrightarrow{k\to\infty}\int g\:{\rm d}\lambda_n\tag4\;\;\;\text{for all }n\in\mathbb N;$$ similarly, $$\int g_k\:{\rm d}\lambda\xrightarrow{k\to\infty}\int g\:{\rm d}\lambda\tag5.$$ Can we conclude? Maybe using that $(\lambda_n)_{n\in\mathbb N}$ is nondecreasing?
 A: By linearity, if $f:E\rightarrow[0,\infty]$ is a non-negative simple
function, we clearly have $\int fd\mu_{n}\rightarrow\int fd\mu$.
Now, let $f:E\rightarrow[0,\infty]$ be a non-negative measurable
function. Choose a sequence $(f_{k})$ of non-negative simple functions
such that $0\leq f_{1}\leq f_{2}\leq\ldots\leq f$ and $f_{k}\rightarrow f$
pointwisely. For each $k,n$, we have
$$
\int f_{k}d\mu_{n}\leq\int fd\mu_{n}\leq\int fd\mu.
$$
(We have used the fact: If $\nu_{1}$ and $\nu_{2}$ are measures
such that $\nu_{1}(A)\leq\nu_{2}(A)$ for all measurable set $A$,
then $\int gd\nu_{1}\leq\int gd\nu_{2}$ for all non-negative measurable
function $g$. We need this fact again in below.)
Note that $\left(\int fd\mu_{n}\right)_{n}$ is an increasing sequence
of non-negative extended real numbers (i.e., possibly taking value
$+\infty)$, so $\lim_{n}\int fd\mu_{n}$ exists (including the case $+\infty$). Letting $n\rightarrow\infty$,
we obtain
$$
\int f_{k}d\mu\leq\lim_{n}\int fd\mu_{n}\leq\int fd\mu.
$$
Further let $k\rightarrow\infty$, by Monotone Convergence Theorem,
then we have $$\int fd\mu\leq\lim_{n}\int fd\mu_{n}\leq\int fd\mu.$$
It follows that $\lim_{n}\int fd\mu_{n}=\int fd\mu$.
