Interchanging limit measure and integral I am reading a book and the author skips a lot of steps in a proof. I don't see if the next result can hold or not and how to prove or disprove it.
Let $\{\mu_\epsilon\}_{\epsilon > 0}$ be a sequence a positive finite measures on $\mathbb{R}$ and $\mu$ a positive finite measure on $\mathbb{R}$ such that for all $A$ Borel set of $\mathbb{R}$, $\lim\limits_{\epsilon \to 0} \mu_\epsilon(A) =\mu(A)$.
Let $f$ be a bounded measurable function of $L^\infty(\mathbb{R})$.
The author seems to use that $\int_{\mathbb{R}} f(x) \mu(dx) = \lim\limits_{\epsilon \to 0} \int_{\mathbb{R}} f(x) \mu_{\epsilon}(dx)$.
I tried to find results like that, but I didn't find any.
Thank you for your help.
 A: If $\mu$ is an infinite measure then $\int f d\mu$ may not make sense for bounded measurable $f$. So it is necessary to assume that $\mu$ is a finite measure. Any bounded measurable function is a uniform limit of simple functions. [ This is easy to see from the usual expression for simple functions that approximate the given function]. From the hypothesis we have $\int f d\mu_{\epsilon} \to \int fd\mu$ for simple functions $f$. So the result follows from triangle inequality if you use the fact that $\mu (\mathbb R) <\infty$ which implies $\mu_{\epsilon} (\mathbb R)$ remains bounded as $\epsilon \to 0$. [Note that it is enough to prove the result for each sequence $(\epsilon_n)$ tending to $0$]. 
A: A similar answer to the answer of master @KaviRamaMurthy. Note that we can decompose $f=f^+-f^-$ where 
$$f^+:=\max\{f,0\},\qquad f^-:=\max\{-f,0\}\tag1$$
Then $f^+,f^-:\Bbb R\to [0,\infty)$, and there are sequences of $\nu$-simple functions such that $\{s_n^+\}\uparrow f^+$ and $\{s^-_n\}\uparrow f^-$ point-wise, and so (by the monotone convergence theorem) we have that
$$\int_{\Bbb R} f^+(x)\nu(dx)=\lim_{n\to\infty}\int_{\Bbb R} s^+_n(x)\nu(dx)\tag2$$
for any finite measure $\nu$ (a similar statement holds for $f^-$). Because the $s_n^+$ are non-negative $\nu$-simple functions we have that
$$\int_{\Bbb R}s^+_n(x)\nu(dx)=\sum_{k=1}^{m_n}c_{k,n}\nu(A_{k,n})\tag3$$
for some measurable sets $A_{k,n}$, some $m_n\in\Bbb N$ and some $c_{k,n}\ge 0$, and so
$$\begin{align}\int_{\Bbb R}s^+_n(x)\mu(dx)&=\sum_{k=1}^{m_n}c_{k,n}\mu(A_{k,n})\\
&=\sum_{k=1}^{m_n}c_{k,n}\lim_{\epsilon\to 0^+}\mu_\epsilon(A_{k,n})\\
&=\lim_{\epsilon\to 0^+}\sum_{k=1}^{m_n}c_{k,n}\mu_\epsilon(A_{k,n})\\
&=\lim_{\epsilon\to 0^+}\int_{\Bbb R}s^+_n(x)\mu_\epsilon(dx)\end{align}\tag4$$
Thus we want to show that
$$\int_{\Bbb R} f^+(x)\mu(dx)=\lim_{n\to\infty}\int_{\Bbb R} s^+_n(x)\mu(dx)=\lim_{n\to\infty}\lim_{\epsilon\to 0^+}\int_{\Bbb R}s^+_n(x)\mu_\epsilon(dx)\\=\lim_{\epsilon\to 0^+}\lim_{n\to\infty}\int_{\Bbb R}s^+_n(x)\mu_\epsilon(dx)=
\lim_{\epsilon\to 0^+}\int_{\Bbb R}f^+(x)\mu_\epsilon(dx)\tag5$$
That is, we want to show that we can exchange the order of the limits in $\rm (5)$. Now set $I_{n,m}:=\int_{\Bbb R}s^+_n(x)\mu_{\epsilon_m}(dx)$ and $\Delta_k I_{k,m}:=I_{k+1,m}-I_{k,m}$ for some arbitrary sequence $\{\epsilon_m\}\downarrow 0$. Then we have that
$$
\lim_{m\to\infty}\sum_{k=0}^\infty\Delta_k I_{k,m}=\lim_{m\to\infty}\lim_{n\to\infty}\sum_{k=0}^n(I_{k+1,m}-I_{k,m})\\
=\lim_{m\to\infty}\lim_{n\to\infty}I_{n,m}\leqslant \sup_m\mu_{\epsilon _m}(\mathbb{R})\|f\|_{\infty }<\infty\tag6
$$
where we used the fact that the sequence $\{\mu_{\epsilon _m}(\mathbb{R})\}$ converges to $\mu(\mathbb{R})$ and so it is bounded. Then the double sequence $\{I_{n,m}\}$ is non-negative and bounded, and so it is also $\{\Delta_k I_{k,m}\}$ because $I_{n,m}$ is increasing respect to $n$, so applying the dominated convergence theorem we find that
$$\lim_{m\to\infty}\lim_{n\to\infty} I_{n,m}=\lim_{m\to\infty}\sum_{k=0}^\infty\Delta_k I_{k,m}=\sum_{k=0}^\infty\lim_{m\to\infty}\Delta_k I_{k,m}=\lim_{n\to\infty}\lim_{m\to\infty} I_{n,m}\tag7
$$
Thus it holds that
$$\int_{\Bbb R} f^+(x)\mu(dx)=\lim_{\epsilon\to 0^+}\int_{\Bbb R}f^+(x)\mu_\epsilon(dx)<\infty\tag8$$
Because a similar statement holds for $f^-$ we are done.
