Let $\mu_n$ be a sequence of finite measures on space $(X,M)$ and $\forall E \in M, \lim_{n \to \infty }\mu_n(E)=\mu(E)< \infty $.. Let $\mu_n$ be a sequence of finite measures on space $(X,M),M-\text{ sigma algebra on X}$ and $\forall E \in M, \lim_{n \to \infty }\mu_n(E)=\mu(E)< \infty $ and let $f$ be a bounded function.
Prove: 
$$\lim_{n\to \infty }\int_{E}fd\mu_{n}=\int_{E}fd\mu$$ 
The idea is to first prove this for simple functions, then positive and lastly for arbitrary functions that satisfy the question. For simple functions it is quite clear. I am interested how would I go about proving for the other two types of functions.
 A: Given $f$ is bounded and measurable, since $|f|\leq M$ a.e. define the simple function
$$g(x):=\sum c_i \chi_{\{f^{-1}(c_i)\}}(x)$$
 where $c_i \in [-M, M]$ and $\cup_i (c_i-\epsilon/2, c_i+\epsilon/2) \supset [-M, M]$, we have $\|f - g\|_\infty \leq \epsilon$. 
Now 
$$\limsup_n \bigg|\int_X f d\mu_n -\int_X f d\mu \bigg| \\
\leq  \limsup_n \bigg|\int_X f d\mu_n -\int_X g d\mu_n\bigg| + \bigg|\int_X g d\mu - \int_X f d\mu \bigg|  \\
\leq \epsilon \lim_n \mu_n (X) + \epsilon \mu(X)$$
A: Because $f$ is bounded, we may suppose without loss of generality that $0\le f(x)\le 1$ for all $x\in X$. In this case $\int_E f\,d\mu_n=\int_0^1\mu_n(E\cap \{x:f(x)>t\})\,dt$ (and likewise for $\int_E f\,d\mu$)). As $n\to\infty$ the integrand $\mu_n(E\cap \{x:f(x)>t\})$ converges to $\mu(E\cap \{x:f(x)>t\})$ for each $t\in[0,1]$. The integrand is moreover bounded above by $\mu_n(E)$, and $\sup_n\mu_n(E)<\infty$ because $\lim_n\mu_n(E)$ exists. It follows that $\lim_n\int_E f\,d\mu_n=\int_E f\,d\mu$ by the Bounded Convergence Theorem.
