Seeking hint for If $f_n>0$ Integrable and $f_n\rightarrow f$ a.e. Then $\lim \mu(f_n)$ exists The question is Let $f_n \ge 0$ Be integrable and $f_n \rightarrow f$ a.e. Then $\lim_{n\rightarrow \infty} \mu(f_n)$ exists
My Proof: Firsly, we know that $f_n\rightarrow f$ And $f_n$ Is measurable , so is $f$. Then, $\inf_{m\ge n}f_m \le f_k$ for some $k\ge n$. Similarly, $\sup_{m\ge n}f_m \ge f_k$. Then, we know that 
$$\mu(\inf_{m\ge n}f_m)\le \mu ( f_k) \le \mu(\sup_{m\ge n}f_m)$$
Further, we know that $\lim_n \inf_{m\ge n} f_m \ge \inf_{m\ge n}f_m$, hence by monotone convergence theorem, we must have $\lim_n \mu(\inf f_n) = \mu(\lim \inf f_n)$. Similarly, we have $\lim_n \mu(\sup f_n) = \mu(\lim\sup f_n)$. But then, since $f_n \rightarrow f$, $\mu(\lim \inf f_n) = \mu (f) = \mu(\lim \sup f_n)$. Then, finally we have 
 $$ \mu(f)=\mu(\inf_{m\ge n}f_m)\le \lim \mu ( f_k) \le \lim\mu(\sup_{m\ge n}f_m) = \mu(f)$$
Hence, $\lim \mu(f_n)$ exists
I think my proof is wrong, but I am not sure where...
 A: I am assuming that $\mu(f) := \int f d\mu$. If that is what you mean, then I guess the problem would be in the phrase "Similarly, we have $\lim_n \mu(\sup f_n) = \mu(\lim\sup f_n)$". For example take the functions $f_n : [0,1]\rightarrow \mathbb{R}$ defined as
\begin{equation*}
\begin{cases}
f_n(x) = n &\text{if } x<\frac{1}{n}\\
0 &\text{otherwise}
\end{cases}
\end{equation*}
For this functions we have $\mu(\sup_{m\geq n} f_m)\geq \mu(f_n) = 1$ and $\lim \sup f_n = 0$ so  it is not true that  $1\leq\lim_n \mu(\sup f_n) = \mu(\lim\sup f_n)=0$. Be careful that the sequence $\sup f_n$ is decreasing so I guess you'll have to replace the hypothesis that the sequence is positive in the monotone convergence theorem by "the sequence is bounded from above".
Moreover, what you claim in the title is false as far as I see. A counterexample can be made by taking the same sequence that I gave but instead of $f_n = n$ you should put a value for each $n$ that makes the integral take alternating values for each $n$. For example $n$ if $n$ is even and $2n$ if $n$ is odd. Hope this last part can be understood. 
