During one of my take home exams I was asked to prove in a finite measure space, convergence in measure is the same as converging almost everywhere. I did not think seriously and wrote a proof receiving full mark. But now I realize this statement cannot be right, because there are easy counter-example. So I am wondering where is wrong in my "proof" below. I remember I examined it several times and thought it was correct.
If $f_{n}$ does not converge to $f$ almost everywhere, there exist some measurable subset $L$ with positive measure such that $\lim f_{n}\not=f$ on $L$. $L$ is measurable because if we let $L_{n}=(\bigcup^{\infty}_{k=n}\{x||f(x)-f_{k}(x)|>0)$ Then $L=\bigcap_{n=1}^{\infty} L_{n}$. In particular this implies $\mu(L_{n})\ge\mu(L)>T>0$ for all $n$ for some real number $T$.
Now we have $L_{n}=(\bigcup^{\infty}_{i=1}(\bigcup^{\infty}_{k=n}\{x||f(x)-f_{k}(x)|\ge\frac{1}{i})$ By our assumption we should have $\lim_{n\rightarrow \infty}\mu(\bigcup^{\infty}_{k=n}\{x||f(x)-f_{k}(x)|\ge \frac{1}{i}\})=0,\forall i\in \mathbb{N}$
Hence in particular for any $\epsilon<\frac{T}{20}$ there is a sequence $N_{j}\rightarrow \infty$ such that $\mu(\bigcup^{\infty}_{k=N_{j}}\{x||f(x)-f_{k}(x)|\ge \frac{1}{i}\})<\frac{\epsilon}{i^{2}2^{j}}$ Then $\mu(L_{N_{1}})< \sum^{\infty}_{i=1}\sum^{\infty}_{j=1}\frac{\epsilon}{i^{2}2^{j}}=\frac{\pi^{2}\epsilon}{6}<\frac{\pi^{2}}{120}T