Prove that this sequence does not converge pointwise almost everywhere. Suppose that $f_n$ is a sequence of real measurable functions on a set $X$ of finite measure, and suppose that there is some $\epsilon$ such that for all $n\geq1$:
$$m(\{x:|f_n(x)-f(x)|\}\geq\epsilon)\geq\epsilon$$
I want to show that this is false: $f_n(x)\rightarrow f(x)$ a.e. on $X$.
I tried to prove the existence of a set $Y\subseteq X$ of positive measure such that $f_n(x)$ is far from $f(x)$ for all $x\in Y$. However, I only obtain sets $Y_n$ that depend on $n$ so I guess I need some stronger argument. Any help is appreciated!
 A: We define the convergence in measure:

A sequence $\{f_n\}$ converges in measure to $f$ if for all $\delta>0$ we have $$\lim_{n\to\infty}m(\{|f_n(x)-f(x)|\geq\delta\})=0.$$

We have the following result:

Let $(X,\mathcal B,\mu)$ a finite measured space. If the sequence $\{f_n\}$ of measurable functions converges almost everywhere to $f$, then it converges in measure to $f$. 

To see this,fix $\delta>0$ and note that $m\left(\bigcap_k\bigcup_{n\geq k}\{x:|f_n(x)-f(x)|\}\right)=0$. Since the sequence $\left\{\bigcup_{n\geq k}\{x:|f_n(x)-f(x)|\}\right\}_k$ is decreasing and we are in a finite measured space we have $m\left(\bigcup_{n\geq k}\{x:|f_n(x)-f(x)|\}\right)$ converges to $0$ and we get the result.
Note that the fact that the space is finite is primordial, since you get a counter-example taking the real line, with the Borel $\sigma$-algebra and Lebesgue measure, taking $f_n$ equal to the characteristic function of $(n,n+1)$. 
Now, to solve the problem note that the sequence $\{f_n\}$ fail to converge in measure.
