Example where convergence in measure does not implies convergence in $L^1$ This is an example from my Real Analysis class:
Consider the space
$(\mathbb{R},L,m)$ where $m$ is Lebesgue measure.
Let $f_n = n\chi_{(0,\frac{1}{n})}$.
So $\int |f_n| dm = \int_{[0,\frac{1}{n}]} n dm = 1$
But $f_n \rightarrow 0$ pointwise, so $m(|f_n - 0| \geq \epsilon) \leq m((0,\frac{1}{n})) = \frac{1}{n} \rightarrow 0$
Thus convergence in measure does not imply convergence in $L^1$. 
My question is: Why is $m(|f_n - 0| \geq \epsilon) \leq m((0,\frac{1}{n}))$?

Because $\{ x \mid f_n(x) \neq 0\} \subset [0,\frac{1}{n}]$ 
