Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable.
For $c_n>0$ such that either $\lim_{n\to \infty}c_n=0$, or $c_n\geq c>0$ for all $n$, and measurable sets $E_n$ with $m(E_n)>0$ consider the sequence $f_n(x):=c_n\mathcal{X}_{E_n}(x).$ Then $f_n$ converges in measure to zero, iff $c_n\to 0$, or $m(E_n)\to 0$ as $n\to \infty$.
My approach:
($\Rightarrow$) Suppose that $f_n\to 0$ in measure. For all $\epsilon >0$, let $E:=\{x: |f_n(x)|\geq \epsilon\}.$ I feel that it is obvious, but here is my reasoning anyway: as $n \to \infty$, $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}c_n\mathcal{X}_{E_n}(x)$, now if $x\in E_n$ then clearly $m(E_n)\to 0$ as $n\to \infty$ and if $x\notin E_n$ then by assumption we will have $c_n\to 0.$
($\Leftarrow$) I just realized this direction is the obvious one! Isn't it? If $c_n\to 0$ or $m(E_n)\to 0$ as $n\to \infty$ then hmmm... I need help!