If $f_n$ is convergence in measure to f=0 on E$\subset R$,whether we have $\lim_{n \rightarrow \infty}m({x\in E:\left| f_n(x) \right| >0})=0$
definition if for any $\theta >0$, $\lim m({x:\left| f_n(x)-f(x) \right|>\theta})=0$,then we say $f_n$ convergence in measure to f
I think we have .but I can’t proof it
No: let $f_n(x):= 1/n$ if $x\in [0,1]$ and $0$ otherwise. For each positive $\varepsilon$, the set $\{x\in\mathbb R\mid \left\lvert f_n(x)\rvert\gt\varepsilon \right\}$ is empty if $n>1/\varepsilon$ hence the convergence in measure of $\left(f_n\right)_{n\geqslant 1}$ holds but $$\{x\in\mathbb R\mid \left\lvert f_n(x)\rvert\gt 0 \right\}=[0,1].$$