Let $f_n:E \to \mathbb{R}$ be a sequence of functions which converges in measure to $f:E \to \mathbb{R}$, where $m(E)< \infty$. (We are working with the Lebesgue measure here.) Let $\eta >0$.
Is it true that $\lim_{n \to \infty} m\{x:|f_n(x)-f(x)||f_n(x)|> \eta\}=0$?
I'm leaning toward saying "yes."
I see that if the $f_n$ functions are uniformly bounded by some positive constant $K$, (I mean if $|f_n(x)| \leq K$ for each $x,n$) we obtain $m\{x :|f_n(x)-f(x)||f_n(x)|> \eta\} \leq m\{x:|f_n(x)-f(x)|> \eta/K\} \to 0$ as $n\to \infty$, and by the sandwich theorem the desired result holds.
However, I am a little stuck if the $f_n$ functions are not uniformly bounded.
Notice the inclusion $$ \{x:|f_n(x)-f(x)||f_n(x)|> \eta\}\subset \{x:|f_n(x)-f(x)||f_n(x)-f(x)|> \eta/2\}\cup \left(\{x:|f_n(x)-f(x)||f(x)|> \eta\}\cap \{x:|f(x)|\leqslant M\}\right)\cup \{x:|f(x)|> M\} $$ valid for all $M$ hence $$ \mu \{x:|f_n(x)-f(x)||f_n(x)|> \eta\}\leqslant \mu\{x:|f_n(x)-f(x)|^2> \eta/2\}+ \mu\{\{x:|f_n(x)-f(x)| > \eta/M\}\}+\mu\{x:|f(x)|> M\}. $$ Conclude by letting the $\limsup_{n\to +\infty}$ and using the fact that $f$ is finite almost everywhere.