# Suppose E is a measurable set and m ( E ) &lt; + <mi mathvariant="normal

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Suppose $E$ is a measurable set and $m\left(E\right)<+\mathrm{\infty }$. $\left\{{f}_{k}\right\}$ is a sequence of non-negative measurable functions. If

prove that ${f}_{k}$ converge to 0 in measure.

If we can prove $\frac{{f}_{k}\left(x\right)}{1+{f}_{k}\left(x\right)}\to 0$ a.e.$x\in E$, then the problem is easy to solve. Is it correct? If so, how to prove that?
Appreciate any hint or help!
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Nathen Austin
MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document
where $E\left(ϵ\right)=\left\{x\in E:{f}_{k}\left(x\right)>ϵ\right\}$. Hence ${\int }_{E}\frac{{f}_{k}\left(x\right)}{1+{f}_{k}\left(x\right)}\mathrm{d}x\ge \frac{ϵ}{1+ϵ}m\left(E\left(ϵ\right)\right)$. This shows that $m\left(E\left(ϵ\right)\right)\to 0$ as $k\to \mathrm{\infty }$ for any $ϵ>0$ which means ${f}_{k}\to 0$ in measure on $E$.
I have used the fact that $\frac{y}{1+y}$ is an increasing function on $\left(0,\mathrm{\infty }\right)$ so $y>ϵ$ implies $\frac{y}{1+y}>\frac{ϵ}{1+ϵ}$.