If $ \lim_{k\rightarrow+\infty}\int_E \dfrac{f_k(x)}{1+f_k(x)}\mathrm{d}x=0, $ prove that $f_k$ converge to $0$ in measure.

Question

Suppose $E$ is a measurable set and $m(E)<+\infty$. $\{f_k\}$ is a sequence of non-negative measurable functions. If $$ \lim_{k\rightarrow+\infty}\int_E \dfrac{f_k(x)}{1+f_k(x)}\mathrm{d}x=0, $$ prove that $f_k$ converge to $0$ in measure.

If we can prove $\frac{f_k(x)}{1+f_k(x)}\rightarrow 0$ 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!

Answer 1

$$ \int_E \dfrac{f_k(x)}{1+f_k(x)}\mathrm{d}x \ge \int_{E(\epsilon)} \dfrac{f_k(x)}{1+f_k(x)}\mathrm{d}x $$ where $E(\epsilon)=\{x \in E: f_k(x) >\epsilon\}$. Hence $ \int_E \dfrac{f_k(x)}{1+f_k(x)}\mathrm{d}x \ge \frac {\epsilon} {1+\epsilon}m(E(\epsilon))$. This shows that $m(E(\epsilon)) \to 0$ as $k \to \infty$ for any $\epsilon >0$ 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 $(0,\infty)$ so $y >\epsilon$ implies $\frac y {1+y}>\frac {\epsilon} {1+\epsilon}$.