Convergence of ${f_n}$ of measurable function

Question

For a sequence {$ {f_n} $} of measurable function in a set $ A $ of finite measures, show that

$\lim_{n\rightarrow \infty} \int_A \frac{|f_n|}{1+|f_n|}dm=0$ iff {$f_n$} converges to zero in measure.

I have a solution with me but I didn't get that proof

It says,

Let $\epsilon>0$ $A_n$={$x\in X: |f_n|>\epsilon $}

then

$\frac{\epsilon}{1+\epsilon}m(A_n)\leq \int_A \frac{|f_n|}{1+|f_n|}dm \leq m(A_n)+\frac{\epsilon}{1+\epsilon}m(A - A_n) $

Form this they says we can yield the desired result

By assuming $\int_A \frac{|f_n|}{1+|f_n|}dm=0 $ it is easy to obtain that $m(A_n)=0$

but I didn't get the converse can any one help me? please

Answer 1

If $f_n$ converges to $0$ in measure, then for all $\epsilon>0, \delta>0$, let $B_n=\{|f_n|>\delta\}$, then $m(B_n)<\epsilon$ for large $n$.

Note $\frac{|f_n|}{1+|f_n|}\le |f_n|$ and $\frac{|f_n|}{1+|f_n|}\le 1$

Hence for large enough $n$, we have

$\int_ A\frac{|f_n|}{1+|f_n|}=\int_ {B_n^c}\frac{|f_n|}{1+|f_n|}+\int_ {B_n}\frac{|f_n|}{1+|f_n|}\le\int_ {B_n^c} |f_n|+\int_ {B_n} 1\le \delta m(B_n^c)+m(B_n)\le \delta m(A)+\epsilon$

Since $\delta, \epsilon$ are arbitrary, we get the result.