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$E$ is a (Lebesgue) measurable set with finite measure and $f_n$ is a sequence of bounded measurable functions on $E$. I'm trying to show that $f_n$ converges to 0 in measure if and only if $\int_E \frac{|f_n|}{1+|f_n|}$ $dm_L$ $\rightarrow 0$ as $n \rightarrow\infty$.

Since $f_n$ converges to $0$ in measure, I know it has a subsequence that converges pointwise $a.e.$ to zero. I think maybe I need to use this to somehow establish that the integral goes to zero. Or do I need to apply any theorem regarding integrals of bounded measurable functions? Also I can't figure out what to do for the opposite direction. Thanks for any help.

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  • $\begingroup$ There is a version of dominated convergence which assumes $f_n$ converges in measure rather than a.e. $\endgroup$ – user251257 Oct 11 '16 at 21:36
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If $f_n\to 0$ in measure, then (by definition) $\lim_{n\to\infty}m(\{x:|f_n(x)|\geq \varepsilon\})=0$ for all $\varepsilon>0$.

To show that $f_n\to 0$ in measure implies $\int_E\frac{|f_n|}{1+|f_n|}\to 0$, fix some $\varepsilon>0$, let $E_n=\{x:|f_n(x)|\geq \varepsilon\}$, and try splitting the integral over $E$ into an integral over $E_n$ and and integral over $E\setminus E_n$.

On the contrary, if $f_n\not\to0$ in measure, then there exists some $\varepsilon>0$ and a subsequence $f_{n_k}$ such that $m(E_{n_k})\geq \delta>0$ for all $k$. Therefore $$ \int_E\frac{|f_{n_k}|}{1+|f_{n_k}|}\geq \int_{E_{n_k}}\frac{|f_{n_k}|}{1+|f_{n_k}|}\geq \frac{\varepsilon}{1+\varepsilon}m(E_{n_k}) \geq \frac{\varepsilon}{1+\varepsilon}\delta $$ for all $k$, using the fact that $t\mapsto \frac{t}{1+t}$ is increasing on $[0,\infty)$.

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Let's say $m_L(E) = B$. Note that $0 \le |x|/(1+|x|) \le \min(1,|x|)$ for all $x$, while $|x|/(1+|x|) > |x|/2$ if $|x| < 1$. If $|f_n| \le \epsilon$ except on a set of measure $\le \delta$, you get $$ \int_E \frac{|f_n|}{1+|f_n|} \le B \epsilon + \delta$$ Conversely, if $ \int_E \frac{|f_n|}{1+|f_n|} \le \eta$, can you bound the measure of the set where $|f_n| > \epsilon$ (where $0 < \epsilon < 1$)?

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