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Suppose that $(S,\Sigma,\mu)$ is a finite measure space. I am trying to show that a sequence of functions $(f_n)$ converges in measure to f iff

\begin{align*} \lim_{n\to \infty} \int \frac{\left\vert {f_n - f} \right\vert}{1 + \left\vert { f_n - f} \right\vert} d\mu = 0. \end{align*}

I have shown if \begin{align*} \lim_{n\to \infty} \int \frac{\left\vert {f_n - f} \right\vert}{1 + \left\vert { f_n - f} \right\vert} d\mu = 0, \end{align*} then $f_n \to f$ in measure. However, I am not sure how to show the reverse direction. I know that since I converge in measure that there exists a subsequence $(f_{n_k})$ that converges $\mu$-a.e. on S to f. So I can apply the monotone convergence theorem to the subsequence and get the desired result. I am not sure though how to show it for the entire sequence.

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Suppose $\int \frac {|f_n-f|}{1+|f_n-f|}d\mu $ does not tend to $0$. Then there exists $\epsilon >0$ and inetgers $n_1<n_2<...$ such that $\int \frac {|f_{n_k}-f|}{1+|f_{n_k}-f|}d\mu >\epsilon$ for all $k$. There is a subsequence of $f_{n_k}$ which converges almost everywhere to $f$. Call this $f_{n_{k_j}}$. By Bounded Convergence Theorem $\int \frac {|f_{n_{k_j}}-f|}{1+|f_{n_{k_j}}-f|}d\mu\to 0$ and this is a contradiction to the choice of $\epsilon$.

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You already have a nice answer, but I'd like to provide another, where we do this directly. For notational convenience, call $E_{n\epsilon}=\{x\in S: |f_n(x)-f(x)|>\epsilon\},$ also, I'm going to call $$d(f_n,f)=\int\limits_S\frac{|f_n(x)-f(x)|}{1+|f_n(x)+f(x)|}d\mu,$$ since (as you likely have shown) it is a metric on the space of equivalence classes of measurable functions on $(S,\Sigma,\mu)$, where the relation is $\mu$-a.e. equality.

Note that if $\mu(S)=0,$ the result is obvious, so suppose not. If $f_n\rightarrow f$ in measure, for any $\epsilon>0,$ there exists $N\in\mathbb{N}$ for which $$\mu\left(E_{n\frac{\epsilon}{2\mu(S)}}\right)<\frac{\epsilon}{2}$$ for all $n\geq N.$ Observe that \begin{align*} d(f_n,f)&=\int\limits_S\frac{|f_n(x)-f(x)|}{1+|f_n(x)+f(x)|}d\mu =\int\limits_{E_{n\frac{\epsilon}{2\mu(S)}}}\frac{|f_n(x)-f(x)|}{1+|f_n(x)+f(x)|}d\mu\\ &+\int\limits_{S\setminus {E_{n\frac{\epsilon}{2\mu(S)}}}} \frac{|f_n(x)-f(x)|}{1+|f_n(x)+f(x)|}d\mu\\ &\leq \int\limits_{E_{n\frac{\epsilon}{2\mu(S)}}}1d\mu+\int\limits_{S\setminus E_{n\frac{\epsilon}{2\mu(S)}}}\frac{\epsilon}{2\mu(S)}d\mu=\mu\left(E_{n\frac{\epsilon}{2\mu(S)}}\right)\\ &+\frac{\epsilon}{2\mu(S)}\mu\left(S\setminus E_{n\frac{\epsilon}{2\mu(S)}}\right)\leq \mu\left(E_{n\frac{\epsilon}{2\mu(S)}}\right)+\frac{\epsilon}{2\mu(S)}\mu(S)\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon, \end{align*} where we have used the obvious inequalities $\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}\leq 1$ for any $x\in S$, $$\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}\leq\frac{\epsilon}{2\mu(S)(1+|f_n(x)-f(x)|)}\leq \frac{\epsilon}{2\mu(S)}$$ for $x\in S\setminus E_{n\frac{\epsilon}{2\mu(S)}}$, and monotonicity of measure.

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