# Convergence in Measure and Limit of an Integral [duplicate]

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.

Suppose $$\int \frac {|f_n-f|}{1+|f_n-f|}d\mu$$ does not tend to $$0$$. Then there exists $$\epsilon >0$$ and inetgers $$n_1 such that $$\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$$.
You already have a nice answer, but I'd like to provide another, where we do this directly. For notational convenience, call $$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.