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Let $(X , \mathcal{A} , \mu)$ a finite measurable space and $ \mathcal{F}$ the of all $\mathcal{A}$-measurable functions $f : X \rightarrow \mathbb{R} $ .For $ f , g \in \mathcal{F}$ we define : $$ d(f , g) = \int_X \frac{| f -g|}{1 +|f - g| } d \mu .$$

So we have :

  1. $d (f , g) = 0 \Rightarrow f = g $ a.e.
  2. $d(f,g) = d(g,f)$
  3. $d(f,g) \leq d (f,h) + d(h,g)$

    We want to show that if $f_n \in \mathcal{F}$ , for $n = 1,2 , \dots$ is such that $d(f_m , f_n) \rightarrow 0$ so there is a $f \in \mathcal{F}$ such that $d(f_n , f) \rightarrow 0$.

My try : as $d(f_m , f_n) \rightarrow 0 $ we have that for any given $ k \in \mathbb{N}$ there is $n_0 \in \mathbb{N}$ which if $m ,n > n_0$ $d(f_m , f_n) < 1/k$, so $|f_m - f_n| < 1/k $ a.e. , so there is $R_k \subset X$ that $|f_m(x) - f_n(x)| < 1/k $ for all $x \in R_k$ and $\mu (X \backslash R_k) = 0$. Taking $ R = \bigcap_k R_k$, and as $(f_n(x))$ is as Cauchy sequence in $\mathbb{R}$ there is $\lim f(x) $for all $x \in R$, so we take $f(x ) = \lim f(x)$, for $x \in R$. But I don't know if will work. Thanks in advance!

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  • $\begingroup$ $d(f_m,f_n)<1/k$ does not imply $|f_m-f_n|<1/k$ a.e.. $\endgroup$ – Luiz Cordeiro Sep 14 '16 at 22:31
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Given $\epsilon>0$, consider the set $A = \{|f_m(x)-f_n(x)|>\epsilon\}$. Then \begin{equation} d(f_n,f_m)>\frac{\epsilon}{1+\epsilon}\mu(A), \end{equation} which implies that $\mu(A)\to 0$ as $m,n\to 0$, i.e. $f_n$ is a Cauchy sequence in measure. Then there exists an $f$ such that $f_n\to f$ in measure.

Cauchy in measure implies convergent in measure.

Again we have for $B=\{|f_m(x)-f(x)|>\epsilon\}$ \begin{equation} d(f_n,f)\leq \frac{\epsilon}{1+\epsilon}\mu(X) + 1 \mu(B), \end{equation} as, $\mu(X)$ is finite, $\epsilon$ is arbitrary and $\mu(B)\to 0$, we are done.

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