TITLE: A pseudometric on the space of the measurable functions is complete
QUESTION [7 upvotes]: I'm working in the following exercise:
Suppose $(X, \mathcal A, \mu)$ is a finite measure space and suppose $\mathcal F$ is the set of all $\mathcal A$-measurable functions $f: X \rightarrow \mathbb R$. For $f, g \in \mathcal F$, let $$d(f, g)=\int_X\frac{|f-g|}{1+|f-g|}d\mu.$$
Show that:

$d(f, g)=0$ if and only if $f=g$ almost everywhere.
$d(f, g)=d(g, f)$
$d(f, g)\leq d(f, h)+d(h, g)$
If $f_n$ is a sequence in $\mathcal F$ and if $f \in \mathcal F$ then $d(f_n, f)\rightarrow 0$ if and only if for every $\delta>0$ the following holds:
$$\lim_{n\rightarrow\infty}\mu(\{x \in X: |f_n(x)-f(x)|\geq\delta\})=0.$$
If $f_n$ is a Cauchy sequence on $\mathcal F$ then there exists $f \in \mathcal F$ such that $d(f_n, f)\rightarrow 0$.

i.e., I must show that this thing is a complete pseudometric space.
I have already done 1. 2. 3. 4., but I'm stuck on 5. I don't know which $f$ should the $f_n$ converge to.

REPLY [2 votes]: Here is a straight forward argument using the following facts:
1) A countable union of nullsets is null
2) a pointwise limit of measurable functions is measurable
Then the argument goes as follows: Let 
$$A(k) := \{ x \in X ~\vert ~\forall ~N \in \mathbb N~\exists ~n,m > N : \vert f_n(x) - f_m(x) \vert > 1/k\}.$$
It is easy to see that $A(k)$ is a nullset for all $k$ since $f_n$ is a cauchy sequence. Therefore also
\begin{align}
A := \bigcup_{k \in \mathbb N} A(k) &= \{x \in X ~\vert ~\exists ~k \in \mathbb N : \forall ~N \in \mathbb N~\exists ~n,m > N : \vert f_n(x) - f_m(x) \vert > 1/k\}\\
&= \{x \in X ~\vert f_n(x) \text{ is not a cauchy sequence}\}
\end{align}
is a nullset by fact 1. Hence $f_n(x)$ is a cauchy sequence for all $x \in A^c$, which has full measure in $X$, and therefore the limit $\lim_{n \to \infty}f_n(x)$ exists for all $x \in A^c$. Now define
$$f(x) = \begin{cases}
\lim_{n \to \infty} f_n(x), ~x \in A^c\\ 69, ~ x \in A
\end{cases}$$
Then $f$ is measurable by fact 2. Also it is not hard to see that $f_n \to f$ with respect to $d$ for $n \to \infty$.