# A pseudometric on the space of the measurable functions is complete

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:

1. $d(f, g)=0$ if and only if $f=g$ almost everywhere.
2. $d(f, g)=d(g, f)$
3. $d(f, g)\leq d(f, h)+d(h, g)$
4. 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: $$\lim_{n\rightarrow\infty}\mu(\{x \in X: |f_n(x)-f(x)|\geq\delta\})=0.$$
5. 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.

## 2 Answers

As shown (e.g.) here Cauchy in measure implies convergent in measure., it suffices to show that $(f_n)_n$ is Cauchy in measure. Note that this also uses point (4) of the properties you have already shown.

To do this, first observe that

1. for $|x-y| \leq 1$, we have $$\min\{1, |x-y|\} =|x-y|\geq\frac{|x-y|}{1 + |x-y|} \geq \frac{|x-y|}{2} = \frac{1}{2} \min \{1 , |x-y|\}.$$
2. for $|x-y| > 1$, we have $$\frac{1}{2} \min \{1, |x-y|\}=\frac{1}{2} = \frac{|x-y|}{|x-y| + |x-y|}\leq \frac{|x-y|}{1 + |x-y|} \leq 1 = \min\{1, |x-y|\}.$$

All in all,

$$\frac{|x-y|}{1 + |x-y|} \asymp \min \{1, |x-y|\}.$$

Hence for $\varepsilon \in (0,1)$, we have

$$d(f_n, f_m) \asymp \int \min\{1, |f_n(x) - f_m(x)|\} d\mu(x)\geq \int_{|f_n -f_m| > \varepsilon} \min\{1, |f_n(x) - f_m(x)|\} d\mu(x) \geq \varepsilon \cdot \mu(\{x \mid |f_n(x) - f_m(x)| \geq \varepsilon\}),$$

which implies

$$\mu(\{x \mid |f_n(x) - f_m(x)| \geq \varepsilon\}) \to 0 \text{ for } n,m\to \infty$$

for all $\varepsilon \in (0,1)$ (and hence for all $\varepsilon > 0$).

Hence, $(f_n)_n$ is Cauchy in measure, so that the post linked above implies that there is a measurable function $f$ such that $f_n \to f$ in measure, which (by property (4)) implies $d(f_n ,f) \to 0$.

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 $n \to \infty$.