Why is the topology of convergence in measure equivalent to this metric here?

Question

I am currently struggeling with the topic of convergence in measure topologies.

Now I read that on the space of measurable function $L^0$ on $[0,1]$ with the Borel sigma algebra and the lebesgue measure, the topology of convergence in measure is equivalent to the one induced by the metric $d(f,g) = \int_0^1 \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|} dx$. Could anybody here show me the relationship between this metric and convergence in measure? I suspect that $X_n \rightarrow X$ in measure iff $d(X_n,X) \rightarrow 0$ , but I cannot prove it.

And another question would be interesting too: Is this metric space complete?

Other thread that I have about this topic.

If anything is unclear, about my question, please let me know.

Answer 1

Yes, this is true. More generally, suppose $\varphi:[0,\infty)\to[0,\infty)$ is a continuous strictly increasing bounded function. Then $f_n\to 0$ in measure iff $\int_0^1 \varphi (|f_n|)\to 0$.

Indeed: for every $\epsilon>0$ we have $$\varphi(\epsilon)\, m(\{ |f_n|\ge \epsilon \}) \le \int_0^1 \varphi (|f_n|)$$ which shows that convergence of integrals implies convergence in measure.

Conversely, let $M=\sup \varphi$. Then for every $\epsilon> 0$ $$\int_0^1 \varphi (|f_n|) \le M\,m(\{ |f_n|\ge \epsilon \}) + \varphi(\epsilon) $$ which shows that convergence in measure implies convergence of integrals.