Correspondence between modes of convergence and metrics If $\mu(X)<\infty$, then the metric $$\rho(f,g)=\int \frac{|f-g|}{1+|f-g|}$$ forms a metric on the space of measurable functions $f,g:X\rightarrow \mathbb{C}$ when we equivalence out a.e. equal function. Then we can show that $f_n \rightarrow f$ in $\rho$ if and only if $f_n \rightarrow f$ in measure.
Can we find a metric for convergence in measure when $\mu(X) = \infty$? I've always thought of convergence as being defined with respect to some metric, so it would be strange to me to see that convergence in measure isn't coming from a metric.
 A: It's not an answer to your question, but I just wanted to address the latter point of yours

I've always thought of convergence as being defined with respect to some metric, so it would be strange to me to see that convergence in measure isn't coming from a metric. 

Actually, convergence is often related to the topology, rather than to the metric. For example, the topology of pointwise convergence of functions on $[0,1]$ is not compatible with any metric. However, if we take a look on how is it introduced we'll see that we require $f_n(x)\to f(x)$ for any $x\in[0,1]$ and the latter convergence is indeed can be characterized by a metric, for example the Eucledian one. What I mean to say is that, although some of the notions of convergence are naturally coming through the metric (a more intuitive notion that a topology, perhaps), when such notions are lifted to another level (from points to function as above) the characterization by a metric may not hold anymore for the new topological space.
The notion of the convergence in measure, I believe, was introduced without bearing in mind some specific metric which would characterize it. In a way, there are certain similarities with the pointwise convergence example. Namely, you use the metric on the real values to talk about the distances $|f_n(x) - f(x)|\geq \delta$, but also you want to know how much points have their images $\delta$-faraway, that is what is $\mu(x:|f_n(x) - f(x)|\geq \delta)$. If you think of $f_n$ as some simpler approximation of $f$, you may use $f_n$ to do e.g. approximate computations for $f$. However, in such a case you would like to assure that regardless of the chosen precision level $\delta$, for any $\varepsilon$ there exists $n$ after which $f_n$ approximate $f$ good enough, that is
$$
  \lim_{n\to\infty}\mu(x:|f_n(x) - f(x)|\geq \delta) = 0,\quad \forall \delta>0.
$$
Although for some cases, e.g. $\sigma$-finite measures, or the counting measure, the convergence in measure in metrizable, I wouldn't be surprised if it's not the case in general. 
You may also want to check out this.
