real analysis - convergence in measure topology

I'm attending a course on measure theory this semester. While proposing different kinds of convergence (in measure, almost everywhere and in $L^{p}$), our professor stressed (and proved) the fact that convergence almost everywhere is not topological, but claimed that convergence in measure is.

As pointed out in a few questions on this site and wikipedia (1, 2, 3), in the case where $(\Omega, \mathcal{F}, \mu)$ is a finite measure space, convergence in measure can be described by a pseudometric (hence a topology). However, I haven't found an answer to why at least a topology should exist in the case where $\mu$ is an arbitrary measure. Wikipedia (3) claims that their pseudometric works for arbitrary measures, but their proposed function can take $\infty$ as a value, which I believe isn't allowed for metrics.

To sum up: let $(\Omega, \mathcal{F}, \mu)$ be a (not necessarily finite) measure space, does there exist a topology $\mathcal{T}$ on the set of measurable functions $f : \Omega \to \mathbb{R}$ such that a sequence of measurable functions $(f_{n})_{n}$ converges to a measurable function $f$ in measure if and only if it converges to $f$ in the topology $\mathcal{T}$? Extra: Is this topology unique?

Thank you for your help! I've had introductory courses in topology (metric spaces), Banach (Hilbert) spaces and now measure theory.

Answer

The convergence in measure is not just induced by a topology, it is in fact induced by a metric! Admittedly, it is not at all obvious how to come up with it, but here it is:

$$d(f,g) := \inf_{\delta > 0} \big(\mu(|f-g|>\delta) + \delta\big)$$
(I found it a while back in this book)
This is, again, in general a $[0,\infty]$-valued metric, but this is not a problem as previously noted because you could just as well use $d':=d\wedge 1$ or $d'':=\frac{d}{1+d}$ to get the same topology.

Just a side note: What is quite neat is that you can know that there must be some metric even without having a specific candidate, because the space of measurable functions with convergence in measure is a first countable topological vector space and those are all metrisable.

EDIT: As to the uniqueness question: It was already noted that convergence of sequences alone does not uniquely determine a topology. Not unless you add other properties. For example there is a unique metrisable/quasi-metrisable/first-countable topology that induces exactly this convergence of sequences.