convergence in measure topology

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.

Yes, defining $d'=d\wedge1$ works fine, as the balls of diameter $<1$ still form a basis for the topology. But as far as the topology induced by $d$ goes, there is nothing wrong with $d$ taking infinite values either.

As for uniqueness, metrizable topologies are completely determined by their convergent sequences, as a subset $S$ is closed iff $S$ includes all limits of convergent sequences in $S$. More general topologies are completely determined by their convergent nets for the same reason. So there could be a different topology defining convergence of sequences in measure, but it would not define convergence of more general nets in measure and it would not be induced by any metric.