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.

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    $\begingroup$ Given the pseudo-metric $d$ defined in source 1, would defining $d'(f, g) = d(f,g) \wedge 1$ work? $\endgroup$ – Bib-lost Jan 2 '16 at 11:02

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.

  • $\begingroup$ In fact, (global) convergence in measure is metrizable on $L^0$. But for a general measure $\mu$, the generated topology need not be a vector space topology, i.e. $L^0$ with the topology of (global) convergence in measure is not a topological vector space. Addition is continuous (and hence turns $L^0$ into a TAG), but scalar multiplication is continuous if and only if $\mu$ is finite [Fremlin, 245Ye], in which case, we just get back to the simpler situation. $\endgroup$ – yadaddy Oct 25 '19 at 14:06

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.


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