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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.

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

2 Answers2

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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.

Johannes Hahn
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    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. – yada Oct 25 '19 at 14:06
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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.

Tristan Bice
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