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The set of measurable functions together with convergence in measure is a completely metrizable abelian topological group

Below we use Bochner measurability and Bochner integral. Let $(X, \mathcal A, \mu)$ be a complete $\sigma$-finite measure space, $(E, | \cdot |)$ a Banach space, $S (X)$ the space of $\mu$-simple ...
Analyst's user avatar
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10 votes
1 answer
366 views

Are all compact subsets of Banach spaces small in a measure-theoretic sense?

Definition. A subset $K$ of a topological group $X$ is called measure-continuous if there exists a $\sigma$-additive Borel probability measure $\mu$ on $X$ such that for every compact subset $C\subset ...
Taras Banakh's user avatar
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3 votes
2 answers
470 views

If $ F(x,\bullet) \in {L^{2}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable?

This question is related to something that I asked yesterday: If $ F(x,\bullet) \in {L^{\infty}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable? Pietro Majer ...
Transcendental's user avatar
5 votes
1 answer
298 views

If $ F(x,\bullet) \in {L^{\infty}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable?

Let $ (X,\Sigma,\mu) $ be a $ \sigma $-finite measure space and $ B $ a Banach space. A function $ f: X \to B $ is said to be strongly $ \mu $-measurable iff it is the almost-everywhere pointwise ...
Transcendental's user avatar
6 votes
0 answers
365 views

Approximating a measurable function from a second-countable, locally compact Hausdorff group to a separable Banach space

Let $ G $ be a second-countable, locally compact Hausdorff group and $ B $ a separable Banach space. We say that a function $ f: G \to B $ is Bochner-measurable if and only if it is the everywhere ...
Transcendental's user avatar