All Questions
5 questions
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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 ...
10
votes
1
answer
366
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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 ...
3
votes
2
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470
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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 ...
5
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1
answer
298
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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 ...
6
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365
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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 ...