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I know this is not a research question, but I searched somewhat thoroughly and could not find the exact answer I want. But I've always wondered the following: suppose that $(X,\mathcal{M})$ is a measurable space and $Y$ is a real topological vector space equipped with the Borel $\sigma$-algebra $\mathcal{B}$. Let $$L^0(X,Y):=\{f:X\to Y\;\mid\;f \text{ is measurable}\}.$$ When is $L^0(X,Y)$ a real vector subspace of $Y^X$? In other words, what are "minimal" assumption needed on $X$ and $Y$ so that measurable functions form a vector space?

Point 1: For example in the proof of the case when $Y=\mathbb{R}$, if $f,g$ are Borel measurable functions then we use the following equality $$\{ f+g < b\} = \bigcup_{r\in\mathbb{Q}} \{f< r\} \cap \{g< b-r\}.$$ to show that $f+g$ is measurable. So we have used the following assumptions

  • $Y$ is ordered.
  • $Y$ has a countable dense set w.r.t that order.

How much can this argument be generalized?

Point 2: When $Y$ is a Banach space, I know Bochner spaces come into play. Is there any result regarding the original question in this case?

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    $\begingroup$ Why the disclaimer? I think this is indeed a research-level question, in the sense that a full answer at the required level of generality is not readily available, I believe. $\endgroup$ Jan 4, 2021 at 11:19
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    $\begingroup$ Note that being measurable is stable under taking scalar multiplication, so the issue is whether $f,g$ measurable implies $f+g$ measurable. In particular the question makes sense for $Y$ an arbitrary topological group (even if for a counterexample one prefers $Y$ to be a Banach space). $\endgroup$
    – YCor
    Jan 4, 2021 at 14:47
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    $\begingroup$ Probably $\mathcal{L}^0$ would be a better notation than $L^0$, since the latter suggest modding out by the subspace of measurable functions that vanish outside a set of measure zero. Also for the discussion it would be useful not to use the same notation ($L^0$ or else) for several distinct senses of "measurable". $\endgroup$
    – YCor
    Jan 4, 2021 at 14:51

1 Answer 1

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The usual thing to do, even when $Y$ is a Banach space, is to define "measurable" in such a way that it works. (A while back I posted a counterexample to the general case. See below.)

Bochner measurable, meaning there exist simple functions $f_n$ that converge a.e. to $f$. "Simple" functions have finite range. In case (i) $X$ is any measurable space and $Y$ is a separable Banach space or in case (ii) $X$ is a perfect measure space and $Y$ is any Banach space, then Bochner measurable is the same as $f$ is "measurable" from $\mathcal M$ to the Borel sets in $Y$. But in general, Bochner measurability is the useful notion.

OR

(works for $Y$ a locally convex space) Weakly measurable, meaning $\phi \circ f$ is measurable for all continuous linear functionals $\phi : Y \to \mathbb R$.


The mentioned counterexammple, is part of my answer HERE It provides two measurable functions $f,g : \Omega \to B$ with $f+g$ not measurable.

$\Omega = T \times T$ where $T$ has power $2^{\aleph_0}$ and and $\Omega$ has the product sigma-algebra $\mathcal F$, with $\mathcal P(T)$ in each factor.

$B = l^2(T)$, a non-separable Hilbert space with orthonormal basis $\{e_t: t \in T\}$. We use the sigma-algebra $\mathcal B$ of Borel sets for the norm topology.

The definitions are $f\big((u,v)\big) = e_u$ and $g\big((u,v)\big) = -e_v$. Then $f,g$ are $\mathcal F$ to $\mathcal B$ measurable, but $f+g$ is not. Details in the link.

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  • $\begingroup$ Thank you for the helpful answer. For clarificartion: - when $Y$ is a separable Banach space, are any assumptions needed on $X$ to show that $L^0(X,Y)$ (in the sense of Bochner measurable) is vector space ? - You said "or $X$ is a perfect measure space". In this case, does it suffice to take $Y$ as a topological vector space? - In any of two cases above, can you outline how we get the vector space structure on $L^0(X,Y)$? Also, it would be also be great if you can remember the counter example! $\endgroup$
    – UserA
    Jan 4, 2021 at 12:17

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