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?



When is the set of measurable functions a vector space?
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 $\phi : Y \to \mathbb R$.




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The mentioned counterexammple, is part of my answer [HERE](https://mathoverflow.net/a/313792/454)
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.


