# When is the set of measurable functions a vector space?

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?

• 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. Jan 4, 2021 at 11:19
• 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).
– YCor
Jan 4, 2021 at 14:47
• 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".
– YCor
Jan 4, 2021 at 14:51

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.

• 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! Jan 4, 2021 at 12:17