# Can the sum of two measurable functions be non-measurable if they are valued in a general normed space instead of $\mathbb{R}$?

It's well known that the sum of measurable functions is measurable, if they are real or complex valued. However, the proofs I've seen heavily rely on the usage of the countable set of rational numbers. Which made me wonder, what happens if we don't have luxury of having a nice, countable and dense subset in the target space?

If $X$ is a normed space and $f,g: \mathbb{R} \rightarrow X$ are Borel functions, can $f + g$ be non-Borel? I think that using $X = C([0,1], \mathbb{R})$ with the supremum norm could be a good idea, since the space isn't seperable(EDIT: It actually is...), so there's nothing quite like $\mathbb{Q}$ in it. I haven't been able to come up with any concrete examples, though.

Also, if one could solve this problem, it'd be natural to ask whether anything changes if we allow the domain to be any measurable space or restrict $X$ to Banach spaces only. Or is separability of $X$ sufficient for $f + g$ to be measurable?

I know that's a lot of questions, so I'll appreciate any help given or some references where I could find out more.

EDIT: I've found the following paper: http://www.math.ucla.edu/~brh6/DirectIntegral.pdf , which proves at the beginning that the sum of Borel functions valued in a separable Banach space is Borel. Which answers a part of the question, but the main part still remains.

EDIT2: $C([0,1])$ was obviously a bad idea, since it's separable by stone-weierstrass theorem. $B(\mathbb{R})$ (space of bounded functions) seems like a much better choice for $X$ and it also higher cardinality than continuum, which might be helpful. I've stumbled upon a result(without proof unfortunately, so I'm not sure about it being correct) claiming that if $X$ has a dense subset of cardinality $\le \aleph_1$, then the sum of Borel functions must also be Borel. I don't know what is the minimal cardinality of a dense subset of $B(\mathbb{R})$, but we might need an even bigger space to find a good example for a non-measurable sum of measurable functions.

• The space of measurable functions can be nonlinear if the domain is only specified to be measurable and the codomain a Banach space with the Borel $\sigma$-algebra. This result is an exercise in Appendix E of the book Measure theory by Donald Cohn. I haven't found anything yet about the case where the domain is $\mathbb R$. – epimorphic May 29 '15 at 17:00

Yes, it is true that the sum of (Borel) measurable functions from $\mathbb{R}$ to $X$ is again measurable.

Case I - $X$ is separable: In this case, $X$ is second countable. In particular, if $S$ is a countable dense subset of $X$ then the collection, $\mathcal{B}$, of open balls of radius $1/n$ ($n\in\mathbb{N}$) centred on points in $S$ is a countable base for the topology of $X$. This enough to show that the map $\mathbb{R}\to X\times X$ given by $x\mapsto(f(x),g(x))$ is Borel. To prove this, note that any open subset of $X\times X$ is a union of products $U\times V$ for $U,V\in\mathcal{B}$. This is necessarily a countable union, so its inverse image under the aforementioned map is a countable union of sets of the form $f^{-1}(U)\cap g^{-1}(V)$ and, hence, is Borel. The map $X\times X\mapsto X$, $(x,y)\mapsto x+y$ is continuous and hence Borel. Composing these maps shows that $\mathbb{R}\to X$, $x\mapsto f(x)+g(x)$ is Borel.

Case II - $X$ is any normed space. In fact, this reduces to case $I$. Let $S$ be the image of $f$, which is a metric space under the subspace topology. Then, $f$ is a Borel measurable function from the separable metric space $\mathbb{R}$ onto $S$. This is enough to imply that $S$ is separable (see problem 10, section 4.2 of R.M. Dudley, Real Analysis and Probability). Similarly, the image of $g$ is separable. Letting $Y$ be the subspace of $X$ spanned by the images of $f$ and $g$, then $Y$ is separable and the measurability of $f+g$ reduces to case I above, so is measurable.

Note: Case II above relies on the fact that $\mathbb{R}$ is a separable metric space, and the result does not hold if we replace the Borel sigma algebra on $\mathbb{R}$ by an arbitrary measurable space.

• Not entirely related, but what about the case that non-Borel measurable functions say f and g such that their sum and product are Borel measurable? Is that possible? – Siddhartha Feb 28 at 5:11

Edit: As OP points out, this solution only works if $X$ is second-countable.

Let $f, g: \mathbb{R} \to X$ be Borel measurable and $h(s) = f(s) + g(s)$. Consider mappings

\begin{align} \mathbb{R} \ni s & \stackrel{h_1}{\mapsto} (s, s) \in \mathbb{R}^2 \\[1ex] \mathbb{R}^2 \ni (s, t) & \stackrel{h_2}{\mapsto} (f(s), g(t)) \in X^2 \\[1ex] X^2 \ni (x, y) & \stackrel{h_3}{\mapsto} x+y \in X. \end{align}

Then

$$s \stackrel{h_1}{\mapsto} (s, s) \stackrel{h_2}{\mapsto} (f(s), g(s)) \stackrel{h_3}{\mapsto} f(s) + g(s)$$

so $h = h_3 \circ h_2 \circ h_1$. But $h_1$ and $h_3$ are continuous hence Borel measurable and $h_2$ is Borel measurable. Thus so is $h$.

• Does $h_2$ being measurable use second-countability of $X$? I've been trying to prove it and can only get it in that case. – Tim Carson May 20 '15 at 19:06
• You got me here. :p I didn't notice that. – Adayah May 20 '15 at 19:35