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Let $M$ be the set of measurable functions in $\mathbb{R^R}$. Now, $\mathbb{R^R}$ has the borelian sigma-algebra associated with its product topology, which allows us to ask the following question :

Is $M$ measurable?

On the positive side, I have two remarks: $M$ is sequentially closed (a pointwise limit of measurable functions is measurable) but alas it is not closed; also, in some models of ZF, $M$ is $\mathbb{R^R}$ so is obviously measurable.

On the negative side, I feel like a borelian subset of $\mathbb{R^R}$ will only give constraints on the value of functions at countably many points, which is probably not enough to make them measurable.

So I don't know, in case it matters, I specify that I am interested in answers in ZFC :)

If it might be useful, I can explain why I was asking this question: I wanted to be sure that functions were almost always non-measurable. So I decided to measure the set of functions $\mathbb R\to \mathbb{R/Z}$ for the Haar measure ($\mathbb{(R/Z)^R}$ is a compact group). Assuming this set is measurable, I know thanks to a Facebook commenter (Dani Spivak) that it must have measure zero, which is intuitively nice but not enough.

So, if the answer is "no" or "it is undecidable", I am also interested in the weaker question:

Is the set of measurable functions $\mathbb R\to\mathbb{R/Z}$ a subset of some measurable subset of $\mathbb{(R/Z)^R}$ that has measure zero?

Henrik supports the community
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Moinsdeuxcat
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1 Answers1

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If we choose the product-$\sigma$-algebra on $\mathbb{R}^\mathbb{R}$, then any measurable set is the pre-image of a projection $\pi_I \colon \mathbb{R}^\mathbb{R} \rightarrow \mathbb{R}^I$, where $I$ is countable set. (This can be easily shown by using the definition of the product-$\sigma$-algebra. Any such projection is measurable. On the other hand, the set $\bigcup_{I \text{ countable}} \pi_I^{-1}(\sigma(\mathbb{R}^I))$ is a $\sigma$-algebra, where $\sigma(\mathbb{R}^I)$ denotes the product-$\sigma$-algebra on $\mathbb{R}^I$.)

Now we can show that any measurable set contains a non-measurable function (in ZFC).

Thus, let $A= \pi_I^{-1}(B)$ be a non-trivial set. Take $f \in A$ and take a non-measurable set $E \subset \mathbb{R}$. If $f$ is non-measurable, we have nothing to prove. If $f$ is mesurable, we can define $g= f + 1_{E \cap I^c}$. Note that $E \cap I^c$ is not measurable and thus $g$ is not measurable. However, we have $g=f$ on $I$ and thus $g \in A$.

In the above proof I have used that non-measurable sets exists, i.e. the axiom of choice was used.

p4sch
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