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Suppose we are working in ZF set theory without choice. An additive function is a function defined over the real line such that $f(x+y)=f(x)+f(y)$. It is known that if every set of reals is measurable, then every additive function is linear. Is the converse true in ZF?

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    $\begingroup$ I assume a non-linear additive function in ZFC looks like this: choose a Hamel basis of $\mathbb R$ as a $\mathbb Q$-vector space, define the map on the basis arbitrarily and extend by $\mathbb Q$-linearity. So without choice we can't produce it. Correct? $\endgroup$ Jan 2, 2021 at 0:37
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    $\begingroup$ @PatrickDaSilva Yes, correct. $\endgroup$
    – user107952
    Jan 2, 2021 at 0:39
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    $\begingroup$ I'm curious about the statement "every set of reals is measurable $\Rightarrow$ every additive function is linear" in ZF. Do you have a proof written up somewhere? Is it heavy? $\endgroup$ Jan 2, 2021 at 0:39
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    $\begingroup$ I'm thinking a bit about continuity as well because the statement "every additive function is linear" can be re-phrased as "every $\mathbb Q$-linear function is $\mathbb R$-linear" because additivity implies $f(nx) = nf(x)$ by induction on $n$, so that $f(ax/b) = \frac {af(x)}{b}$ for any $a/b \in \mathbb Q$. $\endgroup$ Jan 2, 2021 at 0:44
  • $\begingroup$ On continuity, good point worth expanding upon. It is well known that the function solutions to f(a+b) = f(a) + f(b) must satisfy f(x) = cx for some constant c at least on Q (set of rationals). However, George Hamel showed the existence of nowhere-continuous additive functions using the axiom of choice. It is quite easy to prove that if f is also continuous at a single real value, then f is necessarily of the form f(x) = cx. I even wrote about this in a tiny paper: Michael W. Ecker, Adding a Simple Condition to Additive Functions, MathAMATYC Educator, February 2016, Vol. 7, No. 3. $\endgroup$ Jan 5, 2021 at 17:49

1 Answer 1

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No.

We know that "Every set of reals has the Baire property" will also have the same consequence, and that ZF+"Every set of reals has the Baire property" is actually a weaker(!) theory than ZF+"Every set of reals is Lebesgue measurable", as shown by Shelah.

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    $\begingroup$ Can you cite references to those statements? I want to read more about this topic! (i.e. about "Every set of reals has the Baire property $\Rightarrow$ every additive function is $\mathbb R$-linear" and "ZF+Baire" weaker than "ZF+all real subsets are measurable") $\endgroup$ Jan 2, 2021 at 3:19
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    $\begingroup$ @PatrickDaSilva: You can find the first one in my write-up, arxiv.org/abs/2010.15632, and the second is in Shelah's famous "Can you take Solovay's inaccessible away" paper from 1984. $\endgroup$
    – Asaf Karagila
    Jan 2, 2021 at 9:37
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    $\begingroup$ Nice, I didn't know that this also follows from every set of reals having the Baire property. Now of course one could ask the natural follow-up to the OP's question: does all additive functions are linear imply all sets of reals have BP? $\endgroup$ Jan 2, 2021 at 15:19
  • $\begingroup$ @Alex: Yes, that is a natural question. I believe this is not veering too far away from the Banach limits problem mentioned in my write-up. $\endgroup$
    – Asaf Karagila
    Jan 2, 2021 at 15:32
  • $\begingroup$ So ... A standard thing (perhaps using choice) is to prove: if $f$ is additive and Lebesgue measurable, then $f$ is of the form $f(x) =ax$. Does this argument work in ZF only? Lebesgue measurable function means:$\{x : f(x) < \lambda\}$ is Lebesgue measurable set for every real $\lambda$. The proof uses this: if $A$ is a Lebesgue measurable set with positive measure, then $A - A$ contains a neighborhood of $0$. $\endgroup$
    – GEdgar
    Jan 5, 2021 at 22:39

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