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A UC Berkeley prelim exam problem asked whether an additive function $f\colon {\mathbb R} \to {\mathbb R}$, i.e. satisfying $f(x + y) = f(x) + f(y)$ must be continuous. The counterexample involved taking a positive-valued Hamel basis $X$ of $\mathbb{R}$ as a vector space over ${\mathbb Q}$, and then letting $f(x_1) =1$ and $f(x_2)=-1$ for two different $x_1,x_2 \in X$, and letting $f(x)$ be arbitrary for other $x \in X$, and then extending the function to all of ${\mathbb R}$ using the property of additivity. Then a sequence $a_n = {p_nx_1 + q_nx_2}$ could be found with rational $p_n,q_n$ such that $a_n \to 0$ but $\lim \limits_{n\to \infty} f(a_n) \neq 0$, showing discontinuity. But is Zorn's Lemma necessary to produce such an example? In other words, is Zorn's Lemma saying we can find a Hamel basis of ${\mathbb R}$ over ${\mathbb Q}$ equivalent to being able to construct a discontinuous additive function $f\colon {\mathbb R} \to {\mathbb R}$?

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  • $\begingroup$ I doubt that the two are equivalent since we can consider sets much larger than the reals and ask of Zorn holds for them. $\endgroup$ – Baby Dragon Sep 13 '13 at 16:51
  • $\begingroup$ Good point, I'll revise the question to reflect that I'm only talking about Zorn's lemma for the reals as a vector space over rationals here. $\endgroup$ – user2566092 Sep 13 '13 at 16:53
  • $\begingroup$ There was some discussion of similar ideas here. $\endgroup$ – Old John Sep 13 '13 at 16:53
  • $\begingroup$ @OldJohn Thanks for the link, it is indeed related. I'm pushing the envelope further here, asking whether a Hamel basis is needed to prove discontinuous additive functions exist. $\endgroup$ – user2566092 Sep 13 '13 at 16:57
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    $\begingroup$ Dear user, Your question seems to be more or less a duplicate of this one, and in any case the answer there might interest you. This MO question is also essentially the same, and the answer there should also help. Regards, $\endgroup$ – Matt E Sep 13 '13 at 18:10
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Indeed you cannot prove the existence of such discontinuous solution without using some fragment of the axiom of choice. The simplest and most "familiar" way of proving this is indeed impossible is, as the other answers have shown, using the famous Solovay model where every set of real numbers is Lebesgue measurable. One can show that a discontinuous solution can be used to create a non-measurable set, and thus in Solovay's model there are no discontinuous solutions.

However the existence of Solovay's model requires the stronger theory of $\sf ZFC+\exists\kappa\text{ inaccessible}$, whose consistency strength exceeds that of $\sf ZF$ and $\sf ZFC$. One can wonder, if so, what happens if we don't want to believe that large cardinals are consistent?

It turns out that there is an alternative here. Namely, the Baire property. Discontinuous solutions to this functional equation are not only non-measurable, but they also lack the Baire property. Moreover one can show that if we have a homomorphism of Polish groups which is Baire measurable then it is continuous (this is Pettis theorem). The real numbers are of course a Polish group, and this is indeed a homomorphism.

Shelah proved that while Lebesgue measurability of all subsets implies that the theory of $\sf ZFC+\exists\kappa\text{ inaccessible}$ is consistent. However the consistency of the statement "All sets of real numbers have the Baire property" does not require any stronger theory than $\sf ZFC$.

In particular he constructed, from a model of $\sf ZFC$, a model where $\sf ZF+DC$ holds and every set of real numbers have the Baire property. In this model, if so, every homomorphism from $\Bbb R$ to $\Bbb R$ is Baire measurable, and by Pettis theorem continuous.


It should be noted that the existence of a discontinuous solution is provable from the existence of a Hamel basis of $\Bbb R$ over $\Bbb Q$. However it is currently unknown whether the existence of a discontinuous solution implies the existence of a Hamel basis. On the same note, it is simply false that the existence of discontinuous solutions, Hamel basis, non-measurable sets, or sets without the Baire property, or anything else which concentrates on a subset of the real numbers, would imply the axiom of choice.

The reason is that we can easily arrange that the axiom of choice fails in almost any conceivable way, but the real numbers can be well-ordered -- an assumption which is sufficient to prove all the aforementioned objects exist.

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It can be shown that any measurable function $f$ with $f(x+y) = f(x) + f(y)$ is linear (i.e. $f(x) = ax$ for some $a$), and hence continuous (see for example the page on Cauchy's functional equation on MathWorld). So, to give an example of an additive, discontinuous function, you surely need a non-(Lebesgue-)measurable set.

But to construct a set which is not measurable, you unfortunately need the Axiom of Choice which is equivalent to (Kuratowski-)Zorn Lemma; or at least some weaker version of it. If you forget about the Axiom of Choice altogether, then it might be that no non-measurable sets exist. To quote Wikipedia:

In 1970, Solovay demonstrated that the existence of a non-measurable set for the Lebesgue measure is not provable within the framework of Zermelo–Fraenkel set theory in the absence of the Axiom of Choice (...)

Existence of non-measurable set is not (afaik) equivalent to the Axiom of Choice, but you do need some extra piece except for the standard Zermelo–Fraenkel toolbox. Of course, you can bypass the application of (Kuratowski-)Zorn Lemma (e.g. by proving a special case of it ad hoc), but you cannot find the sought counterexample in an "elementary" way.

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  • $\begingroup$ I don't think that existence of non-measurable functions implies the axiom of choice (even for small cardinalities). The implication goes the other way. It's certainly true that existence of non-meas. sets is independent of ZF (without C). $\endgroup$ – user8268 Sep 13 '13 at 17:14
  • $\begingroup$ I need to meditate on this a while, but I have a feeling I'll make it the accepted answer because you're saying that ZF without choice cannot lead to an example, so in a sense this is saying that choice is necessary unless we come up with a substitute, but that essentially answers my question unless someone can come up with some crazy weaker-than-choice property that can still do the trick. $\endgroup$ – user2566092 Sep 13 '13 at 17:26
  • $\begingroup$ @user8268: Thanks for pointing this out. I just edited in a way which does not suggest this implication. What I mean is that AC is normally used in the proof, and that it is "really needed there" (but possibly something weaker will suffice). In particular, I think you only need AC for sets of cardinality up to $\mathcal{c}$ there. $\endgroup$ – Jakub Konieczny Sep 13 '13 at 17:27
  • $\begingroup$ Dear Feanor, You may be interested in the answers linked to in my comment above. Regards, $\endgroup$ – Matt E Sep 13 '13 at 18:15
  • $\begingroup$ @user8268: Dear user, You may be interested in the answers linked to in my comment above, which discuss in more detail the relationships b/w AC, existence of non-measurable sets, and discontinuous $f$ of the kind asked about. Regards, $\endgroup$ – Matt E Sep 13 '13 at 18:16
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You need some choice because there is no such function in Solovay's model, the reason being all such functions are Lebesgue non measurable.

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    $\begingroup$ Solovay's model is constructed using a weakly inaccessible cardinal. So justifying consistency results using Solovay's model is alway relative to the consistency of weakly inaccessible cardinals. $\endgroup$ – William Sep 13 '13 at 17:34
  • $\begingroup$ @William: Dear unnamed and William, These sorts of issues are discussed in the answers linked to in my comment above. Regards, $\endgroup$ – Matt E Sep 13 '13 at 18:15

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