# Measurable Cauchy Function is Continuous

I found this question here Show that $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable and I want to adapt the proof that t.b had suggested. I don't know the concepts of Baire and separable group. Can this be done by simple analysis and measure theory using the result of t.b, maybe by defining A suitably or in some other way.

Thanks for any help.

• Well in context of your original problem you can always adapt the proof by Banach about solving the functional equation if f is measurable which is seemingly lot easier. – smiley06 Apr 12 '13 at 13:40
• Yes,I have done that.But I want to see how this works – Ester Apr 12 '13 at 17:03
• Can't you think of a title more indicative of the content of the question?! – Mariano Suárez-Álvarez Apr 14 '13 at 9:54
• @Mariano Sorry about that.I shall keep it in mind next time around. – Ester Apr 14 '13 at 11:51
• Why not modifying the title of this question? – Did Apr 14 '13 at 12:18

t.b. mentioned the following result:

If $A \subseteq \mathbb{R}$ has positive measure then $A - A = \{a - a' \mid a,a' \in A\}$ is a neighborhood of zero.

Various proofs are discussed in The set of differences for a set of positive Lebesgue measure.

Given this, we can prove that a Lebesgue measurable homomorphism $f \colon \mathbb{R} \to \mathbb{R}$ is continuous as follows:

It suffices to prove that $f$ is continuous at $0$, so we need to show that for every open neighborhood $U$ of $0$, its pre-image $f^{-1}(U)$ is a neighborhood of $0$.

Choose an open set $V \subseteq U$ such that $V - V \subseteq U$. Using an enumeration $(q_n)_{n \in \mathbb{N}}$ of the rational numbers (or any other countable dense subset of $\mathbb{R}$) we have $\mathbb{R} = \bigcup_{n \in \mathbb{N}} (q_n + V)$ and hence also $\mathbb{R} = \bigcup_{n\in\mathbb{N}} f^{-1}(q_n + V)$.

Since $f$ is measurable, the sets $W_n = f^{-1}(q_n + V)$ are measurable, and since $\mathbb{R} = \bigcup_{n=1}^\infty W_n$, at least one of them has positive measure. Therefore $W_n - W_n$ is a neighborhood of $0$ for some $n$. But $f$ is a homomorphism, so $W_{n} - W_{n} \subseteq f^{-1}(V-V) \subseteq f^{-1}(U)$ and we have shown that $f^{-1}(U)$ is a neighborhood of $0$.

• Did you mean V to be an interval? – Ester Apr 14 '13 at 12:03
• Yes, or any open set satisfying $V-V \subseteq U$. I edited it in. – Martin Apr 14 '13 at 12:11
• But did you show that $f^{-1}(U)$ is open? – mnmn1993 Nov 30 '17 at 15:55