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.

  • $\begingroup$ 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. $\endgroup$ – smiley06 Apr 12 '13 at 13:40
  • $\begingroup$ Yes,I have done that.But I want to see how this works $\endgroup$ – Ester Apr 12 '13 at 17:03
  • 1
    $\begingroup$ Can't you think of a title more indicative of the content of the question?! $\endgroup$ – Mariano Suárez-Álvarez Apr 14 '13 at 9:54
  • $\begingroup$ @Mariano Sorry about that.I shall keep it in mind next time around. $\endgroup$ – Ester Apr 14 '13 at 11:51
  • $\begingroup$ Why not modifying the title of this question? $\endgroup$ – 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 $0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.