Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:\mathbb R \rightarrow \mathbb R$, and for every $x,y\in \mathbb R$ we have $f(x+y)=f(x)+f(y)$.

Show that $f$ measurable $\Leftrightarrow f$ continuous.

share|improve this question
4  
Why not add $\Longleftrightarrow$ $f$ is $C^\infty$ –  GEdgar Jun 17 '11 at 13:47
    
A nice proof is given in Herrlich's Axiom of Choice, p.119. –  Martin Sleziak May 5 '13 at 10:31

1 Answer 1

up vote 10 down vote accepted

One implication is trivial. If a function is continuous, then it is measurable. The converse is more tricky.

You can find a very nice proof in the following document. Another proof can be found considering the function $F(x)=\int_0^x f(t)dt$, which is well defined since $F$ is measurable.

Another approach is the following: prove that a discontinuous solution for the functional equation is not bounded on any open interval. It can be shown that for a discontinuous solution the image of any interval is dense in $\Bbb{R}$, and therefore we have problems with the measurability.

share|improve this answer
3  
My preferred proof of this fact is to show first that for every $A$ of positive measure (or better yet: for every non-meager $A$) the set $A - A$ contains an open neighborhood of zero. See here for a proof of this and links to some relevant further elaborations. Using this, one can easily show that a Baire measurable homomorphism from a Baire group to a separable group is continuous (Pettis' theorem). See Kechris, Classical Descriptive Set Theory, Theorem (9.10) for a nice proof. –  t.b. Jun 17 '11 at 8:55
1  
The document you linked to is very nice, thanks for that! –  t.b. Jun 17 '11 at 8:56
    
$F$ is well defined since $f$ is measurable ... what if $f$ is unbounded in every neighborhood of $0$? –  GEdgar Jun 17 '11 at 13:46
    
I am not sure about this argument: It can be shown that for a discontinuous solution the image of any interval is dense in $\Bbb{R}$, and therefore we have problems with the measurability. According to this MO post, there are measurable functions that have dense graph. –  Martin Sleziak Jun 9 '13 at 14:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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