# Show that $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable

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.

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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

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.
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
$F$ 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