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

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    $\begingroup$ 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. $\endgroup$ – t.b. Jun 17 '11 at 8:55
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    $\begingroup$ The document you linked to is very nice, thanks for that! $\endgroup$ – t.b. Jun 17 '11 at 8:56
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    $\begingroup$ $F$ is well defined since $f$ is measurable ... what if $f$ is unbounded in every neighborhood of $0$? $\endgroup$ – GEdgar Jun 17 '11 at 13:46
  • $\begingroup$ 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. $\endgroup$ – Martin Sleziak Jun 9 '13 at 14:02
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Here is a solution using convolution that complements the ones above. We see that $f(x)=f(1)x$ for rational $x$ by first checking the integers, then numbers of the form $1/n$. The hard part is to take care of the irrational numbers. We may suppose without loss of generality that $f(x)=0$ on the rationals (because subtracting off $f(1)x$ doesn't affect the property in the definition).

Fix some $j\in C_c^\infty(\mathbb R)$. By the defining property of $f(x)$, we have the following expression for the convolution: $$(e^{2\pi if(x)}\star j)(x)=e^{2\pi i f(x)}\int j(y)e^{-2\pi i f(y)}\, dy.$$

In particular, the integral is a constant, so $e^{2\pi i f(x)}$ is continuous, since the convolution is continuous (as $j$ is smooth and compactly supported -- this is a standard theorem).

We must be careful, because we cannot conclude from this that $f(x)$ is continuous without further work. The logarithm function is multi-valued, and it is possible for $f(x)$ to 'skip' by multiples of $1$.

However, we can conclude that $f$ must be integer valued. From this we conclude that it is zero everywhere: given $x\in R$, apply the defining property $n$ times to $x/n$ to conclude that $n|f(x)$ for every $n$, which implies $f(x)=0$.

Note that we use the measurability hypothesis to make sure the integral defining the convolution is well-defined. We use a complex exponential instead of $e^f$ (which would make the conclusion easier) because the latter invites concerns about integrability when defining the convolution.

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  • $\begingroup$ How does one prove the integral is a non-null constant? If it equals 0 one cannot conclude as you did... $\endgroup$ – John Do Nov 4 '18 at 9:46

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