Solution of Cauchy functional equation which has an antiderivative

Solution 1:

By the functional equation, it suffices to prove that $f$ is continuous at one point.

The fact that $f$ is of first Baire class is very straightforward: $$ f(x) = \lim_{n \to \infty} \frac{F(x+1/n)-F(x)}{1/n} $$ is a pointwise limit of continuous functions.

Now a function of first Baire class has a comeager $G_\delta$-set of points of continuity. Done.

Indeed, enumerate the open intervals with rational endpoints as $\langle I_n \mid n \in \omega\rangle$. Then $$ f \text{ is discontinuous at }x \iff \exists n\in \omega : x \in f^{-1}[I_n] \setminus \operatorname{int}f^{-1}[I_n] $$ Since $f$ is of first Baire class, $f^{-1}[I_n]$ is an $F_\sigma$ and so is $f^{-1}[I_n] \setminus \operatorname{int} f^{-1}[I_n]$. Therefore we can write $$ f^{-1}[I_n] \setminus \operatorname{int} f^{-1}[I_n] = \bigcup_{k \in \omega} F_{k}^{n} $$ for some sequence $\langle F_{k}^n \mid k \in \omega\rangle$ of closed sets. Observe that $F_{k}^n$ has no interior, so the set of points of discontinuity of $f$ is $$ \bigcup_{n \in \omega} f^{-1}[I_n] \setminus \operatorname{int}f^{-1}[I_n] = \bigcup_{n\in\omega} \bigcup_{k\in\omega} F_{k}^n, $$ a countable union of closed and nowhere dense sets.

Solution 2:

$ \def \R {\mathbb R} $ Define the function $ G : \R ^ 2 \to \R $ by the equation $ G ( x , y ) = F ( x + y ) - F ( x ) - F ( y ) - x f ( y ) $ for all $ x , y \in \R $. Then, as $ F $ is differentiable, $ G $ is differentiable with respect to the first variable. Since $ F ' = f $ and $ f $ is additive, we have $$ \frac { \partial G ( x , y ) } { \partial x } = F ' ( x + y ) - F ' ( x ) - f ( y ) = f ( x + y ) - f ( x ) - f ( y ) = 0 $$ for all $ x , y \in \R $. Hence, $ G $ is constant with respect to the first variable; i.e. there is a function $ g : \R \to \R $ such that $ G ( x , y ) = g ( y ) $ for all $ x , y \in \R $. Now, using the definition of $ G $ and taking a look at the expression for $ G ( y , x ) - G ( x , y ) $, we have $$ x f ( y ) - y f ( x ) = g ( x ) - g ( y ) \tag {*} \label {eqn} $$ for all $ x , y \in \R $. By additivity of $ f $ we have $ f ( 0 ) = 0 $, and thus letting $ y = 0 $ in \eqref{eqn} we get $ g ( x ) = g ( 0 ) $ for all $ x \in \R $. This, together with \eqref{eqn} itself, proves that $$ y f ( x ) = x f ( y ) $$ for all $ x , y \in \R $. In particular, for $ y = 1 $, this gives $ f ( x ) = c x $ for all $ x \in \R $, where $ c = f ( 1 ) $.

You can see that there's no need to mention measurability, as you wanted. We didn't even reduce the problem to the case when we have additional assumptions (like continuity) on $ f $. In the light of the fundamental theorem of calculus, this method is essentially the same as the one here, only avoiding integration.