calculus - How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable

Suppose $f(x)$ is integrable in any bounded interval on $\mathbb R$, and it satisfies the equation $f(x+y)=f(x)+f(y)$ on $\mathbb R$. How to prove $f(x)=ax$?

Answer

Integrate the functional equation with respect to $x$ between 0 and 1. The result is the equation
$$\int_y^{y+1} f(u) du = \int_0^1 f(x) dx + f(y).$$
The integral on the left side exists and is continuous in $y$ because $f$ is locally integrable. Therefore the right side is also continuous in $y$; that is, $f$ is continuous! The rest is clear sailing.