Let $I\subseteq\mathbb{R}$ be an non-degenerate interval and let $f$: $I\longrightarrow\mathbb{R}$ be a function with the property that for every pair $(x_0, y_0)\in I\times\mathbb{R}$ there exists a unique $F$: $I\longrightarrow\mathbb{R}$ with $F'=f$ and $F(x_0)=y_0$.
Must $f$ be continuous?
The claim is false. There exist functions which are differentiable everywhere, but whose derivatives are not continuous. (Example on Math.SE.) Let $F_0$ be such a function, and $f$ its derivative. Given a pair $(x_0,y_0)$, set $$F(x) = F_0(x) - F_0(x_0) + y_0.$$ We then have $$F'(x) = F_0'(x) = f(x)$$ and $$F(x_0) = F_0(x_0) - F_0(x_0) + y_0 = y_0,$$ as required. (Uniqueness holds since antiderivatives are always unique up to a constant offset.)
