Finding a function $f: \mathbb{R} \rightarrow \mathbb{R}$ with this property? > Find a function $f: \mathbb{R} \rightarrow \mathbb{R}$ that is Riemann integrable on every bounded and closed interval, such that the function $$ g: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto \int_{0}^{x} f$$ is not differentiable in at least one point. I was looking for a function $f$ such that $g$ is not continuous in one point (and so a fortiori not differentiable). I wanted to pick $$ f: \mathbb{R} \mapsto \mathbb{R}: x \mapsto - \frac{1}{(x-1)^2} $$ since then $g(1)$ would blow up. But I think this function $f$ is not Riemann integrable on every bounded and closed interval, since it is not even bounded on, say, the interval $[-1,1]$. So anyone have another idea?

Just take the function $f(x)=1$ for $x>0$ and $f(x)=0$ for $x\leq 0$. The integral should have a corner at $x=0$.