Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous but non-differentiable function. If $g: \mathbb{R} \to \mathbb{R}$ is a function that is differentiable almost everywhere on $\mathbb{R}$, or if it is differentiable on the interior of its support, then it is known that $f * g$ is differentiable with $(f*g)' = f*g'$.
Let $\mu = \frac{1}{2} \mathbb{I}_A$ be the density (w.r.t Lebesgue measure) of the uniform distribution on the interval $A = [-1, 1]$. So $\mu$ is differentiable almost everywhere on $\mathbb{R}$, and it is also differentiable on the interior of its support. Thus we should have that$$(f*\mu)' = f*\mu' = 0,$$where the last equality is because $\mu'$ exists almost everywhere and equals $0$ whenever it exists. Alternatively, we know that the above integral is only over the interior of the set $A$ (since $\mu$ is only supported on $A$ and we can discard the measure zero endpoints), and on this interior $\mu'$ always exists and equals $0$.
But surely this is wrong? Where does this "proof" break down? I spelled out the details of my thinking as much as possible. I appreciate your help!