convergence $f\ast \rho_{\epsilon}\to f$ when $\epsilon \to 0$

Let $(\rho_{\epsilon})_{\epsilon\ge 0}$ be mollifiers, i.e. $C^{\infty}(\mathbb{R}^n)$ with $\rho_{\epsilon}\ge 0, \int \rho_{\epsilon} = 1$ and $supp(\rho_{\epsilon})\subset B(0,\epsilon)$. I want to study the convergence $f\ast \rho_{\epsilon}\to f$ when $\epsilon \to 0$. So I can write$$(f\ast \rho_{\epsilon}-f)(x)=\int_{B(0,\epsilon)}\big(f(x-y)-f(y)\big)\rho_{\epsilon}(y)dy.$$ Now I see that if $f$ is uniformly continuous we would have $\lvert f(x-y)-f(y) \rvert < s$ for $y\in B(0,\epsilon)$ and so we have uniform convergence.
Now if I assume only the continuity of $f$, then for each $x$ fixed and each $s$ there is $\epsilon_x$ such that $\lvert f(x-y)-f(y)\rvert<s$ for $y\in B(0,\epsilon_x)$. So we would have pointwise convergence. But what if $f$ is not continuous? I thought that mollifiers kind of smooth out the function and we would still have convergence.