About example of two function which convolution is discontinuous on the "big" set of points
Solution 1:
Take $$f\left(x\right)=\begin{cases} x^{-1/2}, & x\in\left(0,1\right]\\ 0, & \text{otherwise} \end{cases} $$ This function is obviously in $L^1$; note also that $f\ast f$ is $0$ on $(-\infty,0] \cup [2,+\infty)$, $\pi$ on $(0,1]$, and it decays from $\pi$ to $0$ continuously on $[1,2]$. Therefore, $f \ast f(x)$ is everywhere defined, in the sense that $\intop |f(y) f(x-y)| dy < \infty$ for all $x$; it is bounded in $x$, and its only discontinuity is at zero.
Now take $f_a(x) := f(x-a)$, let $\{a_n\}$ be an arbitrary dense countable set, and obverve that:
- $F := \sum_n 2^{-n} f_{a_n} \in L^1$
- $f \ast F = \sum_n f \ast 2^{-n} f_{a_n}$ pointwise, by monotone convergence.
- Moreover, the series converges absolutely in $L^\infty$ norm, thus the limit is continuous outside $\{a_n\}$, and discontinuous at $\{a_n\}$.
Thus we have constructed a convolution which is everywhere defined and discontinuous at a dense countable set.
Now a general remark. Convolution, whenever it is defined, is a limit of continuous functions (by approximation of one of your $L^1$ functions by bounded ones), so it must be of Baire class one. In particular, if it's locally bounded then it must be almost everywhere continuous in the sense of category, so you cannot do much better than in my example.
Now assume that you have a convolution of $f,g \ge 0$. $f \ast g$ is still of Baire class one, but now with values in the extended real line $[0,\infty]$. So if $f \ast g$ is locally unbounded at every point of some interval, then since it must be Baire almost everywhere continuous (again, in the sense of the topology of $[0,\infty]$), there will be a point on that interval where $f \ast g$ is actually infinite, which probably counts as "nonexistence of convolution at that point".
The general case reduces to the case of positive functions: if your convolution exists everywhere, then it would also exist everywhere if we replaced $f,g$ by $|f|,|g|$.