Are $L1$ functions with a.e. finite support a.e. equal to a continuous function?

Question

I was wondering about this: Let $f \in L^1(\Omega)$ and $\Omega\subset \mathbb{R}^n$ be compact, then $f$ is the $L^1$ limit of continuous functions with support in $\Omega$. Egorov's theorem tells us now that this convergence is uniform a.e. . Since $\overline{\Omega}$ is compact, the continuous functions form a Banach space and the limit function (of a subsequence) which is $f$ a.e. is continuous, too. Thus, I would get that any such $L^1$ function is a.e. a continuous function, is this right? Does not seem right, as $L^1$ functions do not need to be bounded. So where is the mistake?