FALSE: if $f,g\in L^1({\mathbb{R}}^n)$, then $f\ast g(x)$ is defined for every $x$

The function $x\mapsto 1\mathrm{/}(\mathrm{\mid }x\mathrm{\mid }(\log (e\mathrm{/}\mathrm{\mid }x\mathrm{\mid }){)}^2)$ is integrable near $0$ by the substitution $u=\log (e\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$, whose tail is a constant multiple of ${\int }^{\mathrm{\infty }}{u}^{-2}du$. Hence $f,g\in L^1(\mathbb{R})$.

At $x=0$ one has [step 1.1, algebra] ${\int }_{\mathbb{R}}\mathrm{\mid }f(-y)g(y)\mathrm{\mid }dy={\int }_{-1\mathrm{/}2}^{1\mathrm{/}2}\frac{dy}{\mathrm{\mid }y{\mathrm{\mid }}^2(\log (e\mathrm{/}\mathrm{\mid }y\mathrm{\mid }){)}^4}\mathrm{.}$ The single point $y=0$ is irrelevant to Lebesgue integrability, so it is enough to inspect the punctured interval $(0,1\mathrm{/}2)$. There the substitution $u=\log (e\mathrm{/}y)$ gives $y=e^{1-u}$ and $dy=-e^{1-u}du$, hence ${\int }_0^{1\mathrm{/}2}\frac{dy}{y^2(\log (e\mathrm{/}y){)}^4}={\int }_{\log (2e)}^{\mathrm{\infty }}\frac{e^{u-1}}{u^4}du=\mathrm{\infty }\mathrm{.}$ Thus $(f\ast g)(0)$ is not defined as an absolutely convergent Lebesgue integral.

Therefore the convolution of two $L^1$ functions need not be defined at every point.