Let $f,g\in L^1(\mathbb{R};\mathbb{R}).$ Define the convolution $f*g:\mathbb{R} \rightarrow \mathbb{R},$ by $(f*g)(x)=\int\limits_{y\in \mathbb{R}}f(x-y)g(y)dy.$
Note that Fubini's theorem and translation invariance of Lebesgue measure implies that \begin{eqnarray} \int\limits_{R}\int\limits_{R} |f(x-y)g(y)|dydx = ||f||_{L^1(\mathbb{R})}||g||_{L^1(\mathbb{R})}. \end{eqnarray} Now, again invoking Fubini's theorem we get, $(f*g)(x)=\int\limits_{y\in \mathbb{R}}f(x-y)g(y)dy < \infty$ for a.e. $x\in \mathbb{R}.$ Furthermore, $f*g \in L^1(\mathbb{R}).$
Question: Do we have $|(f*g)(x)|< \infty $ for all $x\in \mathbb{R}?$ If yes, how to prove it, if not, what are the counter examples?
Let $f,g:\Bbb{R}\to\Bbb{R}$ be defined as \begin{align} f(x)=g(x)= \begin{cases} \frac{1}{\sqrt{|x|}}&\text{if $0<|x|<1$}\\ 0&\text{else} \end{cases} \end{align} Then, $f,g\in L^1(\Bbb{R})$, so $f*g$ is well-defined a.e. However, the convolution at the origin is \begin{align} (f*g)(0)&:=\int_{\Bbb{R}}f(0-y)g(y)\,dy\\ &=\int_{0<|y|<1}\frac{1}{\sqrt{|-y|}}\frac{1}{\sqrt{|y|}}\,dy\\ &=2\int_0^1\frac{1}{y}\,dy\\ &=\infty. \end{align} So, no the convolution need not be finite at every point. (The idea behind this is that the pointwise product of $L^1$ functions need not be $L^1$ again).