Convolution Integral only defined Almost Everywhere? My professor has mentioned in class that if we have two $2\pi$-periodic functions $f$ and $g$ that are both in $L_1(\mathbb{T})$, then
$$(f*g)(t) := \frac{1}{2\pi}\int_{-\pi}^{\pi}f(s-t)g(s)ds$$ is only guaranteed to be defined for almost every $t$.
I thought I followed it at the time, but now that I'm sitting at home looking at the remark, I just don't see it.  If we took $f_{1}$ and $g_{1}$ such that $f=f_{1}$ almost everywhere and $g = g_{1}$ almost everywhere, wouldnt we have
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}f_{1}(s-t)g_{1}(s)ds = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(s-t)g(s)ds$$  ?
...  or was I missing his point completely?  I thought it had something to do with the fact that functions in the $L_{1}(\mathbb{T})$ are actually equivalence classes.
 A: The only way to show that there exists some $t$ such that $(f*g)(t)$ exists is often to consider the integral of the function $|f*g|$. As you know, Fubini tells you that the result is bounded by the product of the integrals of $|f|$ and $|g|$, hence that it is finite. This proves that $f*g$ is integrable, hence finite almost everywhere, but cannot give you the finiteness of $f*g$ at any given point.
By the way, the natural objects here are not functions but equivalence classes of functions coinciding almost everywhere, and one often defines the convolution of these objects rather than the convolution of functions. Then the result is, by nature, such a class and the assertion, for example, that the convolution of the functions $f$ and $g$ is a given function $h$ often simply means that $f*g$ is in the class of $h$, in other words, that $f*g=h$ almost everywhere and it could be rephrased as the fact that $\tilde f*\tilde g$ is in the class of $h$ for every $\tilde f$ in the class of $f$ and every $\tilde g$ in the class of $g$. Sometimes the space of integrable functions is written $\mathcal L_1$ and the space of classes $L_1$.
A: Here is an example. Function 
$$
f(x)=\sum_{n=1}^\infty\frac{\cos n x}{\sqrt n}
$$
is continuous on $\mathbb{T}$ with the exception of $x=0$. Since $\lim_{x\to0}f(x)|x|^{1/2}=\sqrt{{\pi }/{2}}\ \ $ we have $|f(x)|\le C|x|^{-1/2}$ and  $f\in L_{1}(\mathbb{T})$. The convolution $f$ with itself  is 
$$
f*f(x)=\sum_{n=1}^\infty\frac{\cos n x}n=-\log |x|+\frac{x^2}{24}+O(x^3),\quad x\to0,
$$
so it is not defined at the origin.
