Let $f,g:\mathbb R^n\to\mathbb C$ be two Lebesgue measurable functions. Thus $(x,y)\mapsto f(x-y)g(y)$ is also a mesaurable function. To be rigorous, we define let $$f\ast g(x)=\int_{\mathbb R^n}f(x-y)g(y)dy.$$
We say the convolution is defined, if for almost every $x\in\mathbb R^n$, $f(x-\cdot)g$ is a Lebesgue integrable function, hence $f\ast g$ is an almost everywhere finite measurable function.
If we do not give extra restrictions, the convolution will not be associative.
Take $f=\mathbf 1_{\mathbb R_+}-\mathbf1_{\mathbb R_-}$, $g=\mathbf1_{(0,1)}-\mathbf1_{(0,-1)}$ and $h=\mathbf1_{\mathbb R}$ then $f\ast g(x)=\max(0,2-2|x|)$. We get $(f\ast g)\ast h=2$ but $f\ast (g\ast h)=0$.
My question is, what extra (reasonable) assumption do we need to set to make sure the associativity still holds? Basically to make, say Folland's Real Analysis Proposition 8.6(b), works smoothly.
Here I don't want, say, assuming $f,g,h\in L^1$. I also don't want to go into the case of distributions.
Note that in the example above $f,h\in L^\infty$ and $f\ast h$ is not defined (as a measurable function).
My sub-question is, do we have $(f\ast g)\ast h=f\ast(g\ast h)$ if $f\ast h$ is also defined?
Update: a way to cheat is to assume the double integral exists in the beginning and thus Fubini applies automatically: assuming the "triple convolution" is defined, i.e. for a.e. $x$ $$f\ast g\ast h(x)=\iint_{\mathbb R^{2n}}f(x-y-z)g(y)h(z)dydz$$ with the integrand being either nonnegative or $L^1$.
But this is not that what I want, since it goes beyond the binary operation.