Proof of associativity of convolution I intend to prove the associativity of convolution but failed after several trials, i.e.
$(f \ast g) \ast h = f \ast (g \ast h)$
where $(f \ast g)(t) = \int^{t}_{0}f(s)g(t-s)ds $
There are a number of proves considering $(f \ast g)(t) = \int^{\infty}_{-\infty}f(s)g(t-s)ds $. Those did not help since I have a different definition.
Would anyone be able to show the proof here?
Thanks.
 A: Using the following convolution of $f$ and $g$
\begin{align*}
(f\star g)(t)=\int_0^tf(s)g(t-s)ds
\end{align*}

we obtain
\begin{align*}
\color{blue}{((f\star g)\star h)(t)}&=\int_0^t(f\star g)(s)h(t-s)\,ds\\
&=\int_{s=0}^t\left(\int_{u=0}^sf(u)g(s-u)\,du\right)h(t-s)\,ds\\
&=\int\!\!\!\int_{0\leq u \leq s\leq t}f(u)g(s-u)h(t-s)\,du\,ds\\
&=\int_{u=0}^t\int_{s=u}^tf(u)g(s-u)h(t-s)\,ds\,du\\
&=\int_{u=0}^tf(u)\left(\int_{s=0}^{t-u}g(s)h(t-u-s)\,ds\right)\,du\\
&=\int_{u=0}^tf(u)(g\star h)(t-u)\,du\\
&\,\,\color{blue}{=(f\star (g\star h))(t)}
\end{align*}

Note: This corresponds to the discrete case which could be somewhat easier to follow.
\begin{align*}
(f\star g)(n)=\sum_{k=0}^nf(k)g(n-k)
\end{align*}
Here we get
\begin{align*}
((f\star g)\star h)(n)&=\sum_{k=0}^n(f\star g)(k)h(n-k)\\
&=\sum_{k=0}^n\left(\sum_{l=0}^kf(l)g(k-l)\right)h(n-k)\\
&=\sum_{0\leq l \leq k \leq n}f(l)g(k-l)h(n-k)\\
&=\sum_{l=0}^n\sum_{k=l}^nf(l)g(k-l)h(n-k)\\
&=\sum_{l=0}^nf(l)\left(\sum_{k=0}^{n-l}g(k)h(n-k-l)\right)\\
&=\sum_{l=0}^nf(l)(g\star h)(n-l)\\
&=(f\star(g\star h))(n)
\end{align*}
