Proof that convolution is associative using Fubini's theorem. The question is to prove that $((f *g)*h) (x) = (f*(g*h)) (x)\\$
So far, I have the following:
\begin{align*}
    ((f \ast g) \ast h) (x) &= \int^x_0 (f \ast g)(t) h (x-t) dt\\
    &=\int^x_0 \left[\int^t_0 f(h)g(t-h) dh\right] h(x-t) dt\\
    &= \int^t_0 \int^x_0 g(t-h) h(x-t) f(h) dt \quad dh\\
    &= \int^t_0 \int^{\alpha = x - h}_{\alpha = -h} g(\alpha) h(x - h - \alpha) f(h) d\alpha \quad dh, \quad \text{where } \alpha = t - h \Rightarrow d\alpha = dt\\
\end{align*}
I'm stuck here because somehow, I need:
$$\int^{\alpha = x - h}_{\alpha = 0} \text{instead of} \int^{\alpha = x - h}_{\alpha = -h}$$
I'm aware that there are different definitions of convolution that go from $(-\infty, +\infty)$ that might simply this, but I'm wondering how to do for the above bounds.
Is there something I did incorrectly? It's been a while since I've learned Fubini's theorem, so I may be missing up the bounds somehow. Thank you!
 A: We obtain
\begin{align*}
\color{blue}{((f\ast g)\ast h)(x)}&=\int_0^x(f\ast g)(t)h(x-t)\,dt\\
&=\int_{t=0}^x\left(\int_{u=0}^tf(u)g(t-u)\,du\right)h(x-t)\,dt\tag{1}\\
&=\int\!\!\!\int_{0\leq u\leq t\leq x}f(u)g(t-u)h(x-t)\,du\,dt\tag{2}\\
&=\int_{u=0}^x\int_{t=u}^xf(u)g(t-u)h(x-t)\,dt\,du\tag{3}\\
&=\int_{u=0}^x\int_{t=0}^{x-u}f(u)g(t)h(x-t-u)\,dt\,du\tag{4}\\
&=\int_{u=0}^xf(u)\left(\int_{t=0}^{x-u}g(t)h(x-u-t)\,dt\right)\,du\\
&=\int_{u=0}^xf(u)(g\ast h)(x-u)\,du\\
&\,\,\color{blue}{=(f\ast (g\ast h))(x)}
\end{align*}
Comment:

*

*In (1) we introduce a new integration variable $u$. It is crucial to use different names for different objects to not mix up function names ($h$) with variable names.


*In (2) we write the region of integration conveniently as preparation for changing the order of integration in the next step.


*In (3) we change the order of integration.


*In (4) we substitute $t\to t+u$ to start the region of integration with $t=0$. In this substitution $u$ is to be seen as constant, so that we have again $dt$.
