Convolution on the half-line

Remarks

The integral converges for almost every $t$, and $f * g$ is again locally integrable. If $f$ and $g$ are continuous, then $f * g$ is defined everywhere and continuous, with $(f * g)(0) = 0$. Extending $f$ and $g$ by $0$ on $(-\infty, 0)$ turns this into the convolution on the line.

Example: for integers $m, n \ge 0$, the functions $t^m$ and $t^n$ give $$(t^m * t^n)(t) = \int_0^t \tau^m (t - \tau)^n \, d\tau = \frac{m! \, n!}{(m + n + 1)!} \, t^{m + n + 1},$$ by the Beta integral; for $m = n = 0$ this is $1 * 1 = t$.

Non-example: $f(t) = 1/t$ for $t > 0$ is not integrable on $[0, 1]$, so it is not locally integrable on $[0, \infty)$. Indeed $(f * 1)(t) = \int_0^t d\tau / \tau$ diverges for every $t > 0$.