Convolution doubt

Just notice that for $\tau<0$ you have $g(\tau)=0$ while for $\tau>t$ you have $f(t-\tau)=0$. In both cases $f(t-\tau)g(\tau) = 0$. Hence $$ \int_{-\infty}^{\infty} f(t-\tau)g(\tau)\, d\tau = \int_0^t f(t-\tau)g(\tau)\, d\tau. $$

Hint: Use indicator functions for these kind of problems.

Write $f(t)$ as $f(t)1_{\{t \geq 0\}}$ and $g(t)$ as $g(t)1_{\{t \geq 0\}}$. Then the integrand becomes $$ f(t-\tau)g(\tau)1_{\{\tau \leq t,\tau\geq 0\} } = f(t-\tau)g(\tau)1_{\{0 \leq \tau \leq t\} }$$

$\blacksquare$