= Solution
Use the <law of total probability> to average the conditional interval probability over $Y$:
$$
P(z\leq Z\leq z+dz)=\int_{-\infty}^{\infty}g(y)\int_z^{z+dz}f(v-y)\,dv\,dy.
$$
The integrands are nonnegative, so <Tonelli's theorem> permits interchanging the <integrals>. Thus
$$
P(z\leq Z\leq z+dz)=\int_z^{z+dz}\left[\int_{-\infty}^{\infty}g(y)f(v-y)\,dy\right]dv.
$$
Reading off the <probability density> gives the <convolution of probability densities>:
$$
\boxed{h(z)=\int_{-\infty}^{\infty}g(y)f(z-y)\,dy}.
$$
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