= Solution
The <independence of random variables> means that fixing $Y=y$ leaves the <conditional probability density> of $X$ equal to $f$. Given that value of $Y$, $Z$ is the translate $X+y$. Hence
$$
\boxed{P(z\leq Z\leq z+dz\mid Y=y)=\int_{z-y}^{z+dz-y}f(x)\,dx}.
$$
At a continuity point of $f$, the expression is $f(z-y)\,dz+o(dz)$. Since $P(Y=y)=0$, this conditioning uses a <conditional probability density>, rather than division by the probability of the single event $Y=y$.
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