Let $X,Y$ be independent real random variables, with $Y$ having a pdf. The following statement is true if $X$ also has a pdf (it is just the standard convolution integral written as an expectation):
$$f_{X+Y}(z) = E[f_Y(z-X)]$$
However, intuitively speaking, it does not seem like $X$ has to have a pdf for this to hold. Is it true? If so, can you give a reference or a proof?
The statement $$f_{X+Y}(z) = E[f_Y(z-X)]$$ for almost all $z$ is equivalent to $$\int _A f_{X+Y}(z)dz = \int _A E[f_Y(z-X)]dz$$ fro al measurable sets $A$. LHS is $$P( X+Y \in A).$$ By Fubini's Theorem RHS is $$E\left[\int_A f_Y(z-X)\right]dz$$ which is nothing but $$E[P(Y \in A-X)].$$ The desired equality now follows by independence of $X$ and $Y$.