PDF of X+Y when the PDF of X exists

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Let's suppose that $Y$ is the normal distribution and that $X$ is another random variable whose density function may or may not exist.

Does it follow that $Y+X$ has a density function?

I am reading Resnick's book "A probability path" and this comment was made in passing which confused me. It's an essential step in the proof of the Fourier inversion for characteristic functions.

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Yes: as soon as at least one of the random variables $X$ and $Y$ has a density and if $X$ and $Y$ are independent, the sum $X+Y$ has a density.

To understand the magic, assume that $(X,Y)$ is independent, that $Y$ has density $f$ and that $X$ is purely discrete with $P(X=x_n)=p_n$ for every $n$, then $X+Y$ has density $g$ with $$g(x)=\sum_np_n\,f(x-x_n).$$ Without independence, things can go awry: consider $Y$ with a density and $X=-Y$ then $X+Y$ has no density.