Is the PDF $f(y)$ continuous if conditional PDFs $f(y\mid X=x)$, $x\in \mathbb{Z}$ are continuous? $X\in \mathbb{Z}$ and $Y\in \mathbb{R}$ are random variables. Suppose that the conditional PDF $f(y\mid X=x)$ is continuous for all $x\in \mathbb{Z}$. Is the PDF $f(y)$ continuous?
I'm aware that if $X$ can take only finitely many values then $f(y)$ is continuous since $f(y)=\sum_{x}\Pr(X=x)f(y\mid x)$ and the fact that the sum of finitely many continuous functions is also continuous. But I cannot figure out if it holds when $X$ can take infinitely many values.
Thanks in advance.

Let $\mu_n:=f_{X,Y}(\cdot ,n)\cdot \lambda =f_{Y|X}(\cdot |n)\Pr [X=n]\cdot \lambda $, then each $\mu_n\ll \lambda $, where $\lambda $ is the Lebesgue measure in $\mathbb{R}$, and you have that
$$
P_Y=f_Y \cdot \lambda =\sum_{k\in \mathbb{Z}}\mu_k=\sum_{k\in \mathbb{Z}}f_{X,Y}(\cdot ,k)\cdot \lambda \tag1
$$
where $P_Y:=P\circ Y^{-1}$ is the measure induced by $Y$ in $\mathbb{R}$. Then you want to know under what circumstances the continuity of the $f_{X,Y}(\cdot ,n)$ ensures the existence of a continuous version of $f_Y$. One of these circumstances is when $g_n:=\sum_{k=-n}^nf_{X,Y}(\cdot ,k)$ converges locally uniformly, that is, that for every $y\in \mathbb{R}$ there is a $\delta >0$ such that
$$
\lim_{n\to \infty }\sup_{s\in B(y,\delta )}|f_Y(s)-g_n(s)|=\lim_{n\to\infty}\sup_{s\in B(y,\delta )}\left(\sum_{|k|> n}f_{X,Y}(s ,k)\right)=0\tag2
$$
A condition that ensures (2) is
$$
\sum_{k\in \mathbb{Z}}\sup_{s\in B(y,\delta )}f_{X,Y}(s,k)<\infty \tag3
$$
And a much stronger condition, that ensures (3), is the existence of a sequence of constants $\{M_n\}_{n\in\mathbb{Z}}$ such that $\sup_{y\in \mathbb{R}}f_{X,Y}(y,n)\leqslant M_n$ for all $n\in \mathbb{Z}$ and that $\sum_{k\in \mathbb{Z}}M_k<\infty $.
