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.
 A: 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 $.
A: Suppose the values $X$ takes are $(x_n),$ each of which with positive probability. Then, as you already know, the density of $Y$ is $f(y) = \sum\limits_n f(x_n, y)$ where $f(x_n, \cdot)$ satisfies that for each Borel set $\mathrm{B},$
$$
\int\limits_\mathrm{B} f(x_n, y) dy = \mathbf{P}(X = x_n, Y \in \mathrm{B}).
$$
You assume each $f(x_n, \cdot)$ is continuous (note that $f(y \mid x_n) \propto f(x_n, y)$ so continuity is not affected by the proportionality constant). You want to prove that $f$ is also continuous. This will be true under general conditions.
Continuity signifies that for every converging sequence $y_k \to y,$ we have $f(y_k) \to f(y).$ Now, for each $y_k,$ define $u_k = f(\cdot, y_k).$ We know, from assumption, that $u_k \to u$ simply (pointwise), where $u = f(\cdot, y).$ If $u_k \leq v$ where $v$ is integrable (relative to the counting measure), then Lebesgue dominated convergen theorem will guarantee that
$$
f(y_k) =\int u_k dc \to \int u dc = f(y),
$$ where
$$
\int \varphi dc = \sum\limits_n \varphi(x_n)
$$
(and $c$ is the counting measure).
In sum, if $f(x_n, y) \leq a_n$ (uniformly in $y$) and $\sum a_n < \infty,$ then $f$ is continuous.
