Everywhere existence of marginals

Question

Let $f\in L^1(\mathbb{R}^2)$ be a (joint) probability density function which satisfies $f(x,y)>0$ for all $(x,y)\in \mathbb{R^2}$.

What is a necessary and sufficient condition under which the marginal $$ f^y (y)\equiv\int\limits_{\mathbb{R}} f(x,y) dx $$ exists for all $y\in \mathbb{R}$?

Comments:

• because of Fubini, I'm pretty sure $f^y$ exists almost-everywhere in $y$. However, I don't see how continuity or smoothness of $f$ would guarantee existence everywhere.

• I believe that on a compact domain (e.g., $f:[0,1]^2 \to \mathbb{R}$), Lipschitz/Holder continuity of $f$ is sufficient to guarantee existence of the marginals everywhere.

** cross posted from MSE, after a week+ with no comments

Answer 1

Assumptions such as smoothness will not guarantee this. Consider $f(x,y)=\varphi((1+y^2)x)$ with a $\varphi\in C^{\infty}(\mathbb R)$ with $\varphi> 0$, $\varphi(0)=1$, $\int\varphi = 1/\pi$. Then $f\in C^{\infty}(\mathbb R^2)$, $$ \int dy\int dx f(x,y)= \frac{1}{\pi}\int \frac{dy}{1+y^2} =1 , $$ so $f$ is a density, but $f(0,y)=1\notin L^1$.