Does $f(x,y) \space \epsilon \space L^1(\Bbb{R}^2)$ imply that that $f_x(y) \space \epsilon \space L^1(\Bbb{R})$?

Question

My question is as follows: If $f(x,y)$ is a function on $\Bbb{R}^2$ (real or complex valued) and $f_x$ is defined as the function on $\Bbb{R}$ such that $f_x(y) = f(x,y)$ for all $y \space \epsilon \space \Bbb{R} $ then does $f \space \epsilon \space L^1(\Bbb{R}^2)$ imply that $ f_x \space \epsilon \space L^1(\Bbb{R})$ for every $x \space \epsilon \space \Bbb{R}$? I know the conclusion holds for almost all $x$ since $$ \Vert f \Vert_1 = \int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty} \vert f_x(y)\vert dy \lt \infty $$ which implies that $\int_{-\infty}^{\infty} \vert f_x(y)\vert dy \lt \infty $ almost everywhere (i.e. for almost all $x$ in $\Bbb{R}$). I'm just wondering if there is a way to prove the stronger conclusion that $\int_{-\infty}^{\infty} \vert f_x(y)\vert dy \lt \infty $ for all $x$ in $\Bbb{R}$. If the given assumptions on $f$ don't guarantee the stronger conclusion, are there any additional assumptions we can make about $f$ which would? Thanks for any help.