I need to find if the following is true: Let $N>1$, given a function $f \in L^1(\mathbb{R}^N), f > 0$,I want to know if it is true that\begin{equation} \int_{\mathbb{R}^{N-1}} f(x, y)dx \leq \int_{-\infty}^{\infty}\int_{\mathbb{R}^{N-1}} f(x, y)dxdy \; \text{ for almost all } y \in \mathbb{R}.\end{equation}For me, this has a lot of sense at least when using continuous functions, but I can't find a formal proof of it.
Firstly I define another function $F:\mathbb{R} \to \mathbb{R}$ by fixing one of the variables and integrating the others, like:\begin{equation} F(y) = \int_{\mathbb{R}^{N-1}} f(x, y)dx > 0.\end{equation}
Integrating this function, $\displaystyle \int_{-\infty}^{\infty}F(y)dy = \int_{-\infty}^{\infty}\int_{\mathbb{R}^{N-1}} f(x, y)dxdy = \int_{\mathbb{R}^N}f(x)dx = \|f\|_1< \infty$ by Fubini's Theorem, and therefore $F \in L^1(\mathbb{R})$ and, in particular, $F < \infty$ a.e., so the function $F$ is at least well-defined (up to a set of measure zero). This proofs that at least both integrals in my question are defined and then questioning the inequality is not meaningless.
When thinking intuitively, it has full sense that "integrating along one line" results in a smaller value than doing it over every possible lines in the last coordinate, but I cannot come with a proof. In fact, when thinking in this as a $1$ dimensional function ($N=1$), the left hand integral could be though as taking the value of the function at a point, and it is fairly easy to find functions such that $\|f\|_\infty < \int_{-\infty}^{\infty}f(x)dx$, but I don't know if this is just a limiting case.
Edit 1: Fixed a typo in an integral