I have a function $$ f(x):\mathbb{R}^n->\mathbb{R} $$ which is non negative on the whole space.
Can I say the following statment: $$ \forall x_0 \in \mathbb{R}^n \ , \ f(x_0) \leq \int_{\mathbb{R}^n}f(x) \ dx $$ I think its true by intuation: The function is non negative and so integrating is like accumulating mass.
But I couldn't find a proof anywhere online.
Your question can be answered by many counter-examples from probability; just because the probability integrates to $1$ doesn't mean that the density function must be less than one. Consider for example the pdf of the $\chi^2_1$ distribution:
$$f(x)=\begin{cases}0&x\le0\\\frac{\exp\left(-\frac{1}{2}x\right)}{\sqrt{2\pi x}}&x\gt 0\end{cases}$$
We have that:
$$\int_{\Bbb R}f(x)\,\mathrm{d}x=1$$
But $f\to+\infty$ as $x\to 0^+$, and attains values greater than $1$; e.g., $f(10^{-4})\approx40\gt1$.
With regards to intuition, I'll say that a very thin slice is integrated to a very small value - our function here blows up to $\infty$, yet in integration the region where it is large is so slim that the integral is small...