I just read a related question which showed that probability density functions do not always have a limit of zero as a variable approaches infinity because they don't always have a limit at infinity.
Given that a probability density function has a limit at infinity, is that limit always zero? If so, is there a rigorous and formal proof that shows this property?
Update:
I just found another related question, which is a generalization of what I'm asking to the multivariate case. In a mathematical sense, this makes my question a duplicate.
It could be proven by contradiction.
A pdf function must satisfy the following $$f(x)\geq0\quad\forall x\in\Bbb R\tag1$$ $$\int_{-\infty}^{+\infty}f(x)\mathrm dx=1\tag2$$ Suppose $f(x)\geq0$ with $$\lim_{x\to+\infty}f(x)=a>0$$ By definition of the limit, for all $\epsilon\geq0$, there exist an $N\in\Bbb R$ such that if $x>N$ then $\lvert f(x)-a\rvert<\epsilon$
Let's take $\epsilon=\frac a2>0$, then there exist $N_0$ such that if $x>N_0$ then $$\lvert f(x)-a\rvert<\frac a2$$ which implies $$\frac a2<f(x)<\frac {3a}2$$ Now \begin{align} \int_{-\infty}^{+\infty}f(x)\mathrm dx&\geq \int_{N_0}^{+\infty}f(x)\mathrm dx\\&\geq\int_{N_0}^{+\infty}\frac a2\mathrm dx\\&\geq+\infty \end{align} This contradict our second condition. If a function has a limit at infinity that is not $0$, then the function is not a probability densité function.