Limit of Multivariate Probability Density Function as one or more or all variables approach positive or negative infinity

Could someone point out a good source or provide justification as to whether the limit of a multivariate probability density function, as one or more or all variables tends to positive or negative infinity, goes to zero.

$$\lim_{x_{1}\to\pm\infty}f(x_{1},x_{2},...,x_{n})=0$$

$$\lim_{x_{1},x_{2}\to\pm\infty}f(x_{1},x_{2},...,x_{n})=0$$

$$\lim_{x_{1},x_{2},...,x_{n}\to\pm\infty}f(x_{1},x_{2},...,x_{n})=0$$

I have found this related question for the univariate case: Limit of probability density function as random variable approaches +/- infinity

Solution 1:

There is in general no guarantee, that the limits exists. If the limits exist, they all have to be zero.

Proof: Assume that $$\lim_{x_{1}\to\infty}f(x_{1},x_{2},...,x_{n})=a>0$$ then by definition of limit there exist $0<\epsilon<a$ and $\hat x_1>0$ such that $f(x_{1},x_{2},...,x_{n})>\epsilon ~ \forall x_1 > \hat x_1$ and any $x_2,\ldots,x_n$. But then \begin{align} \int_{\mathbb R^n} f(x_{1},x_{2},...,x_{n}) d(x_{1},x_{2},...,x_{n}) & \geq \int_{\hat x_1}^{\infty} \int_{\mathbb R^{n-1} } f(x_{1},x_{2},...,x_{n}) d(x_{2},...,x_{n})d(x_{1} ) \\ & \geq \epsilon . \infty = \infty \end{align} so the you see that measure of the whole space is not $1$ and that is contradiction with $f$ beeing density. Therefore $$\lim_{x_{1}\to\infty}f(x_{1},x_{2},...,x_{n})=0.$$ You can do the others pretty same way.

But as I have stated, there is no guarantee that the limits exist. You can for example take any continuous 1 dimensional density function $f$ and define \begin{align} g(x) & = f(x) &\text{if } x\neq n, n\in \mathbb N, \\ g(x) & = 1 &\text{if }x = n, n\in \mathbb N. \end{align} Then $g$ is density function with countable many incontinuities, but the limit $\lim_{x\rightarrow \infty}g(x)$ does not exist.