Integrability given a growth condition Let $\mathbb{P}$ be a probability distribution over $\mathbb{R}^{p}$ and let $f:\mathbb{R}^{p}\to \mathbb{R}$ be a continuous map. Assume that $f$ satisfies the following growth condition
$$\lim_{R\to+\infty}\mathbb{P}(\{\|\mathbf{x}\|>R\})\cdot\sup_{\|\mathbf{x}\|\le R}|f(\mathbf{x})|=0.$$
Is it true that $f$ is integrable with respect to $\mathbb{P}$, i.e. $\int_{\mathbb{R}^{p}}|f(\mathbf{x})|d\mathbb{P}<+\infty$?
What if $\mathbb{P}$ also admits finite moment of a certain order, that is $\int_{\mathbb{R}^{p}}\|\mathbf{x}\|^{q}d\mathbb{P}<+\infty$ for some $q$?
 A: In general the answer is no.
Let $p=1$. Let $f(x) := \dfrac{|x|}{\log(1+|x|)}$. Then, $f(x)$ is increasing with respect to $|x|$.
Let $\mathbb P$ be the Cauchy distribution, that is, $\mathbb{P}(dx) = \dfrac{1}{\pi (x^2 + 1)} dx$. Then, $\mathbb{P}(|x| > R) \le \dfrac{1}{R}$, and hence,
$\mathbb{P}(|x| > R) \sup_{|x| \le R} |f(x)| \le \dfrac{1}{\log(1+R)} \to 0, R \to \infty$.
On the other hand,
$$ \int_1^{\infty} |f(x)| \mathbb{P}(dx) = \frac{1}{\pi} \int_1^{\infty}  \frac{x}{\log(1+x)}\frac{1}{ (x^2 + 1)} dx \ge \frac{1}{2\pi} \int_1^{\infty} \frac{1}{x \log(1+x)} dx $$
$$ \ge \frac{1}{2\pi} \int_2^{\infty} \frac{1}{x \log x} dx = +\infty, $$
where in the last inequality we used $\int \frac{1}{x \log x} dx = \log \log x$.

We can also construct an example which satisfies $\int_{\mathbb R} |x| \mathbb{P}(dx) < +\infty$. Let $\mathbb{P}(dx) = C \cdot \dfrac{1}{(x^2+1)(1+\log(1+x^2)^2)}dx$, where $C$ is a normalizing constant.
Then, $\int_{\mathbb R} |x| \mathbb{P}(dx) < +\infty$  because $\displaystyle\int_2^{\infty} \frac{1}{x(\log x)^2} dx < +\infty$.
(We can show this by the change-of-variable $y = \log x$.)
Furthermore, there exists a constant $C$ such that
$$\mathbb{P}(|x| > R) \le C \int_R^{\infty} \frac{1}{x^2 (\log x)^2} dx \le \frac{C}{R (\log R)^2}$$
for every $R > 1$.
Let $f(x) := |x|(1+\log(1+x^2))$. Then, $\sup_{|x| \le R} |f(x)| \le R(1+\log(1+R^2))$ and hence $\mathbb{P}(|x| > R) \sup_{|x| \le R} |f(x)| \to 0, R \to \infty$.
Since $\displaystyle\int_2^{\infty} \frac{1}{x(\log x)} dx = +\infty$, we have that $\displaystyle \int_{\mathbb R} |f(x)| \mathbb{P}(dx) = +\infty$.
