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Let $\mu$ be a probability measure on $X \subseteq \mathbb{R}$.

Consider a polynomial function $p_d: \mathbb{R} \rightarrow \mathbb{R}$ of degree $d \in \mathbb{Z}^+$.

I would like to know if the following is true.

$ \int_X |p_d(x)| \mu(dx) < \infty \ \Leftrightarrow \ \int_X |x^d| \mu(dx) < \infty $

In the case the result does work, I'm wondering how it could be extended to the multi-dimensional case $X \subseteq \mathbb{R}^m$, $m \in \mathbb{Z}_{> 1}$.

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    @Adam I thought you got that already settled. Sorry. Interesting question.2012-03-22

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I'm wondering if the following argument could work.

By the Limit Comparison Test we have that

$ \lim_{x \rightarrow \infty} \frac{ | \sum_{i=0}^d a_i x^i |}{ |x^d| } = a_d \in (0, \infty)$

so $| \sum_{i=0}^d a_i x^i |$ is integrable iff $|x^d|$ is integrable.

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    this is nice question!2012-03-23
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Hint: By Hölder's inequality, if $1 \le j \le d$, $\int x^j \ \mu(dx) \le \left(\int 1^q \ \mu(dx)\right)^{1/q} \left(\int |x^d| \ \mu(dx)\right)^{1/p}$ where $p = d/j$ and $1/p + 1/q = 1$.

The multidimensional case is trickier, because e.g. if $\mu$ is concentrated on $\{(x,y) \in {\mathbb R}^2: y = 0\}$, $\int |x^j y| \ d\mu$ will converge for all $j$ while $\int |x^j| \ d\mu$ might diverge for all $j \ge 1$.

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    Well, $Y$ MUST be bounded so there is no need to take it not-compact as we are taking $\int_Y \cdot$2012-03-22