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

Prove that $|| a + B x ||^d$ is integrable iff $||x||^d$ is integrable:

$$ \int_{X} || a + B x ||^d \mu(dx) < \infty \ \Leftrightarrow \ \int_{X} ||x ||^d \mu(dx) < \infty $$

for any $a \in \mathbb{R}^n$, $B \in \mathbb{R}^{n \times m}$, $d \in \mathbb{R}_{> 0}$

Extend the result to any measurable (non constant) function $f(x)$:

$$ \int_{X} || a + B f(x) ||^d \mu(dx) < \infty \ \Leftrightarrow \ \int_{X} || f(x) ||^d \mu(dx) < \infty $$

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The "for all $B$" must be part of the left side. Otherwise e.g. the $B=0$ case would not imply anything useful. –  Robert Israel Mar 16 '12 at 1:49
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2 Answers 2

Is this homework? Hint: $\|a + B f\| \le \|a\| + \|B\| \|f\|$, and for $s, t \ge 0$, $(s+t)^d \le 2^d (s^d + t^d)$.

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Yes, it is. I'm able to prove the part $\Leftarrow$ as $||a+Bf|| \leq ||a|| + ||B||||f||$. I'm trying the other implication. –  Adam Mar 16 '12 at 1:00
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Is $\mu$ supposed to be a continuous probability measure? Else, the result is not true. For example, suppose $\mu$ is the measure on $\mathbf{R}^2$ defined through $\mu((i,0))=C/i^2$ for all integers $i \geq 1$ and $\mu(x)=0$ for all other points $x \in \mathbf{R^2}$. Here $C$ is a normalizing constant.

Then picking $B$ to be the matrix which projects onto the $y$ axis, we have that $\int_{\mathbf{R^2}} ||Bx||^2 = \int_{\mathbf{R^2}} 0 = 0$ but $\int_{\mathbf{R^2}} x^2 = \sum_{i=1}^{\infty} i^2 (C/i^2) = \infty$.

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