# Inequality in measure theory

I'm having trouble proving the following statement: let $\Omega \subset \mathbb{R}^n$ be an open set, let $h: \Omega \mapsto \mathbb{R}$ and $\rho : \Omega \mapsto [0,+\infty)$ be Lebesgue-measurable functions and suppose that $\int_{\Omega} \rho \, d \mu =1$. Prove that, if $\mu$ denotes the Lebesgue measure on $\mathbb{R}^n$, for every $p \in [1,+\infty)$: $$\left( \int_{\Omega} |h| \rho \, d \mu \right)^p \le \int_{\Omega} |h|^p \rho \, d \mu$$ At first sight, I was thinking about Jensen's inequality because $f(x)=|x|^p$ is convex if $p \ge 1$, but the hypothesis $\mu(\Omega)=1$ is not satified, so we cannot apply that, at least not directly. On the other hand, I can't figure out how classical measure theory inequalities (for instance, Holder's) can work here. I would appreciate any hint.

## 1 Answer

This is Jensen's inequality -- let $d \nu = \rho d \mu$. Then, its easy to see $\nu(\Omega) = 1$ and $\nu$ is thus a probability measure, and you have the usual Jensen's inequality with respect to $\nu$.

If you want to think of it in probability terms, $\rho$ is a probability density function by virtue of its definition wrt the lebesgue measure. The left hand side is $E[|h(X)|]^p$ and the right is $E[|h(X)|^p]$ where $X \sim \rho$.

• Thank you very much! I think I was missing the passage $\int_{\Omega} |h| \, d \nu = \int_{\Omega} |h| \frac{d \nu}{d \mu} \, d \mu = \int_{\Omega} |h| \rho \, d \mu$. – GaC Jan 11 '17 at 14:07