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Suppose I have a probability measure $P([a,b]) = \int_a^b{x}{dx}$ and I want to find the expectation of the random variable $X = w^2$, which is by definition:

$E(X) = \int_{\Omega}{w^2P(dw)}$.

I have no idea what to do here.. I know if the probability measure is the Lebesgue measure then you just turn the $P(dw)$ into $dw$ and integrate "normally" but with such measures I am unsure. Any help appreciated.

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    Apologies for the lac$k$ of detail. I just made something up from what I had seen recently but couldn't find the exact details.2011-09-04

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In this particular case, you should only apply that $EX=\int_\Omega\,w^2\,P(dw)=\int_{\Omega}\,w^2\,w\,dw$ by applying the next theorem:

Theorem. Let $(X,\mathcal{A})$ be a measurable space, and $\mu$ a measure on it and $f:X\rightarrow [0,\infty]$ a measurable function. Then \begin{align} \varphi:\mathcal{A}&\rightarrow [0,\infty]\newline A&\mapsto \int_{A}\,f\,d\mu \end{align} is a measure over $(X,\mathcal{A})$, and for every measurable function $g:X\rightarrow [0,\infty]$ we have $\int_{\Omega}\,g\,d\varphi=\int_{\Omega}\,g\cdot f\,d\mu$

Proof: For showing that it is a measure, we only have to show $\sigma$-additivity. So, let $\{E_n\}$ be a denumerable family of disjoint sets of $\mathcal{A}$, then $\varphi\left(\bigcup E_n\right)=\int_{\bigcup E_n}\,f\,d\mu=\int_{\Omega}\,\chi_{\bigcup E_n}\cdot f\,d\mu$ $=\int_{\Omega}\,\sum\chi_{E_n}\cdot f\,d\mu=\sum\int_{\Omega}\chi_{E_n}\cdot f\,d\mu=\sum\int_{E_n}f\,d\mu=\sum\varphi(E_n)$ thanks to Lebesgue's Monotone Convergence Theorem. Finally, the second part is true by the previous part for characteristic functions, so it will be also true for simple functions. And, so by Lebesgue's Monotone Convergence Theorem for arbitrary measurable functions.

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    Thanks. I have studied Lebesgue measure/integration theory only -- not much about general measures but I guess they're not too different.2011-09-05