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If $X$ is a random variable and $h$ a Borel measurable function, let $\mathbb{P}^{x}$ be the distribution of $X$. I now have problems proving that $h(X)\in L^1(\Omega,\mathcal F,\mathbb{P})$ if and only if $h\in L^1(\mathbb{R},\mathcal B,\mathbb{P}^{x})$, in which case $$\mathbb{E}h(X)=\int_{\mathbb{R}}h d\mathbb{P}^{x}.$$

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up vote 1 down vote accepted

We use the definition of Lebesgue integral: first, as the sum of two integrable functions is integrable, and $f$ is integrable if and only if so are $f^+:=\max(f,0)$ and $f^-:=f^+-f$. So we just have to show the result when $h$ is non-negative.

  • If $h(X)\in L^1(\Omega,\mathcal F,\mathbb P)$, then we can find a sequence $\{h_n\}$ of simple functions such that $h_n\uparrow h$ almost everywhere. We get by Fatou's lemma $$\int_{\Bbb R}h(x)d\mathbb P^x\leq \liminf_{n\to +\infty}\int_{\Bbb R}h_n(x)d\mathbb P^x.$$ We can show that for each simple function $s$, we have $\int_{\Bbb R}s(x)d\mathbb P^x=\int_\Omega s(X(\omega))d\mathbb P(\omega)$ (by definition of integration of simple functions and pullback of a measure). Since $h_n\leq h$ for all $x$, $$\int_{\Bbb R}h(x)d\mathbb P^x\leq \liminf_{n\to +\infty}\int_\Omega h_n(X(\omega))d\mathbb P(\omega)\leq \int_{\Omega}h(X(\omega))d\mathbb P(\omega)<\infty$$
  • Conversely, if $h\in L^1(\Bbb R,\mathcal B(\Bbb R),\mathbb P^x)$, we have by a similar argument $$\int_{\Omega}h(X(\omega))d\mathbb P(\omega)\leq \liminf_{n\to +\infty}\int_\Omega h_n(X(\omega))d\mathbb P(\omega)\leq\int_{\Bbb R}h(x)d\mathbb P^x <\infty.$$ This proves the equivalence.

The last two displayed equations show that when one of the two conditions is satisfied, we have equality. (actually, we have equality in $\overline{\Bbb R}$)

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I was wondering if it could also be solved using the standard machinery for $h$, i.e. first show that it holds true for indicator functions, then extending it to nonnegative simple functions etc till you find that it holds true for all functions $h\in L^{1}(\Omega,F, \mathbb{P})$?? – Math Girl Oct 2 '12 at 14:56
Yes, such a method will work (and is not fundamentally different from what I did). – Davide Giraudo Oct 2 '12 at 14:58
And how would I tackle it this way? – Math Girl Oct 2 '12 at 15:16
It's actually more or less similar than what I did. First, show that equivalence holds when $h$ is the characteristic function of a measurable set. Then that it's still the case when $h$ is a simple non-negative function, then for non-negative one. But you have to prove equivalences in each case, so it's probably not shorter. – Davide Giraudo Oct 2 '12 at 15:19

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