# Related Measures

($\Omega, F,P)$ is a measure space, $\ Q \ll P$ ($Q$ is related to $P$) i.e. $\ Q(A)= \int 1_ADdP$ where $\ D=dQ/dP$. Then $X$ is integrable wrt $Q$ if and only if $XD$ is integrable wrt $P$ and $\int XdQ = \int XDdP$. How do I prove this? I can only do it for $X$ a simple Borel function, not a general one.

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## 1 Answer

• Let $X$ integrable wrt $Q$. Write $X=X^+-X^-$, where $X^+$ and $X^-$ are the positive and negative parts of $X$. Let $X_n$ a sequence of step functions which converge pointwise to $X^+$ and $X_n\leq X^+$. We can show by Fatou's lemma that $$\int X^+DdP=\int \liminf_n X_nDdP\leq \liminf_n\int X_nDdP=\liminf_n\int X_ndQ\leq \int X_n^+dQ,$$ hence $DX^+$ is $P$ integrable. Do the same for $X^-$ to get the result. Use the same idea to get the converse, using the fact that $D$ is non-negative.
• You got $\int XdQ=\int XDdP$ for each $X$ simple. By a monotone convergence argument, you have to show that this equality holds if $X\geq 0$ and is integrable. Then you will deduce the general case.
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Ok, so for the second one I can say that for any non negative borel function X there exits a monotone sequence Xi where Xi are simple for all i such that $\ X=Sup_i(X_i)$ then $\int X dQ = \ int sup_iX_i dQ =sup_i \ intX_idQ$ via monotone convergence which is then $\sup_i \ intX_iDdP$ But now what can I do with this? – Rosie May 31 '12 at 21:53
You can write a Borel-measurable function as a difference of two measurable non-negative functions. – Davide Giraudo Jun 2 '12 at 13:02