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When I was reading this definition of product measure from the text book, it says: $A\in \mathcal {F= F_1 \times F_2}$ is an arbitrary set, $A=A_1\times A_2$. Then the product measure is defined by $P(A)=P_1(A_1)P_2(A_2)=\int_{\Omega_2} P_1 (A_{\omega_2})dP_2(\omega_2)$

I just wonder how can it directly get that intergal.

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Sorry, I don't understand your question. –  Michael Greinecker May 10 '13 at 10:16
    
@MichaelGreinecker sorry, what is unclear here? The definition or my qustion? –  Dylan Zhu May 10 '13 at 10:20
    
I am not sur what the question is. You want to know why $P_1(A_1)P_2(A_2)=\int_{\Omega_2} P_1 (A_{\omega_2})dP_2(\omega_2)$? –  Michael Greinecker May 10 '13 at 10:23
    
yes, i dont know how is that integral come from –  Dylan Zhu May 10 '13 at 10:30
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If $A$ is any subset of $\Omega_1\times\Omega_2$, $\omega_1\in\Omega_1$, and $\omega_2\in\Omega_2$, we can build the sections $$A_{\omega_2}=\{\omega_1'\in\Omega_1:(\omega_1',\omega_2)\in A\}\subseteq\Omega_1$$ and $$A_{\omega_1}=\{\omega_2'\in\Omega_2:(\omega_1,\omega_2')\in A\}\subseteq\Omega_2.$$

If $A$ is a rectangle, $A=A_1\times A_2$, then $A_{\omega_2}=A_1$ if $\omega_2\in A_2$ and $A_{\omega_2}=\emptyset$ if $\omega_2\notin A_2$. In particular, the function $P_1(A_\omega)$ takes the value $P_1(A_1)$ on $A_2$ and the value $0$ ouside of $A_2$. Let $1_{A_2}$ be the indicator function of $A_2$ on $\Omega_2$. So we get $$\int_{\Omega_2} P_1 (A_{\omega_2})~dP_2(\omega_2)=\int_{\Omega_2}P_1(A_1) 1_{A_2}~dP(\omega_2)$$ $$=\int_{A_2}P_1(A_1)~dP(\omega_2)=P_1(A_1)P_2(A_2).$$

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