Probability Theory 1에서 가장 중요한 Part이다. 이 내용들을 이해하기 위해 Product measure 파트가 필요하다.
If \((\Omega_1,\mathcal{F}_1, \mu_1)\) and \((\Omega_2\mathcal{F}_2, \mu_2)\) are \(\sigma\)-finite measure spaces and if \(f:\Omega_1\times\Omega_2\rightarrow \bar{\mathbb{R}}\) is and \(\mathcal{F}_1\times\mathcal{F}_2\)-measurable function for which \(\int_{\Omega_1\times\Omega_2}f\mbox{ }d(\mu_1\times\mu_2)\) exists, then \[\begin{eqnarray*} \int_{\Omega_1\times\Omega_2}f(\omega_1, \omega_2)\mbox{ }d(\mu_1\times\mu_2)(\omega_1, \omega_2)&=&\int_{\Omega_1}\left[\int_{\Omega_2}f(\omega_1,\omega_2)\mbox{ } d\mu_2(\omega_2)\right]d\mu_1(\omega_1) \\ &=&\int_{\Omega_2}\left[\int_{\Omega_1}f(\omega_1,\omega_2)\mbox{ } d\mu_1(\omega_1)\right]d\mu_2(\omega_2). \end{eqnarray*}\] Here, \(\int_{\Omega_2}f(\omega_1,\omega_2)\mbox{ } d\mu_2(\omega_2)\) is defined for \(\mu_1\)-almost all \(\omega_1\) and we take its value to be zero(or any other arbitrary constant) for those \(\omega_1\) for which it is not defined.
Similarly, \(\int_{\Omega_1}f(\omega_1,\omega_2)\mbox{ } d\mu_1(\omega_1)\) is defined for \(\mu_2\)-almost all \(\omega_2\) and we take its value to be zero for those \(\omega_2\) for which it is not defined.
In particular, the result holds if \(f\) is either nonnegative or integrable w.r.t \(\mu_1\times\mu_2\), and in the latter case, \(\int_{\Omega_2}f(\omega_1,\omega_2)\mbox{ } d\mu_2(\omega_2)\) is finite for \(\mu_1\)-almost all \(\omega_1\) and \(\int_{\Omega_1}f(\omega_1,\omega_2)\mbox{ } d\mu_1(\omega_1)\) is finite for \(\mu_2\)-almost all \(\omega_2\) and
To set notation, for \(A\subset\Omega_1\times\Omega_2\), let \[ A_{\omega_1}^{(1)}=\{\omega_2:(\omega_1,\omega_2)\in A \}\subset \Omega_2, \] and \[ A_{\omega_2}^{(2)}=\{\omega_1:(\omega_1,\omega_2)\in A \}\subset \Omega_1 \] be the \(\omega_1\) and \(\omega_2\) sections of \(A\), respectively.
\(xy\)-plane에서의 사각형을 생각해보면 쉽다. \((\omega_1,\omega_2)\in A\)에 대해서 \(A_{\omega_1}^{(1)}\)는 \(\omega_2\), 즉 \(y\)축 부분의 Range(Section)로, \(A_{\omega_2}^{(2)}\)는 \(x\)축 부분의 Range(Section)로 연상할 수 있다.
두 set \(A, B\in \Omega_1\times \Omega_2\)에 대해, \(A_{\omega_1}^{(1)}\cap B_{\omega_1}^{(1)}=\phi\) for all \(\omega_1\in \Omega_1\)라면 \(A\cap B=\phi\)이다.
\(\omega_1\) section of a union of subsets of \(\Omega_1\times\Omega_2\) is the union of their \(\omega_1\) sections (\(\Omega_1\times\Omega_2\)의 부분집합들의 합집합에서 \(\omega_1\) sections은 부분집합들의 \(\omega_1\) sections의 합집합과 같다).
If \(A\in \mathcal{F}_1\times \mathcal{F}_2\), then \(A_{\omega_1}^{(1)}\in \mathcal{F}_2\) for all \(\omega_1\in \Omega_1\).
Section은 \((\omega_1,\omega_2)\) 둘 중 하나를 fix하고 한쪽 축만 고려하는 것과 같다.
그러므로 \(A_{\omega_1}^{(1)}\)는 \(\omega_1\)을 fix했기 때문에 \(A_2\in \mathcal{F}_2\)의 문제로 바뀐다.
증명을 위해 필요
If \((\Omega_1,\mathcal{F}_1, \mu_1)\) and \((\Omega_2,\mathcal{F}_2, \mu_2)\) are \(\sigma\)-finite measure spaces, and if \(A\in \mathcal{F}_1\times\mathcal{F}_2\), then
for fixed \(\omega_1\), the function \(\omega_2 \mapsto I_A(\omega_1,\omega_2)\) is \(\mathcal{F}_2\)-measurable;
the function \(\omega_1\mapsto \int_{\Omega_2}I_A(\omega_1,\omega_2)d\mu_2(\omega_2)\) is \(\mathcal{F}_1\)-measurable;
\((\mu_1\times\mu_2)(A)=\int_{\Omega_1}\left[\int_{\Omega_2}I_A(\omega_1, \omega_2)d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\).
1번: \(\omega_1\in \Omega_1\)으로 fix하면, \(A\in \Omega_1\times\Omega_2\)에 대해, \[ I_A(\omega_1,\omega_2)= I_{A_{\omega_1}^{(1)}}(\omega_2) \] 이다. 즉 \(I\left(\omega_2\in A_{\omega_1}^{(1)}\right)\)이고 이는 \(\omega_2\)가 \(A\)의 \(y\)축 영역에서의 범위 안에 들어있는지에 대한 Indicator function을 의미한다. 바로 위의 Lemma에서 \(A_{\omega_1}^{(1)}\in \mathcal{F}_2\)이기 때문에, \(I_{A_{\omega_1}^{(1)}}(\omega_2)\)는 \(\mathcal{F}_2\)-measurable하다.
2번 : For all \(A=A_1\times A_2\in \text{Rect}(\mathcal{F}_1,\mathcal{F}_2)\) \[
I_{A_1\times A_2}(\omega_1,\omega_2)= I_{A_1}(\omega_1)I_{A_2}(\omega_2)=I(\omega_1\in A_1)I(\omega_2\in A_2),
\] by the definition of measurable rectangle(2.9에서 첫번째 정의). Then, \[
\int_{\Omega_2}I_{A_1\times A_2}(\omega_1,\omega_2)\mbox{ }d\mu_2(\omega_2)= I_{A_1}(\omega_1)\int_{\Omega_2}I_{A_2}(\omega_2)\mbox{ }d\mu_2(\omega_2)=I_{A_1}(\omega_1)\mu_2(A_2),
\] which is \(\mathcal{F}_1\)-measurable because \(A_1\in \mathcal{F}_1\)(즉 Field에 대해서는 성립한다).
First, let \(\mu_2\) be a finite measure, and let \(\mathcal{L}\) be the class of \(\mathcal{F}_1\times\mathcal{F}_2\)-sets \(A\) for which 2. holds. Then, \(\mathcal{L}\) is a \(\lambda\)-system:
\(\Omega_1\times \Omega_2\in \text{Rect}(\mathcal{F}_1,\mathcal{F}_2)\subset \mathcal{L}\) \(\mbox{ }\)\(\mbox{ }\) (Field에서는 hold하기 때문).
If \(A\in \mathcal{L}\), then because \(\mu_2(\Omega_2)<\infty\), \[ \mu_2\left((A^c)_{\omega_1}^{(1)}\right)=\mu_2\left(\left(A_{\omega_1}^{(1)}\right)^c\right)=\mu_2(\Omega_2)-\mu_2\left(A_{\omega_1}^{(1)}\right), \] which is clearly a \(\mathcal{F}_1\) measurable function of \(\omega_1\) because \(\mu_2\left(A_{\omega_1}^{(1)}\right)\) is \(\mathcal{F}_1\)-measurable (Linear combination of \(\mathcal{F}_1\) measurable functions). Thus, \(A^c\in \mathcal{L}\).
If \(A_1,A_2,\ldots,\in \mathcal{L}\) are disjoint, then \[ \mu_2\left(\left(\biguplus_{n=1}^\infty A_n \right)_{\omega_1}^{(1)}\right)= \mu_2\left(\biguplus_{n=1}^\infty A_{n, \omega_1}^{(1)}\right)= \sum_{n=1}^{\infty}\mu_2(A_{n, \omega_1}^{(1)}), \] which is \(\mathcal{F}_1\) measurable, being a limit of \(\mathcal{F}_1\) measurable functions.
But since \(\text{Rect}(\mathcal{F}_1,\mathcal{F}_2)\subset \mathcal{L}\) is a \(\pi\)-system generating \(\mathcal{F}_1,\mathcal{F}_2\), it follows from the \(\pi-\lambda\) theorem that \(\mathcal{F}_1\times \mathcal{F}_2\subset \mathcal{L}\), i.e., 2. holds for all \(A\in \mathcal{F}_1\times\mathcal{F}_2\).
Second, let \(\mu_2\) be \(\sigma\)-finite. Let \(B_{2,n}\in \mathcal{F}_2\) satisfy \(B_{2,n}\uparrow \Omega_2\) and \(\mu_2(B_{2.n})<\infty\) for all \(n\ge 1\). Then, \(\mu_{2,n}(A_2):=\mu_{2,n}(A_2\cap B_{2.n})\) defines a finite measure on \(\mathcal{F}_2\), and for any \(A\in \mathcal{F}_1\times\mathcal{F}_2\), \[ \mu_{2,n}(A_{\omega_1}^{(1)})=\mu_2(A_{\omega_1}^{(1)}\cap B_{2,n})\uparrow \mu_2(A_{\omega_1}^{(1)}) \] (좌변이 finite measure값이기 때문에 아무리 increasing한다고 해도 limit 또한 finite값이다).
즉 finite 과 \(\sigma\)-finite한 \(\mu_2\)에 대해 function \(\omega_1\mapsto \mu_2(A_{\omega_1}^{(1)})\), being a limit of \(\mathcal{F}_1\)-measurable functions, is itself \(\mathcal{F}_1\)-measurable.
3번: Define \[ \mu(A)=\int_{\Omega_1}\mu_2(A_{\omega_1}^{(1)})d\mu_1(\omega_1), \mbox{ }\mbox{ }\mbox{ }A\in \mathcal{F}_1\times\mathcal{F}_2. \] Clearly \(\mu\) is nonnegative(Basic properties of Integral, 4 참고), and, since \((\uplus_n A_n)_{\omega_1}^{(1)}=(\uplus_{n} A_{n,\omega_1}^{(1)})\), countable additivity of \(\mu_2\) and monotone conergence for series of nonnegative terms implies that \(\mu\) is countably additive and hence a measure on \(\mathcal{F}_1\times \mathcal{F}_2\): \[\begin{eqnarray*} \mu(\uplus_n A_n)&=& \int_{\Omega_1}\mu_2\left((\uplus_n A_n)_{\omega_1}^{(1)}\right)d\mu_1(\omega_1)\\ &=&\int_{\Omega_1}\mu_2\left((\uplus_{n} A_{n,\omega_1}^{(1)})\right)d\mu_1(\omega_1)\\ &=&\int_{\Omega_1}\sum_n\mu_2\left(A_{n,\omega_1}^{(1)}\right)d\mu_1(\omega_1)\\ &=&\sum_n\int_{\Omega_1}\mu_2\left(A_{n,\omega_1}^{(1)}\right)d\mu_1(\omega_1)=\sum_n\mu(A_n). \end{eqnarray*}\] Further, for \(A=A_1\times A_2\in \text{Rect}(\mathcal{F}_1,\mathcal{F}_2)\) we have \[\begin{eqnarray*} \mu_2\left((A_1\times A_2)_{\omega_1}^{(1)}\right)&=&\int_{\Omega_2} I_{A_1\times A_2}(\omega_1,\omega_2)d\mu_2(\omega_2) \\ &=&I_{A_1}(\omega_1)\mu_2(A_2)\\ \implies\mu(A_1\times A_2)&=&\int_{\Omega_1}I_{A_1}(\omega_1)\mu_2(A_2)d\mu_1(\omega_1)\\ &=&\mu_1(A_1)\mu_2(A_2)=(\mu_1\times\mu_2)(A_1\times A_2). \end{eqnarray*}\] Thus, \(\mu\) and \(\mu_1\times\mu_2\) agree on the \(\pi\)-system \(\text{Rect}(\mathcal{F}_1,\mathcal{F}_2)\), and \(\mu_1\times\mu_2\) is \(\sigma\)-finite on \(\text{Rect}(\mathcal{F}_1,\mathcal{F}_2)\). By the uniqueness theorem for \(\sigma\)-finite measures, \(\mu=\mu_1\times\mu_2\).
Simple function \(f\) \(\rightarrow\) nonnegative function \(f\) \(\rightarrow\) general function \(f\) 순서로 증명할 것이다.
Suppose that \(f=\sum_{i=1}^m a_iI_{A_i}\) is a nonnegative simple function, with \(A_1,\ldots, A_m \in \mathcal{F}_1\times\mathcal{F}_2\). Then, for each fixed \(\omega_1\), the function \[ \omega_2\mapsto f(\omega_1,\omega_2)=\sum_{i=1}^m a_i I_{A_i}(\omega_1,\omega_2) \] is \(\mathcal{F}_2\)-measurable, being a nonnegative linear combination of \(\mathcal{F}_2\)-measurable functions. Also, by linearity of the integral, \[ \int_{\Omega_2} f(\omega_1, \omega_2)d\mu_2(\omega_2)= \sum_{i=1}^m a_i\int_{\Omega_2}I_{A_i}(\omega_1,\omega_2)I_{A_i}(\omega_1,\omega_2)d\mu_2(\omega_2) \] so that the function \[ \omega_1\mapsto \int_{\Omega_2} f(\omega_1,\omega_2)d\mu_2(\omega_2) \] is \(\mathcal{F}_1\)-measurable, being a nonnegative linear combination of \(\mathcal{F}_1\)-measurable functions.
Finally by linearity of the integral and Fubini’s theorem for indicator functions, \[\begin{eqnarray*} \int_{\Omega_1\times\Omega_2} f(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)&=&\int_{\Omega_1\times\Omega_2} \sum_{i=1}^m a_iI_{A_i}(\omega_1,\omega_2) d(\mu_1\times\mu_2)(\omega_1, \omega_2)\mbox{ }\mbox{ }\mbox{ by the definition of }f\\ &=&\sum_{i=1}^m a_i\int_{\Omega_1\times\Omega_2} I_{A_i}(\omega_1,\omega_2) d(\mu_1\times\mu_2)(\omega_1, \omega_2) \\ &=&\sum_{i=1}^m a_i\int_{\Omega_1}\left[\int_{\Omega_2} I_{A_i}(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1) \mbox{ }\mbox{ }\mbox{ by }3 \mbox{ of the above lemma}\\ &=&\int_{\Omega_1}\left[\sum_{i=1}^m a_i\int_{\Omega_2} I_{A_i}(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\\ &=&\int_{\Omega_1}\left[\int_{\Omega_2}\sum_{i=1}^m a_i I_{A_i}(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\\ &=&\int_{\Omega_1}\left[\int_{\Omega_2}f(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\mbox{ }\mbox{ }\mbox{ by the definition of }f. \end{eqnarray*}\]
For \(\mathcal{F}_1\times\mathcal{F}_2\)-measurable \(f\ge 0\), let \(f_n, n\ge 1\) be simple function satisfying \(0\le f_n \uparrow f\). Then, for each fixed \(\omega_1\), the function \[ \omega_2\mapsto f(\omega_1,\omega_2)= \lim_n f_n(\omega_1, \omega_2) \] is \(\mathcal{F}_2\)-measurable, being a limit of \(\mathcal{F}_2\)-measurable functions.
Also, by the monotone convergence theorem, \[ 0\le \int_{\Omega_2}f_n(\omega_1,\omega_2)d\mu_2(\omega_2)\uparrow \int_{\Omega_2}f(\omega_1,\omega_2)d\mu_2(\omega_2) \mbox{}\mbox{ }\mbox{ for all }\omega_1, \] so that the function \[ \omega_1\mapsto \int_{\Omega_2}f(\omega_1,\omega_2)d\mu_2(\omega_2) \] is \(\mathcal{F}_1\)-measurable, being a limit of \(\mathcal{F}_1\)-measurable functions ($ _{_2}f_n(_1,_2)d_2(_2)$이 \(\mathcal{F}_1\)-measurable이기 때문).
Thus by monotone convergence and the result for the simple function again, \[\begin{eqnarray*}
\int_{\Omega_1\times\Omega_2} f(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)&=&\lim_n \int_{\Omega_1\times\Omega_2} f_n(\omega_1,\omega_2) d(\mu_1\times\mu_2)(\omega_1, \omega_2)\mbox{ }\mbox{ }\mbox{ by M.C.T}(0\le f_n\uparrow f, \mbox{ take integral w.r.t } \mu_1\times\mu_2) \\
&=& \lim_n \int_{\Omega_1}\left[\int_{\Omega_2} f_n(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\mbox{ }\mbox{ }\mbox{ }\mbox{ simple function }f_n\\
&=& \int_{\Omega_1}\left[\lim_n\int_{\Omega_2} f_n(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\\
&=&\int_{\Omega_1}\left[\int_{\Omega_2} f(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\mbox{ }\mbox{ }\mbox{ by M.C.T}(0\le f_n\uparrow f, \mbox{ take integral w.r.t } \mu_2).
\end{eqnarray*}\]
Finally, suppose that \(f\) is a general \(\mathcal{F}_1\times\mathcal{F}_2\)-measurable function.
Then, by the result for nonnegative functions, the function \[ \omega_2\mapsto f^+(\omega_1,\omega_2) \] is \(\mathcal{F}_2\)-measurable for each fixed \(\omega_1\), and the function \[ \omega_1\mapsto \int_{\Omega_2}f^+(\omega_1,\omega_2)d\mu_2(\omega_2) \] is \(\mathcal{F}_1\)-measurable, and \[ \int_{\Omega_1\times\Omega_2} f^+(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)=\int_{\Omega_1}\left[\int_{\Omega_2} f^+(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1) \] Of course the analogous results hold for \(f^-\).
Now if \[ \int_{\Omega_1\times\Omega_2} f(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)=\int_{\Omega_1\times\Omega_2} f^+(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)-\int_{\Omega_1\times\Omega_2} f^-(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2) \] exists(well-defined), then at least one of the integrals on the right-hand side must be finite. WLOG, suppose that the negative part is finite. Then, \[ \int_{\Omega_1\times\Omega_2} f^-(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)=\int_{\Omega_1}\left[\int_{\Omega_2} f^-(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1) \] which implies that \[ 0\le \int_{\Omega_2} f^-(\omega_1,\omega_2) d\mu_2(\omega_2) < \infty \mbox{ }\mbox{ }\mbox{ }\mu_1\mbox{-a.e. } \] by the \(5^{th}\) part of the Basic Properties of the Integral.
Thus, \[ \int_{\Omega_2} f(\omega_1,\omega_2)d\mu_2(\omega_2)=\int_{\Omega_2} f^+(\omega_1,\omega_2)d\mu_2(\omega_2)-\int_{\Omega_2} f^-(\omega_1,\omega_2)d\mu_2(\omega_2) \] is defined \(\mu_1\)-a.e., as claimed.
Thus, finally we have \[\begin{eqnarray*} \int_{\Omega_1}\left[\int_{\Omega_2} f(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)&=&\int_{\Omega_1}\left[\int_{\Omega_2} f^+(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1) -\int_{\Omega_1}\left[\int_{\Omega_2} f^-(\omega_1,\omega_2) d\mu_2(\omega_2)\right]d\mu_1(\omega_1)\\ &=&\int_{\Omega_1\times\Omega_2} f^+(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)-\int_{\Omega_1\times\Omega_2} f^-(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2)\\ &=&\int_{\Omega_1\times\Omega_2} f(\omega_1,\omega_2)d(\mu_1\times\mu_2)(\omega_1, \omega_2). \end{eqnarray*}\]
Fubini’s theorem은 두 주어진 measurable spaces에서 \(f\)가 measurable-\(\mathcal{F}_1/\mathcal{F}_2\)일 때(가장 중요) product measure에 대한 indefinite integral이 각 measure에 대한 indefinite integral로 쪼개질 수 있고, 그 순서도 바뀔 수 있다는 것을 보여준다. 즉 \[\begin{eqnarray*} \int_{\Omega_1\times\Omega_2}f(\omega_1, \omega_2)\mbox{ }d(\mu_1\times\mu_2)(\omega_1, \omega_2)&=&\int_{\Omega_1}\left[\int_{\Omega_2}f(\omega_1,\omega_2)\mbox{ } d\mu_2(\omega_2)\right]d\mu_1(\omega_1) \\ &=&\int_{\Omega_2}\left[\int_{\Omega_1}f(\omega_1,\omega_2)\mbox{ } d\mu_1(\omega_1)\right]d\mu_2(\omega_2). \end{eqnarray*}\] 이고, 매우 많이 쓰이기 때문에 반드시 암기하자.