Suppose that \(\mu\) and \(\nu\) are measures on a measurable space \((\Omega, \mathcal{F})\). Then, we say that \(\nu\) is absolutely continuous with respect to \(\mu\), denoted \(\nu \ll\mu\), if for all \(A\in \mathcal{F}\), \[ \mu(A)=0 \mbox{ }\mbox{ }\mbox{ implies }\mbox{ }\mbox{ }\nu(A)=0. \]
매우 중요하다
Let \(\mu\) and \(\nu\) be measures on a measurable space \((\Omega, \mathcal{F})\), where \(\mu\) is \(\sigma\)-finite and \(\nu \ll \mu\). Then, there exists a nonnegative, \(\mathcal{F}\)-measurable function \(f:\Omega \rightarrow \bar{\mathbb{R}}\) such that \[ \nu(A)=\int_A f\mbox{ }d\mu\mbox{ }\mbox{ }\mbox{ for all }A\in \mathcal{F}, \] and the function \(f\) is unique. If in addition \(\nu\) is \(\sigma\)-finite, then \(f\) is finite \(\mu\)-a.e.
즉 \(\nu\)가 absolutely continuous w.r.t the \(\sigma\)-finite measure \(\mu\)일 때, 위를 만족하는 \(f=d\nu/d\mu\)가 존재한다.
여기서 point는 \(f\)이다.
The density \(f\) in the Radon-Nikodym theorem is called the Radon-Nikodym derivative of \(\nu\) w.r.t \(\mu\), and is usually denoted \(d\nu/d\mu\).
지금부터 아래는 Radon-Nikodym을 증명하기 위해 필요한 많은 내용들이므로 간단히 넘어가려 한다.
Suppose that \((\Omega, \mathcal{F})\) is a measurable space. An extended real-valued set function \(\varphi\) defined on \(\mathcal{F}\) is a signed measure if
\(\varphi(\phi)=0\);
\(\varphi\) is countably additive: if \(A=\uplus_nA_n\) with \(A_n\mathcal{F}\), then \(\varphi(A)=\sum_n \varphi(A_n)\)
If \(\varphi\) is a signed measure on \((\Omega, \mathcal{F})\) and \(A\in \mathcal{F}\), then we say that \(A\) is positive set if \[ \varphi(E)\ge 0 \mbox{ }\mbox{ }\mbox{ for all } E\subset A, E\in \mathcal{F}, \] and \(A\) is a negative set if \[ \varphi(E)\le 0 \mbox{ }\mbox{ }\mbox{ for all } E\subset A, E\in \mathcal{F}. \] If \(A\) is both positive and negative, then \(\varphi(E)=0\) for all \(E\subset A, E\in \mathcal{F}\), and in this case \(A\) is called a null set.
Let \(\varphi\) is a signed measure on the measurable space \((\Omega, \mathcal{F})\). Then there exists a positive set \(A\) and a negative set \(B\) such that \(\Omega=A\uplus B(A\cap B=\phi)\).
Every measurable subset of a positive(negative) set is positive(negative).
The union of a countable collection of positive(negative) sets is positive(negative).
Let \(E\in \mathcal{F}\) satisfy \(-\infty<\varphi(E)<0\)(finite, strictly negative).
Then there exists a negative set \(A\subset E\) with \(\varphi(A)<0\).
\(E\)가 negative set들로만 이루어져있다면, trivial하지만 그렇지 않을수도 있다(E의 subset들이 positive set일 경우도 가능).
증명 idea: Let \(n_1\) be the smallest positive integer such that there exists \(E_1\in\mathcal{F}\) satisfying \(E_1\subset E\), \(\varphi(E_1)\ge \frac{1}{n_1}\).
Two measure \(\mu\) and \(\nu\) on a measurable space \((\Omega, \mathcal{F})\) are said to be mutually singular if there exists disjoint sets \(A,B\in \mathcal{F}\) such that \(\Omega=A\uplus B\)(i.e.,\(B=A^c\)) and \(\mu(A)=\nu(B)=0\).
In this case, we also say that \(\mu\) is singular w.r.t \(\nu\) and vice versa, and we write \(\mu \perp \nu\).
If \(\varphi\) is a signed measure on a measurable space \((\Omega, \mathcal{F})\), then there exist measures \(\varphi^+\) and \(\varphi^-\) on \((\Omega,\mathcal{F})\) such that \(\varphi^+\perp \varphi^-\) and \(\varphi=\varphi^+ + \varphi^-\). Moreover, there is only one such pair of mutually singlular measures(unique pair).
Let \(\Omega=A\uplus B\) be a Hahn decomposition for \(\varphi\), so that \(A\in \mathcal{F}\) is a positive set, \(B=A^c\) is a negative set. Define \(\varphi^+(E)= \varphi^+(E\cap A)\) and \(\varphi^-(E)= \varphi^-(E\cap B)\) for \(E\in\mathcal{F}\). Clearly \(\varphi^+\) and \(\varphi^-\) are measures on \((\Omega, \mathcal{F})\). Also, \(\varphi^+(B)=\varphi(A\cap B)=\varphi^-(A)=\varphi(\phi)=0\). Thus, \(\varphi^+\perp\varphi^-\).
Unique pair는 증명 생략.
즉, Jordan Decomposition의 핵심은 measurable space \((\Omega, \mathcal{F})\)에서 정의된 signed measure \(\varphi\)가 mutually singlular한 두개의 positive signed measure \(\varphi^+\)와 negative signed \(\varphi^-\)로 분리될 수 있다는 것이다.
The decomposition \(\varphi=\varphi^+ +\varphi^-\) is called the Jordan decomposition of the signed measure \(\varphi\). The measures \(\varphi^+\) and \(\varphi^-\) are called the positive and negative variations of \(\varphi\). Since \(\varphi\) assumes at most one of the values \(\pm\infty\), at least one of \(\varphi^+\) and \(\varphi^-\) must be finite. If both are finite then \(\varphi\) is called a finite signed measure. The measure \(|\varphi|\) defined by \[ |\varphi|(E)=\varphi^+(E)+\varphi^-(E), \mbox{ }\mbox{ }\mbox{ } E\in \mathcal{F}, \] is called the total variation of \(\varphi\).
증명을 위해 다시 살펴보자
Let \(\mu\) and \(\nu\) be measures on a measurable space \((\Omega, \mathcal{F})\), where \(\mu\) is \(\sigma\)-finite and \(\nu \ll \mu\). Then, there exists a nonnegative, \(\mathcal{F}\)-measurable function \(f:\Omega \rightarrow \bar{\mathbb{R}}\) such that \[ \nu(A)=\int_A f\mbox{ }d\mu\mbox{ }\mbox{ }\mbox{ for all }A\in \mathcal{F}, \] and the function \(f\) is unique. If in addition \(\nu\) is \(\sigma\)-finite, then \(f\) is finite \(\mu\)-a.e.
\(\mu\)와 \(\nu\)가 finite한 case에 대해서만 증명할 것이다.
우선 만약 \(\mu\)가 \(\sigma\)-finite measure on \(\mathcal{F}\)이고 \(\int_A f\mbox{ }d\mu=\int_A g\mbox{ }d\mu\) for all \(A\in\mathcal{F}\)라면, \(f=g\) a.e.라는 Theorem을 기억하자. 때문에 여기서 \(f\)는 Unique하다.
만약 \(\nu\)가 \(\sigma\)-finite하다면, \(A_1,A_2,\ldots\in \mathcal{F}\), \(\Omega=\cup_nA_n\)에 대해 \(\nu(A_n)=\int_{A_n}f\mbox{ }d\mu<\infty\) for all n이다. 이 indefinite integral이 finite이라는 말은 \(A_n\) set에 대해서 \(f\)의 measure값이 finite하다는 의미이므로 \(\mu(A_n\cap \{f=\infty\})=0\) 이다(\(f=\infty\)라면 indefinite integral은 무조건 \(\infty\)이기 때문). 즉 \[ \mu(A_n\cap\{f=\infty\})=0 \mbox{ }\forall n\implies \mu(\{f=\infty\})\le \sum_n \mu(A_n\cap\{f=\infty\})=0. \] 이므로 \(\mu\)는 finite measure이다.
A Good set \(\mathcal{G}\)를 다음과 같이 정의하자.
\[
\mathcal{G}=\left\{ g:\Omega\rightarrow \bar{\mathbb{R}}: \mbox{ g measurable }, g\ge 0, \int_Ag\mbox{ }d\mu \le \nu(A)\mbox{ } \forall A\in \mathcal{F} \right\}.
\]
또한 \(\gamma=\sup\left\{ \int g\mbox{ }d\mu:g\in \mathcal{G}\right\}\)로 정의하자. 그렇다면 good set의 정의에 의해 \(0\in \mathcal{G}\)이고 \(0\le \gamma \le \nu(\Omega)<\infty\)이다.
또한 \(g,h\in \mathcal{G}\)에 대해 \(g\vee h\in \mathcal{G}\)이다. 왜냐하면 \[ \int_A(g\vee h)d\mu=\int_{A\cap[g>h]}g\mbox{ }d\mu+\int_{A\cap[g<h]}h\mbox{ }d\mu\le \nu(A\cap[g>h])+\nu(A\cap[g<h])=\nu(A). \] 또한 \(g_n\in \mathcal{G}\)가 \(\int g_n\mbox{ }d\mu\rightarrow \gamma\)를 만족한다고 하고 \(f_n=\max\{g_1,\ldots,g_n\}\)이라 하자.
그렇다면 \(f_n\)은 increasing function이고 \(0\le f_n \uparrow f\), where \(f=\sup_ng_n\ge 0\) is measurable이다. 여기서 \[ \int f\mbox{ }d\mu \stackrel{\text{M.C.T}}{=}\lim_{n\rightarrow\infty}\int f_nd\mu\stackrel{\text{Def of } f_n}\ge \lim_{n\rightarrow\infty}\int g_nd\mu=\gamma. \] 그리고 \(f_n\in \mathcal{G}\) 이기 때문에 \(\int_A f\mbox{ }d\mu=\lim_{n\rightarrow\infty}\int_A f_nd\mu\le \nu(A)\) for all \(A\in \mathcal{F}\)이다. 결국 \(f\in \mathcal{G}\)이고 \(\int f\mbox{ }d\mu=\gamma\)이다.
Let \(\nu_{ac}(A)=\int_A f\mbox{ }d\mu\), and \(\nu_{s}(A)=\nu(A)-\nu_{ac}(A)\). 여기서 \(f\in \mathcal{G}\)라서, \(\nu_s\)는 nonnegative하다. 또한 \(\nu_s\)와 \(\nu_{ac}\)는 finite measure이고 \(\nu_{ac}\ll \mu\), \(\nu_{s}\ll \mu\)이다.
우리의 목표는 \(\nu_s(\Omega)=0\)을 보여서 \(\nu(A)=\nu_{ac}(A)=\int_A f\mbox{ }d\mu\)을 보이는 것이다.
Contradiction을 위해 \(\nu_s(\Omega)>0\)라 가정하자. 그렇다면 \(\mu\)는 finite measure이기 때문에, \(\mu(\Omega)-k\nu_s(\Omega)<0\)을 만족하는 \(k\)가 존재한다.
여기서 signed measure \(\varphi=\mu-k\nu_S\)라 정의하고 이 signed measure를 통해 \(\Omega=A\uplus B\)로 Positive set \(A\), Negative set \(B=A^c\)로 Hahn decomposition한다고 생각하자(Hahn decomposition은 any signed measure로도 가능하다).
여기서 \(\mu(B)>0\)이다. 왜냐하면 \(\mu(B)=0\)이라면 absolute continuity에 의해 \(\nu_s(B)=0\)이 되고, \(\varphi(B)=\mu(B)-k\nu_s(B)=0\).이 된다. 그렇다면 \(A\)가 positive set이기 때문에 \[ \varphi(\Omega)= \varphi(A\uplus B)= \varphi(A)+\varphi(B)=\varphi(A)\ge 0 \] 인데, \(\varphi(\Omega)=\mu(\Omega)-k\nu_S(\Omega)<0\)로 \(k\)를 잡았으므로 contradiction이다. 때문에 \(\mu(B)>0\)이다.
\(f^*=f+k^{-1}I_B\)라고 하자. B는 \(\varphi\)-negative set이기 때문에, for any \(E\in \mathcal{F}\), \[ \varphi(E\cap B)=\mu(E\cap B)-k\nu_s(E\cap B)\le 0 \implies \frac{1}{k}\mu(E\cap B)\le \nu_s(E\cap B), \] 이고 이를 이용하면 \[ \int_E f^*d\mu=\int_E f\mbox{ }d\mu+\frac{1}{k}\mu(E\cap B)\le \nu_{ac}(E)+\nu_s(E\cap B)\le\nu_{ac}(E)+\nu_s(E)=\nu(E) \] 라서, 결국 \(f^*\in \mathcal{G}\)이다. 하지만 \[ \int f^*d\mu=\int f\mbox{ }d\mu+\frac{1}{k}\mu(B)= \gamma+\frac{1}{k}\mu(B)>\gamma \mbox{ }(\mu(B)>0) \] 이므로 \(\nu_s(\Omega)>0\)에 contradiction이다. 즉 \(\nu(A)=\int_A f\mbox{ }d\mu\)인 f가 존재한다.