Radon-Nikodym定理“推廣”（Ex 2.12.2）

標籤：style   ext   com   get   os   set   

（Foundations of Modern Analysis, Avner Friedman）Problem 2.12.2. The Radon-Nikodym theorem remains ture in case $\mu $ is a $\sigma-$finite signed measure.

Radon-Nikodym theorem is about saying that $\left( {X,a,\mu } \right)$ is a $\sigma-$finite measure, $\mu$ is a measure, $\nu $ is a signed measure and $\nu  \ll \mu $. Then there exists a measurable function $f$ such that $\nu \left( E \right) = \int_E {fd\mu } $. Problem 2.12.2 extended the condition “$\mu$ is a measure” to that “$\mu$ is a signed measure”.

My idea is that, consider the Jordan decomposition ${\mu ^ + }$ and ${\mu ^ - }$ (where $\mu  = {\mu ^ + } - {\mu ^ - }$) over the Hahn decomposition ($A,B$) of $X$. Thus ${\mu ^ + }$ and ${\mu ^ - }$ are both measures, we then can apply Radon-Nikodym theorem. Consider the positive set A, easily, we get $\nu  \ll {\mu ^ + }$, hence there exists a measurable function ${f_1}$ such that \[v\left( E \right) = \int_E {{f_1}d{\mu ^ + }} ,\forall E \subseteq A\]

Similarily, there exists a measurable function ${f_2}$ on B such that $-v\left( E \right) = \int_E {{f_2}d{\mu ^ - }} ,\forall E \subseteq B$. (I am not sure if it‘s true)

Then for all $E \in a$, \[\begin{array}{l}
\nu \left( E \right) = \nu \left( {E \cap A} \right) + \nu \left( {E \cap B} \right)\\
 = \int_{E \cap A} {{f_1}d{\mu ^ + }}  - \int_{E \cap B} {{f_2}d{\mu ^ - }} \\
 = \int_E {{\chi _A}{f_1}d{\mu ^ + }}  - \int_E {{\chi _B}{f_2}d{\mu ^ - }} \\
\mathop  = \limits^{{\rm{(a)}}} \int_E {\left( {{\chi _A}{f_1} + {\chi _B}{f_2}} \right)d{\mu ^ + }}  - \int_E {\left( {{\chi _A}{f_1} + {\chi _B}{f_2}} \right)d{\mu ^ - }} \\
\mathop  = \limits^{{\rm{(b)}}} \int_E {\left( {{\chi _A}{f_1} + {\chi _B}{f_2}} \right)d\mu } \\
 \buildrel \Delta \over = \int_E {fd\mu }
\end{array}\]

where equality (a) is because ${\mu ^ + }$(${\mu ^ - }$) to $B$($A$) is 0; equality (b) is by definition that $\int_E {fd\mu }  = \int_E {fd{\mu ^ + }}  - \int_E {fd{\mu ^ - }} $.