1
$\begingroup$

I am reading Royden Real Analysis Forth edition chapter 19.3 named: "The Kantrovitch Representation Theorem for the dual of $L^\infty$" He gives defenition of bounded finitely additive signed measure $v$ as:

set function on measurable space which is finitely additive and has finite total variation of $v$ over whole space $X$.

I am wondering why does this imply that this is in fact a signed measure. So far, I've derived that $v$ of empty set is zero, and some other facts but I can't show countable additivity.

Is this even true?

Thank you!

$\endgroup$
1
$\begingroup$

Let $\nu$ be the signed finitely additive set function over $\mathfrak M$ and $\|\nu\|$ its total variation. Then $\|\nu\|$ is a bounded positive measure, and we have that $\lvert\nu(A)\rvert\le\|\nu\|(A)$, for all $A\in\mathfrak{M}$.

In order to show that $\nu$ is a signed measure, it suffices to show that $$ \nu(A)=\sum_{n\in\mathbb N}\nu(A_n), $$ whenever $\{A_n\}_{n\in\mathbb N}$ is a sequence of disjoint measurable sets with union $A$.

Let $S_n=A_1\cup\cdots\cup A_n$, $T_n=A\setminus S_n$. Then $$ \nu(A)-\nu(T_n)=\nu(S_n)=\nu(A_1)+\cdots+\nu(A_n). $$ As $\|\nu(T_n)\|\to 0$, then $\nu(T_n)\to 0$. Next observe that $$ \Big|\,\nu(A)-\sum_{n\in\mathbb N}\nu(A_n)\Big|=\Big|\,\nu(S_n)+\nu(T_n)-\sum_{k=1}^n\nu(A_k)-\sum_{k>n}\nu(A_k)\Big| \\ =\Big|\,\nu(T_n)-\sum_{k>n}\nu(A_k)\Big|\le \lvert \nu(T_n)\rvert+\sum_{k>n}\lvert\nu(A_k)\rvert \le \| \nu(T_n)\|+\sum_{k>n}\|\nu\|(A_k) \\=\| \nu(T_n)\|+\|\nu\|\Big(\bigcup_{k>n}A_k\Big)=2\|\nu\|(T_n)\to 0. $$

$\endgroup$
  • $\begingroup$ Dear Yiorgos, I believe that $\|\nu\|$ is a bounded positive measure only if $\nu$ is signed measure. (This guess is based on Royden's textbook). Which is what we actually want to show. Moreover, I believe I've managed to derive that signed finitely additive set function is in fact not always signed measure because there is linear isomorphism of $L^1$ to $L^\infty$ however, it is not surjective. $\endgroup$ – Vadim Dec 30 '15 at 7:00
  • $\begingroup$ So from Kantorovitch Representation Theorem we infer that there is a bounded finitely additive signed measure $\nu$ for which there is no function $g$ from $L^1$ mapping to it. However, if every signed finitely additive set function is a signed measure this would not be true! $\endgroup$ – Vadim Dec 30 '15 at 7:02
  • $\begingroup$ @Vadim If you pay attention to what I show: Every finitely additive set function which HAS BOUNDED VARIATION is a signed measure – I didn't say that every finitely additive measure is a measure! $\endgroup$ – Yiorgos S. Smyrlis Dec 30 '15 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.