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!


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 $$ \Big|\,\nu(A)-\sum_{n\in\mathbb N}\nu(A_n)\Big|=\Big|\,\nu(S_n)+\nu(T_n)-\sum_{k=1}^n\nu(A_k)-\sum_{k>n}\nu(A_k)\Big| \\ =\Big|\,\nu(T_n)-\sum_{k>n}\nu(A_k)\Big|\le \lvert \nu(T_n)\rvert+\sum_{k>n}\lvert\nu(A_k)\rvert \le \| \nu(T_n)\|+\sum_{k>n}\|\nu\|(A_k) \\=\| \nu(T_n)\|+\|\nu\|\Big(\bigcup_{k>n}A_k\Big)=2\|\nu\|(T_n)\to 0. $$

  • $\begingroup$ 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

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