While reading Vector Measure from Diestel's book I find that Considering any Hahn-Banach extension $T$ to $L_{\infty}[0,1]$ of point mass functional on $C[0,1]$ we can construct a measure $F$ defined on all Lebesgue measurable subsets of $[0,1]$ which satisfies $F(E)=T(\chi_E)$ , where $E\subseteq [0,1]$ is Lebesgue measurable. Author said this measure is Finitely Additive but not Countably Additive.
I can not show that this measure fails to be Countably Additive, what I can guess is that if my point mass functional is $\delta_x:C[0,1]\rightarrow \Bbb R,\delta_x(f)=f(x)$ where $x\in [0,1]$ then $||\delta_x||=||T||=1$ and the measure $F$ has the property that $F(E)=0$ if and only if $x\in E$. But I can not prove it. Am I right? Thanks in advance.
Best Answer
Suppose $F$ is countably additive. Then it is absolutely continuous w.r.t. Lebesgue measure $m$. Let $g=\frac {dF} {dm}$. Then $f(x)=Tf=\int_0^{1} f(y)g(y)\, dy$ for all $f \in C[0,1]$ which means $\int_0^{1} f(y)g(y)\, dy=\int_0^{1} f(y)\, d\delta_x (y)$ and this implies $g(y)\, dy =\delta_x(dy)$ which is a contradiction.