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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.

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    $\begingroup$ The basic example of the problem is $\nu(A) = \sum_{ n \in A} \frac{1}{n+1}$ which is a non-negative measure on $\mathbb{N}$, $\mu(A)= \sum_{ n \in A} \frac{(-1)^n}{n+1}$ which is not a signed measure on $\mathbb{N}$. $\endgroup$ – reuns Nov 20 '18 at 5:40
  • $\begingroup$ Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact. $\endgroup$ – Mathlover Nov 20 '18 at 5:44
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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.

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    $\begingroup$ How can I prove $F<<m$. $\endgroup$ – Mathlover Nov 20 '18 at 6:04
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    $\begingroup$ If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{\infty}$, so $T(I_E)=0$. $\endgroup$ – Kavi Rama Murthy Nov 20 '18 at 6:08
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    $\begingroup$ I think you contradict by showing that the continuous linear functionals $f\rightarrow \int_0^1fgdm$ and $f\rightarrow \int_0^1fd\delta_x$ are same on the dense set $C[0,1]$ of $L^1[0,1]$ , hence these functionals are equal on $L^1[0,1]$ , hence $\int _Egdm=\int_Ed\delta_x=\delta_x(E)\implies \delta_x<<m$ and which is impossible - am I right? $\endgroup$ – Mathlover Nov 20 '18 at 7:00
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    $\begingroup$ @UserD You are right, but I just used (without proof ) the well known result that if $\int f\, d\mu=\int f\, d\nu$ for all continuous $f$ then $\mu =\nu$ on the Borel sigma algebra. What you are doing is to provide a proof of this fact. $\endgroup$ – Kavi Rama Murthy Nov 20 '18 at 7:17

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