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.

• 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 $\mathbb{N}$. – reuns Nov 20 '18 at 5:40
• Can you explain in some details how do I prove my $F$ is not countably additive from your stated fact. – Mathlover Nov 20 '18 at 5:44

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.
• How can I prove $F<<m$. – Mathlover Nov 20 '18 at 6:04
• If $E$ has Lebesgue measure $0$ then $I_E$ is the zero element of $L^{\infty}$, so $T(I_E)=0$. – Kavi Rama Murthy Nov 20 '18 at 6:08
• 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? – Mathlover Nov 20 '18 at 7:00
• @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. – Kavi Rama Murthy Nov 20 '18 at 7:17