Hi I found the following exercise in the Dudley's book and I'd like to see if my answer is correct; the last part is what I'm not entirely sure, since I'm not completely familiar with this kind of arguments and there is no examples in the book. Thanks in advanced :)
EX: Let $(X,\mathcal{T})$a topological space and $p\in X$. A punctured nbhd of $p$ is a set $B$ which includes $A\setminus \{p\}$ for some nbhd $A$ of $p$. Let $m(A)=1$ if $A$ is punctured nbhd of $p$ and $m(A)=0$ if is disjoint to some punctured nbhd of $p$.
a) Show that $m$ is defined in an algebra.
b) If $(S,d)$ is a metric space and $\{p\}$ is not open, show that $m$ is finitely additive.
c) If $(S,d)$ is a metric space show that $m$ cannot be extended to a measure.
(a) Let $\mathcal{A}$ the domain of $m$. It's clear that $X\in \mathcal{A}$. Let $A\in \mathcal{A}$, if $A$ is punctured so $X\setminus A\in \mathcal{A}$, if $A$ is disjoint to some punctured $U$, $U\subset X\setminus A$, so $X\setminus A$ is punctured. Let $A_n\in \mathcal{A}$ a finite sequence, we consider two cases: If all $A_n$ are disjoint to some punctured nbhd of $p$, so is the union (consider the -finite-intersection of all the puncture nbhd for which each $A_n$ is disjoint). If $A_k$ is punctured nbhd of $p$, so is the union. In any case the union is in $\mathcal{A}$
(b) It's clear that $m(\varnothing) =0$. Let $A_n\in \mathcal{A}$ a finite disjoint sequence. If $A_1$ is punctured, as the others are disjoint to $A_1$, $m(\bigcup_n A_n)=1=m(A_1)+\sum_{n\not=1}m(A_n)$. If each one is disjoint to some punctured nbhd of $p$, so is the union: let $U_n$ be a punctured nbhd of $p$ s.t., $A_n \cap U_n=\varnothing$. There is an open ball $B(p,r_n)\setminus \{{p\}}\subset U_n$ consider $r= \text{min} \{r_n\}>0$, so the ball $B(r,p)$ is disjoint to each $A_n$ and for instances of the union, i.e., $m(\bigcup_n A_n)=0=\sum_n m(A_n)$.
(c) Let $A_n:=\{x:d(x,p)=1/n\}$ so each $A_n$ is disjoint to some punctured set, consider $B_{(S,d)}(p,1/n)$, but $A=\bigcup_n A_n$ is a punctured nbhd of $p$ and so $$1=m(A)\not=\sum_n m(A_n)=0$$