Dirac measures on the real line Let $p \in R$. 
Define a measure on the set of subsets of $R$ as follows: 
$\delta_p(E) = 1$ if $p \in E$ and $\delta_p(E) = 0$ otherwise for some $p \in R$. 
Question: Suppose we have a measure $m$ defined on the subsets of the real line that takes only the values $0$  and $1$. 
Prove that $m$ is either $0$ everywhere or it is equal to the $\delta_p$ for some $p$. 
My approach : 
Suppose we have $m$ that takes only the values $0$ and $1$. 
By countable additivity $m$ cannot take $1$ on any two disjoint sets. 
So there is only one set call it $X$ on which $m$ can be $1$. 
Now we have to find a $p$ in this $X$ to make it the same as $m$ but I don't see how? 
 A: This is not completely trivial: after all, it is conceivable that $m$ is $0$ at every point without being the zero measure.
Assuming $m\neq 0$, one has to show that $m(\{ p\})=1\;$ for some point $p\in\mathbb R$.  
First, note that it is not possible that every $x\in\mathbb R$ has an open neighbourhood $V_x$ such that $m(V_x)=0$. Indeed, since $\mathbb R$ has a countable basis of open sets, it would follow that $\mathbb R=\bigcup_{n\in\mathbb N} V_n$, where $(V_n)$ is a sequence of open sets such that $m(V_n)=0$ for all $n$; and hence $m=0$ by countable sub-additivity.
So there is some point $p\in \mathbb R$ such that $m(V)>0\;$ for every neighbourhood of $p$; so $m(V)=1\;$ for any such neighbourhood. Taking $V_n=(p-\frac1n, p+\frac1n)$, so that $\bigcap_n\downarrow V_n=\{ p\}$, it follows that $m(\{ p\}) =\lim(V_n)=1$.
Note that the required conclusion does not hold true if you only assume that $m$ is finitely additive: take any nonprincipal ultrafilter $\mathcal U$ on $\mathbb R$, and consider the finitely additive measure $m$ defined by $m(A)=1$ if $A\in\mathcal U$ and $m(A)=0$ if $A\not\in\mathcal U$.
