Real analysis radon measure Show that a Dirac delta measure on a topological space is a Radon measure.
Show that the sum of two Radon measures is also a Radon measure. 
 Please help me.
 A: Dirac measure is a measure $\delta_{x}$ on a topological space $X$ (with a Borel $\sigma$-algebra $B(X)$ of subsets of $X$) defined for a given $x \in X$ and any Borel set $A  \subseteq  X$ by $\delta_{x}(A)=1$ iff $x \in A$. Notice that $\{ x\}$ like $\emptyset$ is compact subset of $X$. For each  $A \in B(X)$ and for $\epsilon>0$ we take under $F_{\epsilon}$ the set $\{x\}$ if $x \in A$ or $\emptyset$ if $x \notin A$. Then we get
$\delta_{x}(A \setminus F_{\epsilon})=0<\epsilon$ which means that $\delta_{x}$ is radon measure(It is obviousthat $F_{\epsilon}\subseteq A$ and $F_{\epsilon}$ is compact).
Now let $\mu_1$ and $\mu_2$ be two Radon measures in $X$. Let $A \in B(X)$. Since $\mu_i$ is radon measure, for $\epsilon>0$  there is a compact subset $F^{(i)}_{\epsilon} \subseteq A$ such that $\mu_i(A \setminus F^{(i)}_{\epsilon})<\frac{\epsilon}{2}$ for $i=1,2$. It is obvious that $C_{\epsilon} :=F^{(1)}_{\epsilon} \cup F^{(2)}_{\epsilon}$ is compact subset of $A$ such that
$$(\mu_1+\mu_2)(A \setminus C_{\epsilon})=\mu_1(A \setminus C_{\epsilon})+\mu_2)(A \setminus C_{\epsilon})\le \mu_1(A \setminus F^{(1)}_{\epsilon})+\mu_2)(A \setminus F^{(2)}_{\epsilon})<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
The latter relation means that $\mu_1+\mu_2$ is Radon measure in $X$.
A: you also can apply the Reisz representation theorem. $\delta(f)=f(x_0)$ is a positive linear functional, then it must be a Radon measure. Second problem is obvious.
