Equal integration of continuous functions gives equality of measures Let $X$ be a compact Hausdorff space and $C(X)$ denotes the space of all continuous functions from $X$ to $\mathbb C$. Let $\mu$ and $\nu$ are two Borel complex measures defined on $X$ such that $\displaystyle \int f d\mu= \displaystyle \int f d\nu$ for all $f \in C(X).$ Then I want to show that $\mu=\nu.$
My approach: Let $\Delta$ be the sigma-algebra of all Borel sets of $X$. In order to show that $\mu=\nu$, we have to show $\mu(U)=\nu(U)$ for all $U\in \Delta$. For that I have to show $\displaystyle \int 1_U d\mu= \displaystyle \int 1_U d\nu$ but indicator functions are not always continuous. So I have to show that, there exists a net $(f_\lambda)_\lambda$ of continuous functions in $X$ such that $~\displaystyle\lim_\lambda f_\lambda = 1_U.$ Then we are done. If my approach is on the right direction please help me to complete this or if I am in wrong direction please  help me to solve this with some different approach. Thank you for your time. Thanks.
 A: In what follows I will assume that both $\mu $ and $\nu $ are regular measures.  With this we will prove that $\mu =\nu $.
For ease of reference I will denote $\mu $ and $\nu $ by $\mu _1$ and $\mu _2$,  respectively.
Fix any measurable set $E\subseteq X$, and any $\varepsilon >0$.  Then for each
$i=1,2$, we may
use regularity to produce a closed set $K_i$, and an open set $U_i$, with
$$
K_i\subseteq E\subseteq U_i,
$$
and such that $|\mu _i|(U_i\setminus K_i) < \varepsilon $.
By replacing both $K_1$ and $K_2$ with $K_1\cup K_2$, as well as replacing both $U_1$ and $U_2$ with  $U_1\cap U_2$, we may suppose
without loss of generality that $K_1=K_2=:K$, and that $U_1=U_2=:U$.
Observing that $K\subseteq U$, we may use Urysohn to find a continuous function $$f:X\to [0,1],$$
such that $f\equiv 1$ on $K$, and $f\equiv 0$ outside $U$.
It then follows that
$$
  \Big|\mu _i(E) - \int _X f\ d\mu _i\Big|$$$$=
  \Big|\int _X (1_E-f)\ d\mu _i\Big|$$$$=
  \Big|\int _{U\setminus K} (1_E-f)\ d\mu _i\Big|\leq
  |\mu _i|(U\setminus K) < \varepsilon .
  $$
Consequently
$$
  \Big|\mu _1(E) - \mu _2(E)\Big|=
  \Big|\mu _1(E) - \int _X f\ d\mu _1  + \int _X f\ d\mu _2 - \mu _2(E)\Big| <2\varepsilon .
  $$
Since $\varepsilon $ is arbitrary we get $\mu _1(E)=\mu _2(E)$.
