Suppose $K$ is a compact and $T_2$ space and $a \in K$. Define the linear map $$E_a:(C(K), ||.||_\infty) \to \Bbb{C}\\f \mapsto f(a)$$
Then $||E_a||= 1$. So by Riesz-Representation Theorem for positive linear functional on $C(K)$ ($\because E_a$ is positive linear functional) there exists a regular positive Borel measure, say, $\mu_a$ that represents $E_a$ as follows: $$E_a(f)= \int f ~\mathrm{d}\mu_a; ~ (f \in C(K))~~~~~~~~~~~~~~~~~~~~~~~~(*)$$ Also here $\mu_a(K)=1$.
From $(*)$, we have $\int f ~\mathrm{d}\mu_a=f(a)$, for all $f \in C(K)$.
My question: Is it true that $\mu_a = \delta_a$; where $\delta_a$ is the Dirac delta measure on the Borel $\sigma$-algebra of $K$?
I assume the definition: for any Borel set $E$, $~\delta_a(E) = 1$ if $a \in E$ and $\delta_a(E) = 0$ if $a\not\in E$.
My attempt. I tried in this way, for a Borel set $E$, since $C(K)$ is dense in $L^1(\mu_a)$, there is a sequence $(f_n)$ in $C(K)$ such that $f_n \to_{L^1(\mu_a)} \chi_E$ and this implies there is a subsequence $(f_{n_k})$ which converges pointwise $\mu_a-$almost everywhere on $K$ to $\chi_E$. Thus (as $\mu_a$ is a probability measure) using DCT we have the following: $$\mu_a(E) =\int\chi_E ~d\mu_a=\int\lim_k f_{n_k} ~d\mu_a=\lim_k\int f_{n_k} ~d\mu_a= \lim_k f_{n_k}(a)=\chi_E(a)$$
This proves my claim. Is the solution correct? and If $K$ was locally compact only then this same argument will prove that the ealuation by $a \in K$ functional on $C_0(K)$ corresponds to the $\delta_a$. Is it right?
Thanks.
EDIT As mentioned in the comments the last equality, $\lim_k f_{n_k}(a)=\chi_E(a)$ is not necessarily true always.