Existence of probability measure, $\lim_{N\to\infty}\frac{1}{N}\sum_{j=1}^N f(x_j)$ Problem statement:

Let $x_j$ be a sequence in $[0,1]$.  Consider  $$\lim_{N\to\infty}\frac{1}{N}\sum_{j=1}^N f(x_j)\hspace 2cm (\star).$$  If ($\star$) exists for all $f\in C[0,1]$ then there is a positive Borel measure $\mu$ on [0,1] with $\mu([0,1])=1$ such that ($\star)=\int_{[0,1]} f d\mu$ for all $f\in C[0,1]$.  Furthermore, if $x_j\to x$ then ($\star$) exists for all $f$, and the above applies.  Which $\mu$ do we get?

Note:
There are other similar questions like this out there, but they don't ask you to show that $\mu([0,1])=1$ and/or ask about a subsequence instead of just a convergent subsequence.
Attempt:
Given that ($\star$) holds for all $f$ I can show that the the map that sends $f$ to ($\star$) is a positive linear functional.  Applying (one version) of the Riez Representation Theorem gives a measure $\mu$ as needed.  However, how do I know $\mu([0,1])=1$?
As for what happens when $x_j\to x$, I suspect since $f(x_j)\to f(x)$ and we are averaging that $(\star)=f(x)$.  

As for $\mu([0,1])=1$, consider the constant function $f(x) = 1$ on $x\in [0,1]$. 
For the second statement, if $f\in C[0,1]$ then $f(x_j) \rightarrow f(x)$. 
In this case, the limit will converge to $f(x)$. 
Thus, the measure is the Dirac measure at the point $x$. 
