existence of a borel probability measure on $[0,1]$ such that $\int f d\mu=\lim_{k\to\infty}\frac {1} {N_k} \sum_{i=1}^{N_k}f(x_i)$ given sequence Hi I'm really suck with this one, i would really appreciate it if any one can help me with this!
prove that for $\{x_i\}\subseteq[0,1]$ there are $1\le N_1<N_2...$ and a probability measure $\mu$ on $([0,1],\cal B_{[0,1]})$ such that for any $f\in C([0,1])$ we have $\int f d\mu=\lim_{k\to\infty}\frac 1 N_k \sum_{i=1}^{N_k}f(x_i)$  
clue: set a dense sequence $\{f_n\}\subseteq C([0,1])$ and use a diagonal argument to find such $N_k$'s then use Riesz Representation theorem 
so i tried different things to find those $N_k$'s but didn't manage to get any result. once i get those it will be easy to show  that limit is a linear functional and then use Riesz Representation theorem.
 A: I'm not sure what you mean by $C_c([0,1])$, but you can use $C[0,1]$.
Let $f_j$ be a dense sequence in $C[0,1]$ (this being a separable Banach space).
The sequence $\frac{1}{n} \sum_{i=1}^n f_1(x_i)$ is bounded, so for some increasing sequence $N(1,1), N(1,2), N(1,3), \ldots$ of positive integers, $\frac{1}{N(1,n)} \sum_{i=1}^{N(1,n)} f_1(x_i)$ converges.
Take a subsequence $N(2,1), N(2,2), N(2,3), \ldots $ of this such that $\frac{1}{N(2,n)} \sum_{i=1}^{N(2,n)} f_2(x_i)$ also converges (i.e. each $N(2,j)$ is some $N(1,k)$ with $k \ge j$), and continue in this way.  The diagonal sequence
$N(1,1), N(2,2), N(3,3), \ldots$ has the property that 
$\frac{1}{N(k,k)} \sum_{i=1}^{N(k,k)} f_j(x_i)$ converges for all $j$.
A: $$f \mapsto \mu_n(f) = \frac{1}{n}\sum_{i=1}^n f(x_i)$$ is a probability measure. So you have a sequence of probability measures $(\mu_n)$ and we want to show that is has a convergent subsequence ( convergence in the weak topology). Take $(f_l)$ a dense family in $C[0,1]$. 
Start with $l=1$. 
The sequence  $(\mu_n(f_1))_{n\in \mathbb{N}}$ is bounded in absolute value by $||f_1||$ and so it has a convergent subsequence indexed by $\mathbb{N}_1 \subset \mathbb{N}$.
The sequence $(\mu_n(f_2))_{n\in \mathbb{N}_1}$ is bounded by $||f_2||$ so it has a convergent subsequence indexed by $\mathbb{N}_2 \subset \mathbb{N}_1$.
The sequence $(\mu_n(f_3))_{n\in \mathbb{N}_2}$ is bounded by $||f_3||$ so it has a convergent subsequence indexed by $\mathbb{N}_3 \subset \mathbb{N}_2$.
$\ldots$
and so on. We get a sequence of infinite subsets of $\mathbb{N}$ 
$$\mathbb{N} \supset \mathbb{N}_1 \supset \mathbb{N}_2 \supset \ldots $$
with the property that 
$$\mu_n(f_l)_{n\in \mathbb{N}_l}$$
is convergent. 
There exists now $\tilde{ \mathbb{N}}$ an infinite subset of $\mathbb{N}$ with the property that 
$$\tilde{ \mathbb{N}} \backslash \mathbb{N}_l$$ is finite for all $l\ge 1$. (so for all $l$, $\tilde{ \mathbb{N}} \subset \mathbb{N}_l$ except for finitely many elements). One way to construct $\tilde{ \mathbb{N}}$ is as follows: at step $l$ add the to the set the smallest element of $\mathbb{N}_l$ that was not chosen before (Cantor diagonal argument).
From before it follows that 
$$\mu_n(f_l)_{n\in \tilde{ \mathbb{N}}}$$
is convergent for all $l$. 
Using the density of $(f_l)_l$ it is easy to show that for every $f \in C[0,1]$ the sequence
$$\mu_n(f)_{n\in \tilde{ \mathbb{N}}}$$
is convergent ( show that it is Cauchy).
Define
$$\mu(f)\colon =\lim_{n\in \tilde{ \mathbb{N}}} \mu_n(f)$$
a positive linear functional on $C[0,1]$. Now use Riesz' theorem. 
