Can we find a unique probability Borel measurable $\mu$ on $\mathcal{C}$ with $\mu(N_s)=\phi(s)$ If $\phi:2^{<\mathbb{N}}\to [0,1]$ satisfies $\phi(\emptyset)=1$ and $\phi(s)=\phi(s^{\widehat{}}0)+\phi(s^{\widehat{}}1)$ for all $s \in 2^{<\mathbb{N}}$.
Can we find a unique probability Borel measurable $\mu$ on $\mathcal{C}=2^{\mathbb{\omega}}$ with $\mu(N_s)=\phi(s)$ where $N_s=\{x\in 2^{\mathbb{N}}: s \subseteq x \}$ for $s \in 2^{<\mathbb{N}}$.
Can anybody give me any hint how one might prove this?. Thanks
 A: The uniqueness is a standard monotone class or $\pi$-$\lambda$ argument.  Suppose $\mu, \mu'$ are two Borel measures satisfying the desired condition.  Let $\mathcal{P} = \{N_s : s \in 2^{<\mathbb{N}}\}$; clearly $\mu, \mu'$ agree for all sets in $\mathcal{P}$.  You can verify that $\mathcal{P}$ is a $\pi$-system, i.e. closed under intersections, and that it generates the Borel $\sigma$-algebra of $2^\omega$.  Then let $\mathcal{L}$ be the collection of all Borel sets on which $\mu, \mu'$ agree.  You can show that $\mathcal{L}$ is a $\lambda$-system, and then the $\pi$-$\lambda$ lemma lets you conclude that $\mathcal{L}$ contains all the Borel sets.
The existence is a special case of the Kolmogorov extension theorem.  There are many different ways to state this theorem but a useful one for us looks like this:

Let $X$ be a standard Borel space.  For each $n$, let $\mu_n$ be a Borel probability measure on the finite product $X^n$.  Suppose that for each $n$ and each Borel set $B \subset X^n$, we have the consistency condition $\mu_n(B) = \mu_{n+1}(B \times X)$.  Then there is a Borel probability measure on the infinite product $X^\omega$ which extends all the $\mu_n$: for each $n$ and each Borel $B \subset X^n$, we have $\mu_n(B) = \mu(B \times X \times X \times \cdots)$.

You can find a proof in these lecture notes of mine (Theorem 10.8), in the "special" case $X = I = [0,1]$ (the general case follows because every standard Borel space is measurably embedded in $[0,1]$).  
Of course we want to apply the theorem with $X = 2$.  For each $n$, define the measure $\mu_n$ on subsets of $2^n$ by $\mu_n(A) = \sum_{s \in A} \phi(s)$.    The conditions you have given on $\phi$ will guarantee that each $\mu_n$ is a probability measure and that the Kolmogorov consistency condition is satisfied.  Now the measure $\mu$ supplied by Kolmogorov's theorem is your desired measure.  (I will leave it to you to verify the details.)
As a general rule, any time you want to produce a measure on an infinite dimensional space, you should look to this theorem.  It's by far the most commonly used tool for this purpose.
A: EDIT: Nate Eldrige's answer is probably better than mine. I'm leaving mine here anyway in case anyone wants a hint how to do it "by hand"...
I take it $s^j$ is the same as what one might (less correctly but perhaps more intuitively) write $(s,j)$? Assuming so:
Seems not entirely trivial, but fairly routine. Say $A$ is the algebra (not sigma-algebra) generated by the $N(s)$. Show that $A$ is equal to the set of all finite disjoint unions of the $N(s)$. That tells you what $\mu(E)$ should be for $E\in A$. Of course you need to show that $\mu$ is well-defined.
Now the slightly tricky part is showing that $\mu$ is a premeasure on $A$. Meaning that if it happens that a set in $A$ is a countable disjoint union of sets in $A$ then the measures add up. You easily reduce this to the case of one of the $N(s)$ being a countable disjoint union of other $N(s)$'s; the tricky part is that that can be a more complicated union than you may realize at first.
(Although it's less tricky than I thought; less tricky than various similar things: Say $N(s)$ is the disjoint union of the $N(s_j)$. Now $N(s)$ is compact, being a closed subset of $2^\omega$, and each $N(s_j)$ is open...)
Now a theorem of Caratheodory shows that $\mu$ extends to a measure on $\mathcal A$, the sigma-algebra generated by $A$. Show that $\mathcal A$ is the Borel algebra.
