Constructing a measure on an uncountable union of disjoint sets I have an intuition that a measure as in the follows does not exist, but I cannot prove it formally. Any hints will be appreciated.
Consider an uncountable sequence of sets 
$$S_a=\{a,a+1\}$$
for $a\in[0,1)$. Let there exist for each $a$ a probability that $s_a=a$, for $s_a\in S_a$, with the notation
 $$p_a=\text{Prob}[s_a=a].$$
Consider the following set 
$$S=\cup_{a\in [0,1)}S_a\equiv [0,2)$$
Let it have a Borel $\sigma$-algebra coming from the subspace topology on $[0,2)$. Does there exists a measure $\mu\in \Delta(S)$ such that
$$\mu(s_a=a\mid S_a)=p_a$$
 for any $S_a$?
 A: You will have this situation in any continuous measure case$^2$:
$$\mu(\{a\}|S_a)=\frac{\mu(\{a\}\cap S_a)}{\mu(S_a)}=\frac{\mu(\{a\})}{\mu(S_a)}=\frac00$$
If you give actual weight to certain points, then you have a discrete measure. You can have linear combinations of discrete measures and continuous measures. But you still have the problem with the continuous measure.
If you have a strictly discrete measure you just end up with a countable set of points with a positive probability$^1$. Say S is this set.
If you have the measure defined on all the $S_a$ such that:
$$\sum_{a\in S}\mu(S_a)=1$$
Then you can just define $\mu(\{a\}):=\mu(S_a)p_a$. Then:
$$\mu(s_a=a|S_a)=\frac{\mu(\{a\})}{\mu(S_a)}=\frac{\mu(S_a)p_a}{\mu(S_a)}=p_a$$
And $\mu(\{a+1\})=\mu(S_a)-\mu(S_a)p_a=\mu(S_a)(1-p_a)$ thus
$$1=\sum_{a\in S}\mu(S_a)=\sum_{a\in S}\big[\mu(\{a\})+\mu(\{a+1\})\big]$$ But constructing discrete measures is easy.
$^1$ There can not be more than a countable set of points with a positive probability. 
Proof: Consider $A_n=\{a\in [0,2): \mu(\{a\})>1/n\}$ it is clear that any $A_n$ contains less than n elements, which means all $A_n$ are finite. But 
$$\bigcup_{n\in \mathbb{N}}A_n=\{a\in [0,2): \mu(\{a\})>0\}$$
And a countable union of finite sets is countable.
$^2$ You can generalize this notion of conditional probability/conditional expected value. In that case such a measure would probably exist. 
If you are familiar with this: https://en.wikipedia.org/wiki/Conditional_expectation. 
Think about a random variable which has a distribution over [0,2) and then consider $Y=mod_1(X)$ You will end up with a distribution of Y over [0,1). And $p_a=P(X=a|X\in S_a)=P(X=a|Y=a)=\mathbb{E}[1_{\{a\}}(X)|Y=a]$ would be a possible definition. You end up with P(X=a|Y) being a continuous distributed random variable. 
As an example consider: $X=1/3\times 1_{[0,1)}+2/3\times 1_{[1,2)}$ then $Y=mod_1(X)$ is uniform over [0,1] distributed. And $p_a=1/3\ \forall a\in [0,1)$
A: If I understood the question correctly, such probability measure $\mu$ does exist provided that $p_a$, as a function of $a$, is (Borel) measurable. 
Just set $f(x) = p_x$, $x\in [0,1)$ and $f(x) = 1-p_x$, $x\in [1,2)$;
$$
\frac{d\mu}{d\lambda}(x) = c f(x)\text{ with }c = \left(\int_0^2 f(x) dx\right)^{-1}.
$$
The measure $\mu$ is as required. Indeed, denoting $X$ the corresponding random variable, we have for almost all $a\in A:=\{x\in [0,1)\mid p_x\notin\{0,1\}\}$,
$$
\frac{\mathrm P (X = a\mid X\in S_a)}{\mathrm P (X = a+1\mid X\in S_a)} = \frac{\frac{d\mu}{d\lambda}(a)}{\frac{d\mu}{d\lambda}(a+1)} = \frac{p_a}{1-p_a},
$$
so for almost all $a\in A$,
$$
\mathrm P (X = a\mid X\in S_a) = p_a,\ \mathrm P (X = a+1\mid X\in S_a)= 1-p_a.
$$
Also, clearly, for almost all $a\in [0,1)$ such that $p_a = 0$,
$$
\mathrm P (X = a\mid X\in S_a) = 0,
$$
and for almost $a\in [0,1)$ such that $p_a = 1$,
$$
\mathrm P (X = a\mid X\in S_a) = 1.
$$
