A basic question on probability measure Can we have a probability measure in $[0,1]$ such that it assigns $0$ to every singleton but for other sets it assigns either $0$ or $1$. Hint Enough.  
 A: No: $S_0 := [0, 1]$ has measure $1$. One of $[0, \frac{1}{2}]$ or $[\frac{1}{2}, 1]$ has measure $1$ and the other has measure $0$. Let $S_1$ be the one with measure $1$. Inductively define $S_i$ to be the half of $S_{i - 1}$ with measure $1$. Show $\bigcap_i S_i$ is a singleton with measure $1$.
A: You seem to be looking for a two-valued probability measure that is defined for all subsets of $[0,1]$. The set $[0,1]$ has the same cardinality as the set of infinite sequences of $0$'s and $1$'s. It is more convenient to work with that set. 
Let $A_n$ be the set of sequences that are $0$ at $n$, and let $B_n$ be the set of sequences that are $1$ at $n$. Then one of $A_n$ or $B_n$ has measure $1$. Call it $C_n$. The intersection of the $C_n$ has measure $1$. But this intersection is a singleton. 
A: I'm not quite sure but only answering because you said a hint is enough.
Let $\mu$ be that measure. Define $f:[0,1]\to[0,1]$ as $f(x)=\mu\left([0,x]\right)$. Then $f$ is an increasing function, $f(0)=0$ (because $\mu$ assigns 0 to every singleton) and $f(1)=1$ (because it's a probability measure). 
Consider $x_1=\inf\{x\in[0,1]:f(x)=1\}$ and $x_2=\sup\{x\in[0,1]:f(x)=0\}$. Of course, $x_2\le x_1$ but can this be a strict inequality, $x_2<x_1$?
