text: Q:
  
  A probability measure that takes only the values 0 and 1 but not a point mass
  
  Exactly the same question as here: A probability measure that takes only the values $0$ and $1$ but is not a point mass
  Let $\Omega=\{1,2,3,4,5,6\}$, and let $\mathcal{F}=\sigma \left(\{1,2,3,4\},\{3,4,5,6\}\right).$
  Can we show that there exists a probability $P$ on $\Omega$ such that $P(A)$ is zero or one for all $A\in\mathcal{F}$, yet $P$ is not a point mass?
  In A probability measure that takes only the values $0$ and $1$ but is not a point mass, it is explained that it can be solved using the fact that the $\sigma$-algebra cannot differentiate between 1 and 2, but I still do not get it.
  
  A:
  
  A probability measure on a measurable space $(\Omega,\mathcal{F})$ is a function
  $\mu\colon \mathcal{F} \to [0,1]$ satisfying the axioms of a measure, and such that $\mu(\Omega)=1$.
  Here, you have 
  $$\begin{align}
  \mathcal{F} &= \sigma(\{1,2,3,4\},\{3,4,5,6\})\\
  &= \{\emptyset,\Omega,\{1,2,3,4\},\{3,4,5,6\},
  \{5,6\},\{1,2\},\{1,2,5,6\}, \{3,4\}\}
  \end{align}$$
  the last equality using the definition of a generated $\sigma$-algebra. (I think I did not miss any.)
  Therefore, to define your probability measure $\mu$ you only need to assign (consistent) values to the sets of $\mathcal{F}$. You can see that in this case, one can define $\mu$ by 
  $$
  \mu(\{1,2\})=1
  $$
  and the value assigned to the other sets following from properties of a measure:
  \begin{align}
  \mu(\emptyset) &= 0\\
  \mu(\Omega) &= 1\\
  \mu(\{1,2,3,4\} )&= 1\\
  \mu(\{3,4,5,6\} )&= 0\\
  \mu(\{5,6\})&=0\\
  \mu(\{1,2\})&=1\\
  \mu(\{1,2,5,6\})&=1\\
  \mu(\{3,4\})&=0
  \end{align}
  This fully defines $\mu$, yet one can see $\mu$ is not a "point mass." It does not put probability mass $1$ on any singleton, as there is no singleton in the $\sigma$-algebra with mass $1$. (Nor any singleton, for that matter.)
  In particular, indeed $\mu$ "does not differentiate between $1$ and $2$": both points are always together in any event they belong to, and neither of them has a separate well-defined probability mass. $\{1,2\}$ is an event with probability mass $1$, but neither $\{1\}$ nor $\{2\}$ is an event.
  
  