Im a little confused with regards to the difference between a probability measure and the outer measure.
So firstly to set the scene, given $\Omega = \{ 0,1\}$ we have $\mathcal{F} = \{\emptyset, \{0\},\{1\},\Omega\}$ so firstly clearly $\mathcal{F}$ is a $\sigma$ - algebra.
now on this $\Omega$ we define $\mu(0) = a$ and $\mu(1) = b$ for $\mu$ to be a valid measure on $(\Omega,\mathcal{F})$ we need $\mu(\emptyset) = 0$ and we requre $\mu$ to be countably additive. so in our case we require
$\mu(\Omega) = \mu(0) + \mu(1) = a + b$
The measure $\mu$ is then a probability measure if $a+b=1$
Here is my confusion we know from the definition of the outer measure that $m^*(\{x\})=0$ for $x \in \mathbb{R}$ so since $\mu(0) = a$ and $\mu(1) = b$ does that mean $\mu$ is not an outer measure do we not require $\mu(0) = \mu(1) = 0$ for $\mu$ to be an outer measure, since the outer measure of any singleton is $0$.