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Consider $\mathbb{R}$ and the collection $\mathcal{F}$ which consists of all sets $F \subset \mathbb{R}$ for which either $F$ or $F^c$ is countable. I proved that $\mathcal{F}$ is a $\sigma$-algebra. Let $\mu$ be counting measure on $\mathcal{F}$ and let $\nu(A) = 0$ if $A$ is countable, and $\nu(A) = 1$ if $A$ is uncountable. Explain why $\nu$ is a measure on $\mathcal{F}$.

I am stuck with the explaining part, as $A = [0,1]$ and $B = [2,3] $ gives us: $\nu(A \cup B) = 1$ and $\nu(A) + \nu(B) = 2$, while they should be equal for $\nu$ to be a measure. What do I fail to see?

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$A$ and $B$ are not in $\mathcal F$.

They are not measurable, so this should come as no surprise that their "measures" are not additive -- the same is true for standard Lebesgue measure: you can have disjoint non-measurable sets which have full outer measure (indeed, even with respect to Lebesgue measure, you can find pairwise disjoint families of such sets of size continuum).

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    So every disjoint union of sets of $\mathcal{F}$ contains at most one uncountable element. – 2017-02-13
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    Yes, because the complement of any uncountable element must be countable. – 2017-02-13
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the problem is that $\nu$ is a measure in $\mathcal{F}$, not everywhere