A (counter)-example in probability theory.

Question

Let $\mathcal{D}$ the set of dyadic intervals (open on the right and closed on the left) in $[0,1)$ (i.e. $I\in\mathcal{D}$ if and only if $\exists n\in\mathbb{N_0}, \exists k\in\{1,...,2^n\}, I=[\frac{k-1}{2^n},\frac{k}{2^n})$). I'm wondering if there exists a sequence $(a_n)_{n\in\mathbb{N}}\subset[0,1)$ such that $$\forall I\in\mathcal{D},\exists \lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N\chi_I(a_n)=:\mathbb{P}(I)\in[0,1],$$ but such that there exists a sequence of mutually disjoint dyadic intervals $(I_k)_{k\in\mathbb{N}}\subset\mathcal{D}$ whose union is a dyadic interval $I\in\mathcal{D}$ and $$\mathbb{P}(I)\neq\sum_{k=1}^\infty\mathbb{P}(I_k).$$ Notice that if the disjoint sequence is replaced by an arbitrary finite disjoint family $I_1,...,I_n$, the fact that equality holds in the previous formula is immediate from the definition of $\mathbb{P}$, but, for the denumerable case, a switch of limits problem pops up, so I suspect that such a "pathological" example could exists.

Answer 1

Let $(a_n)$ be any sequence in $[0,1)$ converging to $1$. Then $\mathbb{P}(I)$ is defined for any dyadic interval $I$, and is $1$ if $\sup I=1$ and is $0$ otherwise. This fails to be countably additive, since it vanishes on each of $[0,1/2),[1/2,3/4),[3/4,7/8),\dots$ but is $1$ on their union.