Using a probability measure, P, defined on uncountable sets to construct a probability measure, P' on singleton P-null sets

Question

Let $\Omega$ be an uncountable set and $(\Omega, \mathcal{F},P)$ be a probability space built on $\Omega$.

Let $S \subset \{A \in \mathcal{F}: P(A)=0,\;|A|=1\}:|S|<\infty$ be a finite subset of the class of singleton P-null sets in $\mathcal{F}$.

I am trying to use $S$ to construct a new probability space from $(\Omega, \mathcal{F},P)$: $(S,\sigma(S),P')$

Since $\sigma(S)$ is discrete, I can define a potential probability measure for $P'$ on the above space by specifying its values for each sample point in $S$: $P'(s_i)=\lim\limits_{n\rightarrow \infty} \frac{P(Q_{in})}{ P \left(\bigcup\limits_{j \in \{1...|S|\}}Q_{jn}\right) }$, where $Q_{in} \subset \mathcal{F}:Q_{ij}\supset Q_{ik}\;\forall( j\leq k)$ and $\lim \inf Q_{in} = s_i\in S$

Note that $P'$ can be extended to $\sigma(S)$ via additivity.

Question 1: Under what conditions does $P'$ exist?

Question 2: What additional conditions are needed for the above to be true if $|S| = \infty$?

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My thinking so far

The use of the limit in $P'$ requires a metric space $(\Omega,d)$, such that $P(B_r(s) :=\{\omega \in \Omega: d(s,\omega)<r \})>0 \;\forall(s\in S, r>0)$, which allows the above limit to define a probability measure on $\sigma(S)$. However, I'm not sure if this condition is sufficient, necessary or both?

I initially developed the above notions by working with the simple probability space $([0,1],\mathcal{B}([0,1]),P(A\subset \mathcal{B}([0,1]) = \lambda(A))$. If $|S|=M$, then we have a set of $M$ distinct points in $[0,1]$. We can define $Q_{in} := [(s_i - \frac{s_i}{n}),(s_i+\frac{1-s_i}{n})] \;\forall i:s_i \in S$. This results in a possible candidate for $P'$:

$P'(s_i) = \lim \limits_{n\rightarrow \infty} \frac{\lambda([(s_i - \frac{s_i}{n}),(s_i+\frac{1-s_i}{n})])}{ \lambda \left(\bigcup\limits_{j \in \{1...|S|\}}[(s_i - \frac{s_j}{n}),(s_j+\frac{1-s_j}{n})]\right) }$.

The numerator is easy to calculate for all $n$; however, the $Q_{in}$ are not initially disjoint so the denominator is complicated at the beginning of the sequence. However, since the elements of $S$ are countable and distinct, $\exists n_0: Q_{in}\cap Q_{jn} = \emptyset \; \forall (i\neq j, n>n_0)$. Therefore, we can restrict analysis of the above limit to $n>n_0$ without loss of generality. The benefit of doing so is that the union of the $Q_{in}$ becomes a disjoint union, and we can get a simple formula for the denominator. Specifically,

$P'(s_i) = \lim \limits_{n\rightarrow \infty} \frac{\lambda([(s_i - \frac{s_i}{n_0+n}),(s_i+\frac{1-s_i}{n_0+n})])}{ \lambda \left(\biguplus\limits_{j \in \{1...|S|\}} [(s_i - \frac{s_j}{n_0+n}),(s_j+\frac{1-s_j}{n_0+n})]\right) } = \frac{1/(n_0+n)}{M/(n_0+n)} = \frac{1}{M}\; \forall s_i \in S$

Thus, in this very simple example, $P'$ converges at $n_0<\infty$.

It seems straightforward to extend this to a countable set on the domain of a non-uniform distribution function F (e.g., gaussian). The denominator is guaranteed to be $\leq 1$ since F is a distribution; therefore, the denominator will always be a subset of the domain of F and the numerator, being a subset of the union in the denominator, will always be $\leq$ the denominator. Therefore, it seems like this is quite general, since all random variables map to the real numbers (hence we have a metric space).

I am not sure if I am missing some, possibly pathological case, where you cannot define $P'$ from $P$

Answer 1

The condition you have is not sufficient. Although the question has some wording issues, I can answer it as I think you intended it.

Suppose $(\mathbb{R},\mathcal{B},P)$ is our probability triple, where $P([0,\frac{1}{2^n}])=4^{-\lceil\frac{n}{2}\rceil}$. In this case, there can be no limit defining $P'(\{0\})$ if the sets you are using in your limit are balls size 2^{-n}$.

A simpler counterexample (that you may be more or less happy with):

$P$ has probability density function $f(x)=\left\{\begin{array}\,e^{-x}&x>0\\0&x<0\end{array}\right.$. In this case, depending on whether the sets tending down to zero are to the left or the right, you will get different answers for $P'(\{0\})$

Note: You can always define some $P'$ based on some specific sequence $(Q_{in})_{i=1,n=1}^{i=|S|,n=\infty}$ using your definition; it just isn't unique or useful. The way to define it in general is to first choose any sequence $Q_{in}$ with $P(Q_{in})>0$ for all $i$ and $n$, and use Bolzano-Weierstrauss to find a convergent subsequence of the sequence of $|S|$-tuplets $P(Q_{in})$, and name this sequence (Q^*_{in}). Then define $Q^{*,1}_{in}=Q^*_{in}$ and define P'_1:{{s_1}}\rightarrow [0,1] by $P'_1({s_1})=1$.

Inductively define $P'_k:\{\{s_i\}|i\le k\}\rightarrow [0,1]$ by choosing an $s_i$ with $P'_{k-1}(s_i)\ne 0$ and a subsequence $Q^{*,k}_{in}$ of $Q^{*,k-1}_{in}$ s.t. $\frac{P(Q^{*,k}_{in})}{P(Q^{*,k}_{kn})}$ has a limit in $[0,\infty]$. Then we can define $P'_k(s_i)$ just by keeping everything in the appropriate ratios. $P'_{|S|}$ will be a probability distribution defined by your limit from the sequence of sets $(Q^{*,|S|}_{in})_{i=1,n=1}^{i=|S|,n=\infty}$