$J$ is a semialgebra of subsets of a sample space $\Omega$ and $\mathbb{P} : J \to [0,1]$, with $\mathbb{P} (\Omega)=1.$

If $\{D_j\}$ is a finite collection of disjoint elements of $J$, such that $\bigcup_ {j \leq n} D_j \in J$, then $$\mathbb{P} \left (\bigcup_{j \leq n} D_j \right)=\sum_ {j \leq n}\mathbb{P}(D_j). \qquad (1)$$

Further, for any collection $\{A_n\}$, where the $A_i$ are all finite unions of elements of $J$, such that $A_{n+1} \subseteq A_n$ and $\bigcap_{n} A_n=\emptyset$, then $$\lim_{n \to \infty} \mathbb{P}(A_n)=0.$$

Show that $\mathbb{P}$ also satisfies (1) for countable collections of disjoint sets $\{D_n\}$.

[Edit: question removed as counterexample was not applicable, as pointed out by Eric in comments.]

I am also wondering if there is any more general significance to the conditions placed on the collection $\{A_n\}$?

I have broken the proof into two parts:

Lemma 1: $\mathbb{P}$ is countably subadditive: $$\mathbb{P} \left( \bigcup_{j} D_j \right) \leq \sum_{j} \mathbb{P} (D_j).$$


Note that we assume $$ \bigcup_{j\geq n} D_j \in J,$$ for all $n\geq 1$.

If we let $$A_n= \Big( \bigcup_{j} D_j \Big) \setminus \Big(\bigcup_ {j \leq n} D_j \Big),$$ then it is not hard to show that the collection $\{A_n\}$ satisfies the given criteria to ensure that $$\lim_{n \to \infty} \mathbb{P}(A_n)=0.$$

We also have that $$A_n = \bigcup_{j >n} D_j.$$

Putting these together, for any $\epsilon>0$ we have:

$$ \mathbb{P} \left(\bigcup_ {j \leq n} D_j \right)+\mathbb{P}(A_n)= \mathbb{P} \left( \bigcup_{j} D_j \right) \leq \sum_ {j \leq n} \mathbb{P} (D_j) + \epsilon \quad (n\geq N),$$

where $N$ is some fixed integer. Thus, we can conclude

$$\mathbb{P} \left( \bigcup_{j} D_j \right) \leq \sum_{j} \mathbb{P} (D_j).$$


Lemma 2: $\mathbb{P}$ is countably superadditive: $$\mathbb{P} \left( \bigcup_{j} D_j \right) \geq \sum_{j} \mathbb{P} (D_j).$$


Let $A,B \in J$ such that $A \subseteq B$. Since $J$ is a semialgebra, we can write $$A^c=E_1 \dot \cup E_2 \dot \cup \cdots \dot \cup E_k,$$ for disjoint $E_1, E_2, \cdots, E_k \in J$.

By the finite additivity property of $\mathbb{P}$:

$$\mathbb{P} (B) = \mathbb{P} (A) + \mathbb{P} (B \cap E_1) + \cdots + \mathbb{P} (B \cap E_k) \geq \mathbb{P} (A).$$

So, for any positive integer $n$, we have

$$\mathbb{P}\left( \bigcup_{j} D_j \right ) \geq \mathbb{P} \left( \bigcup_{j\leq n} D_j \right) = \sum_{j \leq n} \mathbb{P} (D_j).$$

Since this result holds for all $n\in \mathbb{N}$, we conclude that $$\mathbb{P} \left( \bigcup_{j} D_j \right) \geq \sum_{j} \mathbb{P} (D_j).$$



The statement you have proved is not the same as the statement in the question you linked! In your statement, the $A_n$ are required only to be finite unions of elements of $J$. In the other question, they are required to be elements of $J$. George's counterexample does not satisfy your stronger hypotheses.

Your proof appears to me to be correct.

  • $\begingroup$ Ah, I should have seen that (especially as that was the first place I looked for an error in my proof)! Thanks for the reply. $\endgroup$ – Moed Pol Bollo Jul 21 '18 at 4:35

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