Proposition about continuity of measures and sigma-additivity, need a proof.

Question

I have the following in my notes:

Proposition: Let $\{E_j\}_{j=1}^{\infty}$ be a monotonically decreasing family of sets, such that $\bigcap_{j=1}^{\infty} E_j = \emptyset$, then $\lim_{j\to\infty} \mathbb{P}(E_j) = 0$

Where $\mathbb{P}$ denotes a probability measure.

I have an attempt of proof of this in my notes but I'm struggling to understand that, and would like to see if I can find something around. Too bad, I don't know if this proposition is known under some common name and therefore I don't know how to look for it on the web. Any suggestion of where to find something about it would be really appreciated. Thanks.

EDIT: Here is my attempt. Let's call $O_j$ the part between $E_j$ and $E_{j+1}$, then $O_j$ and $O_i$ are disjoint $\forall i,j$ such that $i\ne j$. The union $\bigcup_{j=n}^{\infty} O_j$ can be re-written in terms of the sets, which are such that the $E_j \supset E_{j+1}$ and therefore $O_j = E_j \setminus E_{j+1}$ which implies that

$\bigcup_{j=n}^{\infty} O_j = \bigcup_{j=n}^{\infty} E_j \setminus E_{j+1}$

But then again because $E_j \supset E_{j+1}$ you can write that

$\bigcup_{j=n}^{\infty} O_j = E_n \setminus \bigcap_{j=n}^{\infty}E_j = E_n$

because as hypotesis we had that $\bigcap_{j=1}^{\infty} E_j = \emptyset$

Passing to the probabilities we have that

$\mathbb{P}(E_n) = \mathbb{P}(\bigcup_{j=n}^{\infty} O_j) = \mathbb{P}(E_j \setminus E_{j+1}) $

Using the subadditivity of the probability measure and the fact that $O_j$ are pairwise disjoint we can write that

$\mathbb{P}(E_n) = \sum_{j=n}^{\infty} \mathbb{P}(E_j \setminus E_{j+1}) = \mathbb{P}(E_n) - \lim_{j\to\infty} \mathbb{P}(E_j)$

Which means that $\mathbb{P}(E_n) = \mathbb{P}(E_n) - \lim_{j\to\infty} \mathbb{P}(E_j) \to \lim_{j\to\infty} \mathbb{P}(E_j) = 0$

Answer 1

Your proof looks fine to me. I would just word a few steps differently: At the start you reason that since $E_j \supset E_{j+1}$ we have that $O_j = E_j \setminus E_{j+1}$

However, I would just start out the proof by defining $O_j := E_j\setminus E_{j+1}$, since otherwise the beginning of the proof would be a little bit confusing. If you have to present your proof to someone, I would just explain the equality $$\bigcup_{j=n}^\infty O_j = E_n \setminus \bigcap_{j=n}^\infty E_j = E_n$$ in a bit more detail, by first saying that

$$\bigcup_{j=n}^k O_j = E_n \setminus E_{k+1} = E_n \setminus \bigcap_{j=1}^{k+1}E_{j}$$ and then passing to the limit

$$\bigcup_{j=n}^\infty O_j = E_n \setminus \bigcap_{j=1}^\infty E_j = E_n$$

In the last line, I think you made a typo. It should be $$\lim_{n \rightarrow \infty}\mathbb{P}(E_n)= \lim_{n\rightarrow \infty} [\mathbb{P}(E_n)-\lim_{j \rightarrow \infty}\mathbb{P}(E_j)] = 0$$