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I've followed the following lemma, i.e. Borel-Cantelli lemma.

Borel-Cantelli

Let $A_i \in \mathcal{A}, i \in \mathbb{N}$. Then, $\sum_{i = 1}^{\infty} \mathbb{P} (A_i) \Rightarrow \mathbb{P}(\cap_{n \in \mathbb{N}} \cup_{m \geq n} A_m) = 0$

In order to prove the Borel-Cantelli lemma, I followed the way given below:

Since

$\cup_{m \geq n} \ A_m \searrow \cap_{n \in \mathbb{N}} \cup_{m \geq n} A_m$ as ${n \rightarrow \infty}$

the continuity from above of $\mathbb{P}$ implies that $\mathbb{P}(\limsup_{n \rightarrow \infty} A_n) = \lim_{n \rightarrow \infty} \mathbb{P}(\cup_{m \geq n } A_m) \leq \lim_{n \rightarrow \infty} \sum_{m=n}^{\infty} \mathbb{P} (A_m) = 0$

since $\sum_{m=1}^{\infty}\mathbb{P}(A_m) < \infty$.

I'm stuck at the last part:

$\sum_{m=1}^{\infty}\mathbb{P}(A_m) < \infty$ implies that $\lim_{n \rightarrow \infty}\sum_{m=n}^{\infty} \mathbb{P} (A_m) = 0$

I can guess the result from (a kind of) insight, but I cannot find some "clear" ways to understanding the last part where I'm stuck.

Anyone can answer my question? :)

Jinrae Kim
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1 Answers1

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Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that $\sum_{n=1}^{\infty} a_n$ is convergent with sum $S$. Then, we claim that $\lim_{n\to \infty} \sum_{k=n}^{\infty} a_k=0$.

Well, $\sum_{k=n}^{\infty} a_{k}=S-\sum_{j=1}^{n-1} a_j,$ and since $\lim_{n\to\infty}\sum_{j=1}^{n-1} a_j=S,$ we get that

$$ \lim_{n\to\infty}\sum_{k=n}^{\infty} a_{k}=S-S=0 $$

WoolierThanThou
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