# Last step in proving a fundamental probability inequality

I am stuck on the last step in proving Boole's inequality. So I wan to prove that, for events $$A_1,A_2,A_3,...$$ we have

$$P\left(\cup_{i=1}^{\infty}A_i\right) \le \sum_{i=1}^{\infty}P(A_i)$$

Now we do this by creating a sequence $$B_i$$ of disjoint events from the original sequence $$A_i$$. A really good outline of the argument can be found here (see the answer by kccu):

Could someone explain Boole's inequality proof to me?

So $$B_1 = A_1, B_2 = A_1^c \cap A_2, B_3 =A_1^c \cap A_2^c \cap A_3,...$$ Now $$B_i \subseteq A_i$$ and by monotonicity we get that $$P(B_i)\le P(A_i)$$.

However, I don't understand how we can leap to saying that $$\sum_{i=1}^{\infty}P(B_i)\le \sum_{i=1}^{\infty}P(A_i)$$. Also do we need to worry whether or not $$\sum_{i=1}^{\infty} P(A_i)$$ is finite or infinite?

• If you know for sure that $a_i \leq b_i$, then $\sum_i a_i \leq \sum_i b_i$. Commented Jan 14 at 11:17
• Do I need to know whether or not both sums $\sum_{i}b_i$ and $\sum_{i}a_i$ converge or diverge? Commented Jan 14 at 11:26
• Good point. I think the usual proof involves first showing that $\mathbb{P}\left(\bigcup_{i = 1}^n A_i \right) \leq \sum_{i = 1}^n \mathbb{P}(A_i)$, and then sending $n \to \infty$. Maybe someone else can say more. Commented Jan 14 at 11:32
• Is it the fact that $P(B_i)\le P(A_i)$ holds for every $i$ and so $\sum_{i=1}^{\infty} P(B_i) \le \sum_{i=1}^{\infty} P(A_i)$? Do I need to worry or even care about whether these sums are finite or infinite? Commented Jan 14 at 11:55

If you know that $$\mu(B_i)\le\mu(A_i)$$ for $$i=1,2,\cdots$$ then you also know $$\sum_{j=1}^n\mu(B_j)\le\sum_{j=1}^n\mu(A_j)$$ for $$n=1,2,\cdots$$. Call these two partial sum sequences $$b_n$$ and $$a_n$$. This is just a straightforward real analysis exercise now: if $$b_n\le a_n$$ for all $$n$$, then $$\limsup_{n\to\infty}b_n\le\liminf_{n\to\infty}a_n$$. In this case $$\lim_{n\to\infty}b_n$$ exists, because by the $$\sigma$$-additivity axiom we know: $$\sum_{j=1}^\infty\mu(B_j)=\mu\left(\bigcup_{j=1}^\infty B_j\right)=\mu\left(\bigcup_{j=1}^\infty A_j\right)$$Also $$a_n$$ is a monotonic sequence so its limit exists (but is potentially $$+\infty$$) therefore: $$\mu\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{n\to\infty}b_n\le\lim_{n\to\infty}a_n=\sum_{j=1}^\infty\mu(A_j)\in\overline{\Bbb R}$$We must potentially allow that sum to be infinite.
So the real point for you is this: given two real sequences $$b_\bullet$$ and $$a_\bullet$$ with $$b_n\le a_n$$ for all $$n$$, and suppose these sequences converge, show that $$\lim_{n\to\infty}b_n\le\lim_{n\to\infty}a_n$$. Hint: suppose the negation and obtain a contradiction. You can try to show the version with $$\liminf$$ and $$\limsup$$ if you wish as well.
• @Ditherer Only in the instance $\sum_{j=1}^\infty\mu(A_j)=+\infty$. The theorem isn't very interesting in that case anyway, since the inequality is obvious. Commented Jan 14 at 12:11