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?

Thanks in advance!

  • $\begingroup$ If you know for sure that $a_i \leq b_i$, then $\sum_i a_i \leq \sum_i b_i$. $\endgroup$
    – Ekene E.
    Commented Jan 14 at 11:17
  • $\begingroup$ Do I need to know whether or not both sums $\sum_{i}b_i$ and $\sum_{i}a_i$ converge or diverge? $\endgroup$
    – Ditherer
    Commented Jan 14 at 11:26
  • $\begingroup$ 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. $\endgroup$
    – Ekene E.
    Commented Jan 14 at 11:32
  • $\begingroup$ 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? $\endgroup$
    – Ditherer
    Commented Jan 14 at 11:55

1 Answer 1


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.

  • $\begingroup$ Thank you for your reply. So do we need to consider the extended real numbers in this scenario? $\endgroup$
    – Ditherer
    Commented Jan 14 at 12:04
  • 1
    $\begingroup$ @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. $\endgroup$
    – FShrike
    Commented Jan 14 at 12:11

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