Question

Prove Boole’s inequality:

$P\left(\underset{i=1}{\overset{\infty }{\cup }}{A}_i\right)\le \sum _{i=1}^{\infty }P\left(A_i\right)$

Short answer

Proof by mathematical induction:

Assume that equality stands for some$n$, and it follows that inequality stands for$n+1$.

Step-by-step solution

Given Information.

Prove that for all events$A_1,A_2,...A_n$.

$P\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\le \sum _{k=1}^{n}P\left(A_k\right)$

Explanation.

Proof by mathematical induction:

For$n=1$the statement is true

$P\left(A_1\right)\le P\left(A_1\right)$

Hypothesis:$P\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\le \sum _{k=1}^{n}P\left(A_k\right)$this is true for any$n$event.

Then for any $n+1$events:

$P\left(\underset{k=1}{\overset{n+1}{\cup }}{A}_k\right)=P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cup A_{n+1}\right]\text{associative law}$

&$$\begin{gathered} =P\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)+P\left(A_{n+1}\right)-P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cap A_{n+1}\right] \\ \le \sum _{k=1}^{n}P\left(A_k\right)+P\left(A_{n+1}\right)-P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cap A_{n+1}\right] \\ \le \sum _{k=1}^{n+1}P\left(A_k\right)-P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cap A_{n+1}\right] \\ \le \sum _{k=1}^{n+1}P\left(A_k\right) \end{gathered}$$

$P\left(\stackrel{n+1}{\cup }{A}_k\right)\le \sum _{k=1}^{n+1}P\left(A_k\right)$

This proves the statement, by the principle of mathematical induction.