Can this probability be shown by using the properties of Lebesgue integration

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(Grimmett and Stirzaker - Probability and Random Processes - Exercise 1.3.5)

I am studying Lebesgue integration in parallel to probability theory, and my question is:

Can the following be shown by invoking the convergence properties of Lebesgue integrals instead?

Let $A_r \ r \ge 1$ be events such that $\mathbb P (A_r) =1 \ \forall r$. Show that $\mathbb P (\bigcap_{r=1}^{\infty} A_r ) =1$

Here is the solution provided in the textbook (which does not invoke Lebesgue integration, hence my question):

$\mathbb P (\bigcap_r^{\infty} A_r ) = \lim_{ n \to \infty} P (\bigcap_{r=1}^{n} A_r ) = \lim_{ n \to \infty} [ 1-P (\bigcup_{r=1}^{n} A_r )^c ] \ge 1 - \lim_{ n \to \infty} [ \sum_{r=1}^{n}P ( A_r^c ) ]=1-0=1 $
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By countable additivity, which is a property a measure has by definition, $$P\{\cup_{r \geq 1}A_r^\mathsf{c}\} \leq \sum_{r\geq 1} P\{A_r^\mathsf{c}\}$$

The RHS is $0$. Combining this with De Morgen's law, you get to the result. It is important that $r$ belongs to a countable set.

Alternatively, you can define the following sequence of simple functions $$f_n = \mathbf{1}_{\cup_{r\geq1}^nA_r^\mathsf{c}}$$ and $f = \mathbf{1}_{\cup_{r\geq1}A_r^\mathsf{c}}$. Note that $0\leq f_n \nearrow f$. By definition of Lebesgue integral (or the monotone convergence theorem) $\int f_n \rightarrow \int f$.