[Math] Show that continuity of probability for increasing sequence of sets implies the same for a decreasing sequence of sets

continuityelementary-set-theoryprobabilityprobability theory

I have the following definitions of continuity of the probability function:

  1. For an Increasing Sequence of Sets:Suppose that $B_{1} \subseteq B_{2} \subseteq B_{3} \subseteq \cdots$. Then, $P\left(\lim_{n \to \infty}B_{n} \right) = P\left( \cup_{i=1}^{\infty} B_{i}\right) = \lim_{n \to \infty}P(B_{n})$.
  2. For a Decreasing Sequence of Sets: Suppose that $\cdots \subseteq B_{3} \subseteq B_{2} \subseteq B_{1}$. Then, $P\left( \lim_{n \to \infty} B_{n}\right)= P\left( \cap_{i=1}^{\infty}B_{i}\right) = \lim_{n \to \infty}P(B_{n})$.

I need to show that 1 implies 2.

Thus far, what I have done is, I have taken the decreasing sequence $$\cdots \subseteq B_{3} \subseteq B_{2} \subseteq B_{1}$$
and since $B_{n}^{c} = \cup_{i=1}^{n-1}B_{i}$ and $B_{n-1}^{c}\subseteq B_{n}^{c}$ (where the $^{c}$ denotes complementation), I can rewrite this sequence as $B_{1}^{c} \subseteq B_{2}^{c} \subseteq B_{3}^{c} \subseteq \cdots$. Letting $B_{i}^{c} = C_{i}$ $\forall i$, we have that $C_{1} \subseteq C_{2} \subseteq C_{3} \subseteq \cdots$ is an increasing sequence of sets.

So, I can apply 1 to this sequence to obtain $P(\lim_{n \to \infty} C_{n} ) = P\left(\cup_{i=1}^{\infty}C_{i} \right) = \lim_{n \to \infty} P(C_{n})$.

Then, applying DeMorgan's Laws and the complementation law to the middle part, I obtain $$P\left(\cup_{i=1}^{\infty}C_{i} \right) = P\left(\cup_{i=1}^{\infty} B_{i}^{c} \right) = P\left(\cap_{i=1}^{\infty}B_{i} \right)^{c} = 1 – P\left(\cap_{i=1}^{\infty}B_{i} \right) $$

Then, since $P\left( \cup_{i=1}^{\infty}C_{i}\right) = \lim_{n \to \infty}P(C_{n}$, I have that $1 – P\left( \cap_{i=1}^{\infty}B_{i}\right) = \lim_{n \to \infty}P(C_{n}) $, or $$ 1 – P\left( \cap_{i=1}^{\infty}B_{i}\right) = 1 – \lim_{n \to \infty}P(B_{n}) $$

Finally, my question is, how do I turn $\mathbf{P\left(\lim_{n \to \infty}C_{n}\right)}$ into $\mathbf{1 – P\left(\lim_{n \to \infty}B_{n}\right)}$?

I know that $P\left(\lim_{n \to \infty}C_{n}\right) = P\left(\lim_{n \to \infty}B_{n}^{c}\right)$, but I can't figure out how to get any further. Indeed, I certainly don't know what $\lim_{n \to \infty}B_{n}^{c}$ is, and by pulling the $P$ inside the limit, I'd be using exactly what it is I'm trying to prove.

Thank you for your time and patience.

Best Answer

If $A_1, A_2,\dots$ are sets then we can define $\limsup A_n$ and $\liminf A_n$.

$\lim_{n\to\infty}A_n$ is only defined if these sets are equal and this with:$$\lim_{n\to\infty}A_n=\limsup A_n=\liminf A_n$$

You allready proved that: $$P(\bigcup_{n=1}^{\infty} C_n)=1-P(\bigcap_{n=1}^{\infty} B_n)\tag1$$

If $C_n$ is increasing then $\lim_{n\to\infty} C_n$ is defined and equals $\bigcup_{n=1}^{\infty} C_n$.

If $B_n$ is decreasing then $\lim_{n\to\infty} B_n$ is defined and equals $\bigcap_{n=1}^{\infty} B_n$.

So $(1)$ can be rewritten as:$$P(\lim_{n\to\infty}C_n)=1-P(\lim_{n\to\infty}B_n)\tag2$$

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