Probability with Measure

1.7 Measures and limits

In this section we return to the consideration of arbitrary measure spaces \((S, \Sigma , m)\). Let \((A_{n})\) be a sequence of sets in \(\Sigma \). We say that it is increasing if \(A_{n} \subseteq A_{n+1}\) for all \(\nN \), and decreasing if \(A_{n+1}\sw A_n\). When \((A_n)\) is increasing, it is easily seen that \((A^c_n)\) is decreasing.

When \((A_n)\) is increasing, a useful technique is the disjoint union trick whereby we can write \(\bigcup _{n=1}^{\infty }A_{n} = \bigcup _{n=1}^{\infty }B_{n}\) where the \(B_{n}\)s are all mutually disjoint by defining \(B_{1} = A_{1}\) and for \(n > 1, B_{n} = A_{n} - A_{n-1}\). e.g. \(\R = \bigcup _{n=1}^{\infty }[-n, n]\) and here \(B_{1} = [-1, 1], B_{2} = [-2, -1) \cup (1, 2]\) etc.

Lemma 1.7.1 Let \(A_n\in \Sigma \) for all \(n\). It holds that:
- 1. If \((A_{n})\) is increasing and \(A = \bigcup _{n=1}^{\infty }A_{n}\) then \(m(A) = \lim _{n \rightarrow \infty } m(A_{n}).\)
- 2. If \((A_{n})\) is decreasing and \(A = \bigcap _{n=1}^{\infty }A_{n}\), and \(m(A_1)<\infty \), then \(m(A) = \lim _{n \rightarrow \infty } m(A_{n}).\)

Proof: We will prove the first claim here. The second claim can be deduced from the first, which is for you to do in Problem 1.7. We use the disjoint union trick and (M2) to find that

\begin{align*} m(A) & = m\left (\bigcup _{n=1}^{\infty }B_{n}\right ) = \sum _{n=1}^{\infty }m(B_{n}) = \lim _{N \rightarrow \infty }\sum _{n=1}^{N}m(B_{n}) = \lim _{N \rightarrow \infty }m\left (\bigcup _{n=1}^{N}B_{n}\right ) = \lim _{N \rightarrow \infty } m(A_{N}). \end{align*} Here we use that \(A_{N} = B_{1} \cup B_{2} \cup \cdots \cup B_{N}\). ∎

Lemma 1.7.2 (Union bound) If \((A_{n})\) is an arbitrary sequence of sets with \(A_{n} \in \Sigma \) for all \(\nN \) then

\[ m\left (\bigcup _{n=1}^{\infty }A_{n}\right ) \leq \sum _{n=1}^{\infty }m(A_{n}).\]

Proof: From Problem 1.4, we have \(m(A_{1} \cup A_{2}) + m(A_{1} \cap A_{2}) = m(A_{1}) + m(A_{2})\) from which we deduce that \(m(A_{1} \cup A_{2}) \leq m(A_{1}) + m(A_{2})\). By induction we then obtain for all \(N \geq 2\),

\[ m\left (\bigcup _{n=1}^{N}A_{n}\right ) \leq \sum _{n=1}^{N}m(A_{n}).\]

Now define \(X_{N} = \bigcup _{n=1}^{N}A_{n}\). Then \(X_{N} \subseteq X_{N+1}\) and so \((X_{N})\) is increasing to \(\bigcup _{n=1}^{\infty }X_{n} = \bigcup _{n=1}^{\infty }A_{n}\). By Lemma 1.7.1 we have

\begin{align*} m\left (\bigcup _{n=1}^{\infty }A_{n}\right ) & = m\left (\bigcup _{n=1}^{\infty }X_{n}\right ) = \lim _{N \rightarrow \infty }m(X_{N}) = \lim _{N \rightarrow \infty } m\left (\bigcup _{n=1}^{N}A_{n}\right ) \leq \lim _{N \rightarrow \infty }\sum _{n=1}^{N}m(A_{n}) = \sum _{n=1}^{\infty }m(A_{n}). \end{align*} ∎