Probability with Measure

3.3 Measurable functions and open sets \((\Delta )\)

In this section we prove the remaining part of Lemma 3.1.4, in particular that part (1) of that lemma is equivalent to the other parts. We will also prove Lemma 3.2.1. Note that this section is marked with a \((\Delta )\), meaning that it is off-syllabus for those taking MAS31002 and is independent reading for those taking MAS61022. Our arguments will use open subsets of \(\R \). For purposes of this course we work from the following definition.

Definition 3.3.1 A set \(O\sw \R \) is open if for every \(x \in O\) there is an open interval \(I\sw \R \) containing \(x\), with \(I \subseteq O\).

Some of you will have seen open sets in more general contexts e.g. metric or topological spaces. We won’t use those more general contexts within this course, but if you are familiar with metric spaces you will know that some of the results in this section are true in greater generality than we include here.

It follows immediately from Definition 3.3.1 that every open interval in \(\R \) is an open set. We might ask what other kinds of open subset we can find within \(\R \). The following result gives a surprisingly clear answer, a consequence of which is that all open subsets of \(\R \) are Borel sets.

Proposition 3.3.2 Every open set \(O\) in \(\R \) is a countable union of disjoint open intervals.

Proof: Note that a ‘countable union’ includes the case where we only need finitely many intervals. Let us first note that if \(O_i\) are opens sets for all \(i\in I\) then (even if \(I\) is uncountable) the set \(O=\cup _i O_i\) is open. See Exercise 3.7 for a proof of this fact.

For \(x \in O\), let \(I_{x}\) be the union of all open intervals containing \(x\) for which \(I_x\sw O\). Then \(I_x\) is open. Also, \(I_x\) is an interval, because if \(a<b<c\) with \(a,c\in I_x\) then there are open intervals \((a-\eps _1,x+\eps _2)\) and \((x-\eps _3,c+\eps _4)\) within \(I_x\) and \(b\) is within their union, so \(b\in I_x\).

If \(x,y \in O\) and \(x \neq y\) then \(I_{x}\) and \(I_{y}\) are either disjoint or identical. To see this, note that if \(I_x\cap I_y\) is non-empty then \(I_x\cup I_y\) is a non-empty open interval contained within \(O\), which implies \(I_x\cup I_y\) is also contained within both \(I_x\) and \(I_y\). Thus \(I_x=I_y\).

However, there can only be countably many different \(I_x\), because we can only fit at most countably many (non-empty) disjoint open intervals within \(\R \). We now select a rational number \(r(x)\) in every distinct \(I_{x}\) and rewrite \(O\) as the countable disjoint union over intervals \(I_{x}\) labelled by distinct rationals \(r(x)\). ∎

Lemma 3.3.3 Let \(\mc {O}=\sigma (O\sw \R \-O\text { is open})\) be the \(\sigma \)-field generated by the open subsets of \(\R \). It holds that \(\mc {O}=\mc {B}(\R )\).

Proof: Recall that, by Definition 1.4.2, \(\mc {B}(\R )\) is the smallest \(\sigma \)-field that contains the open intervals \((a,b)\) for \(-\infty \leq a<b\leq \infty \). It follows immediately that \(\mc {B}(\R )\sw \mc {O}\), because \(\mc {O}\) is a \(\sigma \)-field containing all the such intervals and \(\mc {B}(\R )\) is the smallest such \(\sigma \)-field. To see the reverse inclusion, note by Proposition 3.3.2, every open set is an element of \(\mc {B}(\R )\). Thus \(\mc {O}\sw \mc {B}(\R )\), because \(\mc {B}(\R )\) therefore contains any \(\sigma \)-field that contains the open sets, and \(\mc {O}\) is the smallest such \(\sigma \)-field. ∎

Remark 3.3.4 \((\star )\) In advanced textbooks on measure theory, Lemma 3.3.3 is usually used as the definition of the Borel \(\sigma \)-field, because open sets make sense in a more general context than open intervals. In particular this definition makes sense for all metric spaces, and more generally for all topological spaces.

Lemma 3.3.5 Let \(f:S \rightarrow \R \). Then \(f\) is measurable if and only if \(f^{-1}(O) \in \Sigma \) for all open sets \(O\) in \(\R \).

Proof: We will prove the forwards and backwards implications in turn. Suppose first that \(f\) is measurable. Let \(O\sw \R \) be open, and note that Proposition 3.3.2 implies that \(O\in \mc {B}(\R )\). Hence, by Definition 3.1.1 we have \(f^{-1}(O)\in \Sigma \), as required.

For the reverse implication, suppose instead that \(f^{-1}(O)\in \Sigma \) is for all open \(O\sw \R \). Let \(\mc {A}=\{E \subseteq \R \- f^{-1}(E) \in \Sigma \}\). We will first show that \(\mc {A}\) is a \(\sigma \)-field, by checking (S1)-(S3).

(S1): \(\R \in {\cal A}\) as \(S = f^{-1}(\R )\).
(S2): If \(E \in {\cal A}\) then \(E^{c} \in {\cal A}\) since \(f^{-1}(E^{c}) = f^{-1}(E)^{c} \in \Sigma \).
(S3): If \((A_{n})\) is a sequence of sets in \(\cal A\) then \(\bigcup _{\nN }A_{n} \in {\cal A}\) since \(f^{-1}\left ( \bigcup _{n}A_{n}\right ) = \bigcup _{n}f^{-1}(A_{n}) \in \Sigma \).

We are now ready to finish the proof. By our assumption, \(O\in \mc {A}\) for all open \(\mc {A}\sw \R \). Writing \(\mc {O}=\sigma (O\sw \R \-O\text { is open})\) as in Lemma 3.3.3, by Lemma 1.2.6 we have that \(\mc {O}\sw \mc {A}\), because \(\mc {A}\) is a \(\sigma \)-field containing all the open subsets and \(\mc {O}\) is the smallest such \(\sigma \)-field. By Lemma 3.3.3 we thus have \(\mc {B}(\R )\sw \mc {A}\). By definition of \(\mc {A}\), this gives that \(f^{-1}(E)\in \Sigma \) for all \(\mc {B}(\R )\), so \(f\) is measurable. ∎

Proof of Lemma 3.1.4, part \(2\) implies part \(1\): With Lemma 3.3.5 we can finish the proof of Lemma 3.1.4. Using the notation from that lemma, assume that part 2 holds. From what we have already proved of Lemma 3.1.4, part 4 therefore also holds. By Proposition 3.3.2 we may write any open set \(O\) as \(O=\cup _n(a_n,b_n)\) for some \(-\infty \leq a_n<b_n\leq \infty \), where the union is countable. Hence,

\[f^{-1}(O)=\bigcup _n f^{-1}((a_n,b_n))=\bigcup _n f^{-1}((-\infty ,b_n))\cap f^{-1}((a_n,\infty )).\]

The right hand side of the above is in \(\Sigma \) by parts 2 and 4, which means that \(f^{-1}(O)\in \Sigma \) for any open set \(O\sw \R \). From this, Lemma 3.3.5 gives that \(f\) is measurable. ∎

The above completes the proof of Lemma 3.1.4, as promised from Section 3.1. We now move on to the proof of Lemma 3.2.1, starting with a proposition that links continuous functions to open sets.

Proposition 3.3.6 A mapping \(f:\R \rightarrow \R \) is continuous if and only if \(f^{-1}(O)\) is open for every open set \(O\) in \(\R \).

Proof: First suppose that \(f\) is continuous. Choose an open set \(O\) and let \(a \in f^{-1}(O)\) so that \(f(a) \in O\). Then there exists \(\eps > 0\) so that \((f(a) - \eps , f(a) + \eps ) \subseteq O\). By definition of continuity of \(f\), for such an \(\eps \) there exists \(\de > 0\) so that \(x \in (a - \de , a + \de ) \Rightarrow f(x) \in (f(a) - \eps , f(a) + \eps )\). But this tells us that \((a - \de , a + \de ) \subseteq f^{-1}((f(a) - \eps , f(a) + \eps )) \subseteq f^{-1}(O)\). Since \(a\) is arbitrary we conclude that \(f^{-1}(O)\) is open.

Conversely, suppose that \(f^{-1}(O)\) is open for every open set \(O\) in \(\R \). Choose \(a \in \R \) and let \(\eps > 0\). Then since \((f(a) - \eps , f(a) + \eps )\) is open so is \(f^{-1}((f(a) - \eps , f(a) + \eps ))\). Since \(a \in f^{-1}((f(a) - \eps , f(a) + \eps ))\) there exists \(\de > 0\) so that \((a-\de , a + \de ) \subseteq f^{-1}((f(a) - \eps , f(a) + \eps ))\). From here you can see that whenever \(|x - a| < \de \) we must have \(|f(x) - f(a)| < \eps \). But then \(f\) is continuous at \(a\) and the result follows. ∎

Proof of Lemma 3.2.1: Let \(f:\R \rightarrow \R \) be continuous and \(O\) be an arbitrary open set in \(\R \). By Proposition 3.3.6 \(f^{-1}(O)\) is an open set in \(\R \). Hence, by Proposition 3.3.2 we have \(f^{-1}(O)\in {\cal B}(\R )\) for all open \(O\sw \R \). Lemma 3.3.5 gives that \(f\) is measurable. ∎