Probability with Measure

3.3 Measurable functions and open sets $(\mathrm{\Delta })$

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 $(\mathrm{\Delta })$, 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 $\mathbb{R}$. For purposes of this course we work from the following definition.

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 $\mathbb{R}$ is an open set. We might ask what other kinds of open subset we can find within $\mathbb{R}$. The following result gives a surprisingly clear answer, a consequence of which is that all open subsets of $\mathbb{R}$ are Borel sets.

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 $I_x\subseteq O$. 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 $(a-{ϵ}_1,x+{ϵ}_2)$ and $(x-{ϵ}_3,c+{ϵ}_4)$ within $I_x$ and $b$ is within their union, so $b\in I_x$.

If $x,y\in O$ and $x\ne y$ 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 $\mathbb{R}$. 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 $r(x)$.   ∎

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

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

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

• (S1): $\mathbb{R}\in \mathcal{A}$ as $S=f^{-1}(\mathbb{R})$.

• (S2): If $E\in \mathcal{A}$ then $E^c\in \mathcal{A}$ since $f^{-1}(E^c)=f^{-1}(E{)}^c\in \mathrm{\Sigma }$.

• (S3): If $(A_n)$ is a sequence of sets in $\mathcal{A}$ then $\bigcup _{n\in \mathbb{N}}{A}_n\in \mathcal{A}$ since $f^{-1}\left(\bigcup _n{A}_n\right)=\bigcup _n{f}^{-1}(A_n)\in \mathrm{\Sigma }$.

We are now ready to finish the proof. By our assumption, $O\in \mathcal{A}$ for all open $\mathcal{A}\subseteq \mathbb{R}$. Writing $\mathcal{O}=\sigma (O\subseteq \mathbb{R};O\text{ is open})$ as in Lemma 3.3.3, by Lemma 1.2.6 we have that $\mathcal{O}\subseteq \mathcal{A}$, because $\mathcal{A}$ is a $\sigma$-field containing all the open subsets and $\mathcal{O}$ is the smallest such $\sigma$-field. By Lemma 3.3.3 we thus have $\mathcal{B}(\mathbb{R})\subseteq \mathcal{A}$. By definition of $\mathcal{A}$, this gives that $f^{-1}(E)\in \mathrm{\Sigma }$ for all $\mathcal{B}(\mathbb{R})$, 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 $-\mathrm{\infty }\le a_n<b_n\le \mathrm{\infty }$, where the union is countable. Hence,

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

The right hand side of the above is in $\mathrm{\Sigma }$ by parts 2 and 4, which means that $f^{-1}(O)\in \mathrm{\Sigma }$ for any open set $O\subseteq \mathbb{R}$. 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.

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 $ϵ>0$ so that $(f(a)-ϵ,f(a)+ϵ)\subseteq O$. By definition of continuity of $f$, for such an $ϵ$ there exists $\delta >0$ so that $x\in (a-\delta ,a+\delta )\Rightarrow f(x)\in (f(a)-ϵ,f(a)+ϵ)$. But this tells us that $(a-\delta ,a+\delta )\subseteq f^{-1}((f(a)-ϵ,f(a)+ϵ))\subseteq f^{-1}(O)$. 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 $\mathbb{R}$. Choose $a\in \mathbb{R}$ and let $ϵ>0$. Then since $(f(a)-ϵ,f(a)+ϵ)$ is open so is $f^{-1}((f(a)-ϵ,f(a)+ϵ))$. Since $a\in f^{-1}((f(a)-ϵ,f(a)+ϵ))$ there exists $\delta >0$ so that $(a-\delta ,a+\delta )\subseteq f^{-1}((f(a)-ϵ,f(a)+ϵ))$. From here you can see that whenever $|x-a|<\delta$ we must have $|f(x)-f(a)|<ϵ$. But then $f$ is continuous at $a$ and the result follows.   ∎

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