Comment: Rudin 4.8 Theorem 将作为之后一个重要结论的引理。这一小节重要的问题是，为什么 random elements (recall the preface of Chap. 18) 要被定义为一个波莱尔可测映射（Borel-measurable map）？为什么我们要关注波莱尔集 (Borel set) 以及波莱尔可测函数 (Borel-measurable function)？波奇酱的看法是，它们有着相比其他集合和函数更为优越的性质（这将在 18.2 Lemma 中揭晓），使得我们之后会接触的一些重要概念（如期望、概率）被良定义（well-defined）。在证明 18.2 Lemma 前，我们首先给出一些关键定义。
18.1 Definition. The Borel σ\sigmaσ-field on a metric space D\mathbb{D}D is the smallest σ\sigmaσ-field that contains the open sets (and then also the closed sets). A function defined relative to (one or two) metric spaces is called Borel-measurable if it is measurable relative to the Borel σ\sigmaσ-field(s). A Borel-measurable map X:Ω↦DX: \Omega \mapsto \mathbb{D}X:Ω↦D defined on a probability space (Ω,U,P)(\Omega, \mathcal{U}, \mathrm{P})(Ω,U,P) is referred to as a random element with values in D\mathbb{D}D.
接下来我们证明本小节最重要的引理，它揭示了波莱尔可测性（Borel measurability）极其良好的性质。
18.2 Lemma. A continuous map between metric spaces is Borel-measurable.
Proof A map g:D↦Eg: \mathbb{D} \mapsto \mathbb{E}g:D↦E is continuous if and only if the inverse image g−1(G)g^{-1}(G)g−1(G) of every open set G⊂EG \subset \mathbb{E}G⊂E is open in D\mathbb{D}D (recall Rudin 4.8 Theorem). In particular, for every open GGG the set g−1(G)g^{-1}(G)g−1(G) is a Borel set in D\mathbb{D}D. By definition, the open sets in E\mathbb{E}E generate the Borel σ\sigmaσ-field. Thus, the inverse image of a generator of the Borel sets in E\mathbb{E}E is contained in the Borel σ\sigmaσ-field in D\mathbb{D}D. Because the inverse image g−1(G)g^{-1}(\mathcal{G})g−1(G) of a generator G\mathcal{G}G of a σ\sigmaσ-field B\mathcal{B}B generates the σ\sigmaσ-field g−1(B)g^{-1}(\mathcal{B})g−1(B), it follows that the inverse image of every Borel set is a Borel set. ■\blacksquare■
为什么其他的可测集合不具有这样的良好性质呢？且看勒贝格可测集合（Lebesgue-measurable set）的一个例子。
Royden & Fitzpatrick Proposition 2.21 Let φ\varphiφ be the Cantor-Lebesgue function and define the function ψ\psiψ on [0,1][0,1][0,1] by
Then ψ\psiψ is a strictly increasing continuous function that maps [0,1][0,1][0,1] onto [0,2][0,2][0,2], and maps a measurable set, a subset of the Cantor set, onto a nonmeasurable set.
Proof The function ψ\psiψ is continuous since it is the sum of two continuous functions and is strictly increasing since it is the sum of an increasing and a strictly increasing function. Moreover, since ψ(0)=0\psi(0)=0ψ(0)=0 and ψ(1)=2,ψ([0,1])=[0,2]\psi(1)=2, \psi([0,1])=[0,2]ψ(1)=2,ψ([0,1])=[0,2]. For O=[0,1]∼C\mathcal{O}=[0,1] \sim CO=[0,1]∼C, we have the disjoint decomposition
which ψ\psiψ lifts to the disjoint decomposition
We note that Vitali's Theorem tells us that ψ(C)\psi(\mathbf{C})ψ(C) contains a set WWW, which is nonmeasurable. The set ψ−1(W)\psi^{-1}(W)ψ−1(W) is measurable and has measure zero since it is a subset of the Cantor set. The set ψ−1(W)\psi^{-1}(W)ψ−1(W) is a measurable subset of the Cantor set, which is mapped by ψ\psiψ onto a nonmeasurable set. ■\blacksquare■
因此即使ψ\psiψ是连续的，它仍有可能会将一个勒贝格可测集映射到一个不可测集（nonmeasurable set）上，这对我们之后定义期望和概率是致命的（recall our definition of random elements）。
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