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Spring 2012 - Problem 2

measure theory

Let XX and YY be topological spaces and X×YX \times Y the Cartesian product endowed with the product topology. B(X)\mathcal{B}\p{X} denotes the Borel sets in XX and similarly, B(Y)\mathcal{B}\p{Y} and B(X×Y)\mathcal{B}\p{X \times Y}.

  1. Suppose f ⁣:X→Y\func{f}{X}{Y} is continuous. Prove that E∈B(Y)E \in \mathcal{B}\p{Y} implies f−1(E)∈B(X)f^{-1}\p{E} \in \mathcal{B}\p{X}.
  2. Suppose A∈B(X)A \in \mathcal{B}\p{X} and E∈B(Y)E \in \mathcal{B}\p{Y}. Show that A×E∈B(X×Y)A \times E \in \mathcal{B}\p{X \times Y}.
Solution.

  1. Let F={E∈B(Y)∣f−1(E)∈B(X)}\mathcal{F} = \set{E \in \mathcal{B}\p{Y} \mid f^{-1}\p{E} \in \mathcal{B}\p{X}}. Since ff is continuous, f−1(U)f^{-1}\p{U} is open for any open set U⊆XU \subseteq X, hence Borel. Thus, F\mathcal{F} contains the open sets, so it suffices to show that F\mathcal{F} is a σ\sigma-algebra:

    XX is open, so X∈FX \in \mathcal{F}.

    Given E∈FE \in \mathcal{F}, if x∈(f−1(E))cx \in \p{f^{-1}\p{E}}^\comp, then f(x)∈Ecf\p{x} \in E^\comp, so x∈f−1(Ec)x \in f^{-1}\p{E^\comp}. Conversely, if x∈f−1(Ec)x \in f^{-1}\p{E^\comp}, then f(x)∈Ecf\p{x} \in E^\comp. But if x∈f−1(E)x \in f^{-1}\p{E}, then f(x)∈Ef\p{x} \in E, so by contrapositive, we see x∈(f−1(E))cx \in \p{f^{-1}\p{E}}^\comp. Thus, (f−1(E))c=f−1(Ec)\p{f^{-1}\p{E}}^\comp = f^{-1}\p{E^\comp}, and since B(X)\mathcal{B}\p{X} is a σ\sigma-algebra, it follows that f−1(Ec)∈B(X)  ⟹  Ec∈Ff^{-1}\p{E^\comp} \in \mathcal{B}\p{X} \implies E^\comp \in \mathcal{F}.

    Finally, given {En}n⊆F\set{E_n}_n \subseteq \mathcal{F},

    x∈f−1(⋃n=1∞En)  ⟺  ∃n:f(x)∈En  ⟺  ∃n:x∈f−1(En)  ⟺  x∈⋃n=1∞f−1(En),\begin{aligned} x \in f^{-1}\p{\bigcup_{n=1}^\infty E_n} &\iff \exists n : f\p{x} \in E_n \\ &\iff \exists n : x \in f^{-1}\p{E_n} \\ &\iff x \in \bigcup_{n=1}^\infty f^{-1}\p{E_n}, \end{aligned}

    so f−1(⋃n=1∞En)=⋃n=1∞f−1(En)f^{-1}\p{\bigcup_{n=1}^\infty E_n} = \bigcup_{n=1}^\infty f^{-1}\p{E_n}. Since B(X)\mathcal{B}\p{X} is a σ\sigma-algebra, it follows that f−1(⋃n=1∞En)∈B(X)f^{-1}\p{\bigcup_{n=1}^\infty E_n} \in \mathcal{B}\p{X}, so ⋃n=1∞En∈F\bigcup_{n=1}^\infty E_n \in \mathcal{F}.

    This shows that F\mathcal{F} is a σ\sigma-algebra containing the open sets, so by definition of the Borel sets, B(Y)⊆F\mathcal{B}\p{Y} \subseteq \mathcal{F}, which completes the proof.

  2. Since X×YX \times Y has the product topology, the coordinate maps πX ⁣:X×Y→X\func{\pi_X}{X \times Y}{X} and πY ⁣:X×Y→Y\func{\pi_Y}{X \times Y}{Y} are continuous. Thus, by (1), πX−1(A)=A×Y∈B(X×Y)\pi_X^{-1}\p{A} = A \times Y \in \mathcal{B}\p{X \times Y} and πY−1(E)=X×E∈B(X×Y)\pi_Y^{-1}\p{E} = X \times E \in \mathcal{B}\p{X \times Y}. Thus, because B(X×Y)\mathcal{B}\p{X \times Y} is closed under finite intersections,

    A×E=(A×Y)∩(X×E)∈B(X×Y),A \times E = \p{A \times Y} \cap \p{X \times E} \in \mathcal{B}\p{X \times Y},

    which completes the proof.