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Fall 2016 - Problem 5

measure theory

Let $X = C\p{\br{0,1}}$ be the Banach space of real-valued continuous functions on $\br{0,1}$ equipped with the norm

\norm{f} = \max_{x \in \br{0,1}} \abs{f\p{x}}.

Let $\mathcal{A}$ be the Borel $\sigma$ -algebra on $X$ .

Show that $\mathcal{A}$ is the smallest $\sigma$ -algebra on $X$ that contains all sets of the form

S\p{t, B} = \set{f \in X \mid f\p{t} \in B},

where $t \in \br{0, 1}$ and $B \subseteq \R$ is a Borel set in $\R$ .

Solution.

Let $\mathcal{S}$ denote the $\sigma$ -algebra generated by the $S\p{t, B}$ . For $t \in \br{0, 1}$ , let $\func{e_t}{X}{\R}$ be the evaluation map, i.e., $f \mapsto f\p{t}$ . Observe then that

\abs{e_t\p{f} - e_t\p{g}} = \abs{f\p{t} - g\p{t}} \leq \norm{f - g},

so $e_t$ is continuous for any $t$ . Since $e_t^{-1}\p{B} = S\p{t, B}$ , it follows that $\mathcal{S} \subseteq \mathcal{A}$ .

Conversely, first observe that

\begin{aligned} \norm{f} \leq r &\iff \max_{x \in \br{0,1}} \abs{f\p{x}} \leq r \\ &\iff \sup_{q \in \Q \cap \br{0,1}} \abs{f\p{q}} \leq r \\ &\iff f \in \bigcap_{q \in \Q \cap \br{0,1}} e_q^{-1}\p{\br{-r, r}}. \end{aligned}

Thus,

\cl{B\p{0, r}} = \bigcap_{q \in \Q \cap \br{0,1}} e_q^{-1}\p{\br{-r, r}} \in \mathcal{S}

and because translation $g \mapsto f + g$ is a continuous map $X \to X$ , it follows that $\cl{B\p{f, r}} \in \mathcal{S}$ for any $f \in X$ and $r > 0$ . Thus,

B\p{f, r} = \bigcup_{n=1}^\infty \cl{B\p{f, r - \frac{1}{n}}},

so open balls are in $\mathcal{S}$ as well. Finally, because $X$ is separable (e.g., by Stone-Weierstrass), the topology on $X$ is generated by a countable family of open balls, and so $U \in \mathcal{S}$ for any open $U \subseteq X$ . It follows that $\mathcal{A} \subseteq \mathcal{S}$ , so we have equality.