Analysis Problem of the Day 20

Today’s problem comes as Problem 3 on UCLA’s Spring 2017 Analysis Qual:

Problem 3: Let $\mathcal{F}$ be a $\sigma$ -algebra on $C([0,1])$ with the sup-norm such that for all $x \in [0,1],$ the map $f \to L_x(f) = f(x)$ is $\mathcal{F}$ -measurable. Show that $\mathcal{F}$ is the Borel $\sigma$ -algebra $\mathcal{B}$ on $C([0,1]).$

Solution: Notice that the condition that all of the evaluation maps are $\mathcal{F}$ -measurable is equivalent to stating that for $B \subset \mathbb{R}$ Borel, $F^x_B := \{f \in C([0,1]): f(x) \in B\} \in \mathcal{F}.$ Let $C$ be a closed set. Then, $F^x_C$ is clearly a closed set in $C([0,1]),$ since $f_n(x) \in C, f_n \to f$ in $C([0,1])$ implies $f_n(x) \to f(x),$ i.e. $f(x) \in C.$ Moreover, it is easy to check that $(F^x_B)^c = F^x_{B^c}$ and $\bigcap_{i=1}^\infty F^x_{B_i} = F^x_{\bigcap_{i=1}^\infty B_i}.$ But since the Borel $\sigma$ -algebra on $\mathbb{R}$ is generated by open (equivalently, closed) sets, it follows that

$\langle \{F^x_C: C \text{ closed}, x \in [0,1]\}\rangle = \{F^x_{\langle C \rangle}: x\in[0,1]\} = \{F^x_{\mathcal{B}}: x \in [0,1]\} = \mathcal{F},$

where $\mathcal{B}$ is the Borel $\sigma$ -algebra on $\mathbb{R}.$ In other words, $\mathcal{F}$ is generated by closed sets and is therefore contained in the Borel $\sigma$ -algebra $\mathcal{B}$ on $C([0,1]).$

Conversely, it suffices to show that that an arbitrary topological basis element $U \subset C([0,1])$ is contained in $\mathcal{F}.$ Without loss of generality, we many assume that $U$ is a closed ball, i.e. $U=\{g \in C([0,1]): \|f-g\|_\infty \leq \epsilon\}$ for some $f \in C([0,1].$ Let $X=\{x_i\}$ be a dense countable subset of $C([0,1]).$ Now, I claim that $U = \bigcap_{i=1}^\infty F^{x_i}_{[f(x_i)-\epsilon,f(x_i)+\epsilon]},$ which is $\mathcal{F}$ measurable. Indeed, if $\|f-g\|_\infty \leq \epsilon,$ then $g(x_i) \in [f(x_i)-\epsilon,f(x_i)+\epsilon]$ for all $i.$ Conversely, if $|f(x)-g(x)|>\epsilon$ for some $x \in [0,1],$ this is a contradiction since there exists a sequence $x_{i_j} \to x$ and $|f(x_{i_j})-g(x_{i_j})| \leq \epsilon.$ Thus, we conclude that $U \in \mathcal{F},$ and since closed balls generate the Borel $\sigma$ -algebra on $C([0,1]),$ it follows that $\mathcal{F} = \mathcal{B}.$

Leave a Reply Cancel reply