How to use cylinder set to generate $\sigma$-algebra generated by a subset of dual space?

Question

I meet a problem when reading Analysis in Banach Space (Volume I). On page 2 the authors assume that $X$ is a separable Banach space. For a given subset $Y\subset X^*$, they denote $\sigma(Y)$ as the smallest $\sigma$-algebra in $X$ for which every $x^*\in Y$ is measurable (where in my opinion this means $\bigcup_{x^*\in Y,B\in\mathscr{B}(\mathbb{K})}(x^*)^{-1}(B)\subset\sigma(Y)$, here $\mathscr{B}(\mathbb{K})$ means Borel sets on $\mathbb{K}$). Then they claim that $\sigma(Y)$ is generated by the collection $\mathscr{C}(Y)$ of all sets of the form $$ \{x\in X:(\langle x_1^*,x\rangle,\cdots,\langle x_n^*,x\rangle)\in B\} $$ with $n\ge1,x_1^*,\cdots,x_n^*\in Y,B\in\mathscr{B}(\mathbb{K}^n)$, which I cannot figure out why.

I've tried the following two ways to prove this claim:

• If $X^*$ is also separable, then the claim is a direct result after taking the countable dense subset of $Y$. But we only know $X$ is separable, which may not lead to the separability of $X^*$ or $Y$.

• I have proved that $\mathscr{C}(Y)$ is a $\pi$-system, and I have tried to use Dynkin's $\pi$-$\lambda$ theorem, showing that $\sigma(Y)$ is the smallest $\lambda$-system containing $\mathscr{C}(Y)$. However, since $\lambda$-system may not be $\sigma$-algebra, I think we should start with a $\lambda$-system $\mathcal{L}$ containing $\mathscr{C}(Y)$ and to show that $\sigma(Y)\subset\mathcal{L}$, which I have no idea how to continue.

So how to prove the claim above? The authors said it's "easy" but I can't agree :( I'll be glad for any reply!

Answer 1

By definition, the $\sigma$-algebra $\sigma (Y)$ on $X$ is generated by sets of the form \begin{equation} \{x\in X : \langle x^{*}, x\rangle \in B\} \end{equation} with $x^{*}\in Y$ and $B\in \mathscr{B}(\mathbb{K})$, while the $\sigma$-algebra $\mathscr{C}(Y)$ on $X$ is generated by sets of the form \begin{equation} \{x\in X : (\langle x_{j}^{*}, x\rangle )_{j=1}^{n} \in B\} \end{equation} with $n\in\mathbb{N}$, $x_{j}^{*}\in Y$ for each $j\in \{1, \ldots , n\}$ and $B\in \mathscr{B}(\mathbb{K}^{n})$. Clearly we have $\sigma (Y) \subseteq \mathscr{C}(Y)$. For the other inclusion, let $n\in\mathbb{N}$, $x_{1}^{*}, \ldots , x_{n}^{*}\in Y$ and $B\in \mathscr{B}(\mathbb{K}^{n})$. Let $e_{1}, \ldots , e_{n}$ be the standard basis for $\mathbb{K}^{n}$. Then for each $j\in \{1, \ldots , n\}$ the map $\xi \mapsto \xi e_{j}$ from $\mathbb{K}$ into $\mathbb{K}^{n}$ is measurable and the map $x_{j}^{*}$ is $\sigma (Y)$-measurable, so the composition map $x\mapsto \langle x_{j}^{*}, x\rangle e_{j}$ from $X$ into $\mathbb{K}^{n}$ is $\sigma (Y)$-measurable. Hence the sum of such maps $x\mapsto \sum_{j=1}^{n} \langle x_{j}^{*}, x\rangle e_{j}$ from $X$ into $\mathbb{K}^{n}$ is $\sigma (Y)$-measurable and its inverse image under $B$ is the $\sigma (Y)$-measurable set \begin{equation} \{x\in X : \sum_{j=1}^{n} \langle x_{j}^{*}, x\rangle e_{j} \in B\} = \{x\in X : (\langle x_{j}^{*}, x\rangle )_{j=1}^{n} \in B\} . \tag{1} \end{equation} As $\mathscr{C}(Y)$ is the $\sigma$-algebra on $X$ generated by sets of the form $(1)$, we obtain the inclusion $\mathscr{C}(Y) \subseteq \sigma (Y)$.

Note that the separability of $X$ was not used anywhere in the above argument.

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Relevant Definitions

Take a Banach space $X$ [over the field $\mathbb{K}$] and $Y\subseteq X^{*}$. There are a few ways we can obtain a $\sigma$-algebra on $X$. The first is to consider the $\sigma$-algebra generated by all the [norm] open subsets of $X$, that is we take the smallest $\sigma$-algebra on $X$ that contains all of the open subsets of $X$. This $\sigma$-algebra is denoted by $\mathscr{B}(X)$ and is called the Borel $\sigma$-algebra on $X$. The second is to consider the $\sigma$-algebra generated by $Y$, that is we take the smallest $\sigma$-algebra on $X$ that makes every $x^{*}\in Y$ measurable, or equivalently is the smallest $\sigma$-algebra generated by sets $(x^{*})^{-1}(B)$ with $x^{*}\in Y$ and $B\in \mathscr{B}(\mathbb{K})$. We denote this $\sigma$-algebra by $\sigma (Y)$. The relation between these $\sigma$-algebras is that $\sigma (Y) \subseteq \mathscr{B}(X)$ in general [because every $x^{*}\in X^{*}$ is $\mathscr{B}(X)$-measurable] and Proposition 1.1.1 in the book provides some conditions on $Y$ and $X$ that imply $\sigma (Y) = \mathscr{B}(X)$.