Showing that $G$ is sequentially closed in $X^*$ with the $\sigma(X^*,X)$ topology.

Question

Let be $X$ Banach space. Let $Y$ be a linear subspace of $X^*$ such that $Y$ is dense for $\sigma(X^*,X)$. Let $\sigma(Y)$ be the sigma algebra on $X$ generated by sets of the form $$\{x\in X:(\langle x, x_1^*\rangle,\ldots,\langle x, x_n^*\rangle)\in B\},\;\;\;\;(1)$$ for all $n\geq 1$ where $x_1^*,\ldots,x_n^*\in Y$ and $B$ is a Borel subset of $\mathbb{R}^n$. Let $$G=\{x^*\in X^*: \text{ the map } x\to\langle x,x^*\rangle \text{ is }\sigma(Y)\text{-measurable}\}.$$ I want to show that $G$ is sequentially closed in $\sigma(X^*,X)$.

I would try to show that if $\{x_n^*\}$ is a sequence in $G$ such that $x_n^*\stackrel{*}{\rightharpoonup} x_0^*\in X^*$ then $x_0^*\in G$. For this I would need to show that for every Borel subset $B$ of $\mathbb{R}$, the set $(x_0^*)^{-1}(B)\in\sigma(Y)$. Now, weak$^*$ convergence implies that for all $x\in X$ we have $\langle x,x_n^*\rangle\to\langle x, x_0^*\rangle.$ But how to proceed from here?

I'm finding it difficult to write $(x_0^*)^{-1}(B)$ as countable unions/intersections of sets of the form (1). Any hints?

Answer 1

Your approach is already a good start.

Here is a sketch for the rest.

There is, however, a simplification: If you want to show that a map $f:X\to \mathbb R$ is measurable, it suffices to check that the preimages of sets of the form $(-\infty,c)$ for $c\in\mathbb R$ are measurable. Thus you need to check the measurability of the set $$ A_c:=(x_0^*)^{-1}((-\infty,c))=\{x \in X : \langle x,x_0^*\rangle < c\} $$ is measurable. For this, one can show the equivalence $$ \langle x,x_0^*\rangle < c \Leftrightarrow \exists N_x\in\mathbb N \forall n\geq N_x : \langle x,x_n^*\rangle < c. $$ Using this equivalence, one can rewrite $A_c$ as a countable union of a countable intersection of measurable sets.

If I am not mistaken, this method should work for all sigma algebras on $X$.

Answer 2

You can repeat the argument in measure theory that shows that pointwise limits of measurable functions are measurable.

Assume for now that $x_n^*$ is real.

First, let $s_0^*=\sup_nx_n^*$, in the sense that $s_0^*(x)=\sup_nx_n^*(x)$. This is measurable: $$ (s_0^*)^{-1}(a,\infty)=\{x:\ s_0^*(x)>a\}=\bigcup_n\{x:\ x_n^*(x)>a\}=\bigcup_n(x_n^*)^{-1}(a,\infty)\in\sigma(Y). $$ As $\inf M=-\sup(-M)$, you also get that the infimum is measurable.

Next you go for $\limsup$: you have $\limsup_n x_n^*=\inf_m\sup_{n>m}x_n^*$, so measurable. Now, if $x_n^*\to x_0^*$, you have $$x_0^*=\lim_nx_n^*=\limsup_nx_n^*, $$ so measurable.

In the general case, $x_n^*=\operatorname{Re}x_n^*+i\operatorname{Im}x_n^*$ is a linear combination of measurable, so measurable.