# Why is $\mathcal{E}(X)=\mathcal{E}(X,X^*)$?

According to a course about $\sigma$-agebras in infinite dimensional space they said that it is easy to see that :

$$\mathcal{E}(X)=\mathcal{E}(X,X^*)$$

where:

• $X$ is separable real Banach space.
• $\mathcal{E}(X)$ is the $\sigma$-algebra generated by cylindrical sets, i.e., the sets of the form $$C=\{x\in X:\ (f_1(x),...,f_n(x))\in C_0\}$$ where $f_1,...,f_n\in X^*$ and $C_0\in \mathcal{B}(\mathbb{R}^n)$
• $\mathcal{E}(X,X^*)$ the $\sigma$-algebra generated by $X^*$ on $X$ i.e. the smallest $\sigma$-algebra such that all the functions $f\in X^*$ are measurable.

Question:

How did they prove such equality which seems to me not obvious ?!

• Unless I'm missing something, that actually seem obvious: the $\sigma$-algebra you defined make all the linear form measurable, and conversely, any $\sigma$-algebra that makes the linear form measurable have to contain those cylindrical sets... – Simon Henry Nov 18 '15 at 12:44
• I agree with Simon that this is rather obvious, but let me give a hint for the part that might be slightly less obvious. Consider the family of subsets $C_0\subseteq\mathbb R^n$ such that the corresponding $C$ (as in your definition of $\mathcal E(X)$) is in $\mathcal E(X,X^*)$. Check that this family is a $\sigma$-algebra containing the basic open sets in $\mathbb R^n$ (rectangular boxes) and therefore containing all the Borel sets in $\mathbb R^n$. – Andreas Blass Nov 18 '15 at 12:52
• A somewhat less obvious fact (though still not too hard to prove) is that both $\sigma$-algebras equal the Borel $\sigma$-algebra on $X$. In particular, the norm and weak topologies have the same Borel sets. – Nate Eldredge Nov 18 '15 at 14:47
• houda's question is OK even when $X$ is non-separable, but Nate's isn't. – Gerald Edgar Nov 18 '15 at 15:25