Q: 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 ?!
Thank you for your time.
