This is a follow-up question to the following ones:
I'm reading the proof the following theorem:
If $X$ is a separable, metrizable locally convex space, $(\Omega,\Sigma,\mu)$ is a $\sigma$-finite measure space, and $f:\Omega\to X$ is weakly measurable, then $f$ is strongly measurable.
The first part of the proof shows that $f$ is measurable:
We can restrict ourselves to the case of finite measures. We prove first that $f$ is measurable. To do this it suffices to show that $$ f^{-1}(x+V)\in\Sigma $$ whenever $x\in X$ and $V$ is a neighborhood of $0$ and a barrel. Replacing $f$ by $f-x$, we may also assume that $x=0$. Choose points $y_n\in X\setminus V$, and open convex neighborhoods $V_n$ of $0$ so that $$ X\setminus V=\bigcup_{n=1}^\infty(y_n+V_n). $$ Choose next functional $\varphi_n\in X^*$ such that $$ \Re\varphi_n(y_n+v_n)>\Re\varphi_n(v)\tag{*} $$ whenever $v_n\in V_n$ and $v\in V$. ($\color{red}{Question:}$ Why $\varphi_n$ exists? Are we using some separation theorem here?) Then the set $$ f^{-1}(x+V)=\bigcap_{n=1}^\infty\{\omega\in\Omega:\Re \varphi_n(f(\omega))\leq\sup_{v\in V}|\varphi_n(v)|\}\tag{**} $$ is a countable intersection of sets in $\Sigma$. ($\color{red}{Question:}$
I'm quite sure that $\Re f(\omega)$ is a typo, which might be $\Re \varphi_n(f(\omega))$.Could anyone explain how to get (**)?I don't see it has anything to do with (*).)
Could somebody help me with the questions indicated in the proof?
Also, I'm wondering if anyone could come up with a cited reference for the proof of this theorem.
[Updated:] One more typo fixed. It is quite clear now that one can use (*) to show (**).