Let $C$ denote the σ-field on $C[0,1]$ generated by all open balls : the open $\epsilon$-ball around $f$ is the set $\{g ∈ C : ρ(f, g) < \epsilon\}$
Denote $\forall t∈[0,1], π_t : C[0,1] →\Bbb R$ to be the evaluation map (also called the projection map), defined as $π_t(f)=f(t),t∈[0,1]$,
(1) Show that $C$ is the σ-field generated by $\{π_t,t∈[0,1]\}$.
(2) Show also that a map $φ:(Ω,A)→(C[0,1],C)$, where $(Ω,A)$ is any measurable space,is measurable if and only if $π_t◦φ$ is measurable for all $t∈[0, 1]$ .
My attempt :
(1) By showing that any closed ball is in $\{π_t,t∈[0,1]\}$ , I have been able to show $C \subseteq \sigma\{π_t,t∈[0,1]\}$. Was able to show the other way thanks to Generating the Borel $\sigma$-algebra on $C([0,1])$
(2) Could do one way : if $\varphi$ is measurable then it's obvious but how to show the other way i.e. $π_t◦φ$ is measurable for all $t∈[0, 1] \implies \varphi$ is measurable .
Note that the trick we normally use for doing such proofs for projections won't work, since $[0,1]$ is uncountable! Can I use $\Bbb Q \cap [0,1]$ somehow?
Thanks in Advance for help!
What you have to show is that $E \in \sigma (\pi_t:0\leq t \leq 1\}$ implies $\phi ^{-1}(E) \in \mathcal A$. Since the collection of all sets $E$ in $C[0,1]$ such that $\phi ^{-1}(E) \in \mathcal A$ is a sigma algebra (by a simple verification) it is enough to consider the special case when $E=\{\omega: (\pi_{t_1} (\omega),\pi_{t_2} (\omega),\cdots,\pi_{t_n} (\omega)) \in A$ is in $\mathcal A$ for any Borel set $A$ in $\mathbb R^{n}$ (and $n,t_1,t_2,\cdots,t_n)$ are arbitrary. But $$\phi^{-1}(E) =(\phi\circ \pi_{t_1},\phi \circ\pi_{t_2},\cdots ,\phi \circ\pi_{t_n})^{-1}(A)$$ which belongs to $\mathcal A$ because $(\phi\circ \pi_{t_1},\phi \circ\pi_{t_2},\cdots ,\phi \circ\pi_{t_n})$ is a measurable map from $\Omega$ into $\mathbb R^{n}$.