The Borel $\sigma$-algebra of a metric space is the smallest $\sigma$-algebra with respect to which all continuous functionals are measurable

Question

I'm trying to solve Theorem I.1.7 in Parthasarathy's textbook Probability Measures on Metric Spaces.

Let $(E, d)$ be a metric space and $\mathcal B(E)$ its Borel $\sigma$-algebra. Then $\mathcal{B}(E)$ coincides with $\mathcal A$ defined as the smallest $\sigma$-algebra with respect to which all continuous functionals on $E$ are measurable.

Could you verify if my attempt is fine?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Maybe I am missing something, but wouldn't that imply that for every Banach space $X$, $\mathcal B(X)=σ(X^*)$. This fails for non separable Banach spaces https://mathoverflow.net/questions/24432/borelx-sigmax-for-x-non-separable. — Evangelopoulos Foivos, Apr 21 '22 at 16:05
@EvangelopoulosF. I'm not sure :v This theorem is also mentioned as Prop 1.3 in this lecture note. I hope that I quote the theorem correctly. — Analyst, Apr 21 '22 at 16:50
@EvangelopoulosF. OP is talking about all continuous functionals, i.e. not necessarily linear. — Michh, Apr 21 '22 at 17:45

Analyst · Accepted Answer · 2022-04-22T07:39:27.867

Lemma: Let $f:X \to Y$ and $C$ be some collection of subsets of $Y$. Then $$f^{-1}[\sigma(C)]=\sigma(f^{-1}[C]).$$

Let $\mathcal O$ be the standard topology of $\mathbb R$. Then $\sigma(\mathcal O)$ is the Borel $\sigma$-algebra of $\mathbb R$. Let $\mathcal C(E)$ the space of all all continuous functionals on $E$. Let $$ f^{-1}[\sigma(\mathcal O)] := \{f^{-1}(B) \mid B \in \sigma(\mathcal O)\} \quad \forall f \in \mathcal C(E). $$

Then $\mathcal A$ is the $\sigma$-algebra generated by \begin{align} \bigcup_{f \in \mathcal C(E)} f^{-1}[\sigma(\mathcal O)]. \end{align}

We have \begin{align} \mathcal A &= \sigma \left ( \bigcup_{f \in \mathcal C(E)} f^{-1}[\sigma( \mathcal O )] \right ) \\ &= \sigma \left ( \bigcup_{f \in \mathcal C(E)} \sigma( f^{-1}[\mathcal O] ) \right ) \quad \text{by our Lemma} \\ & \subseteq \sigma \left ( \sigma \left ( \bigcup_{f \in \mathcal C(E)} f^{-1}[\mathcal O] \right) \right ) \\ &= \sigma \left ( \bigcup_{f \in \mathcal C(E)} f^{-1}[\mathcal O] \right) \\ &\subseteq \mathcal B(E). \end{align}

Because metric space is perfectly normal. For each closed subset $F$ of $E$, there is a continuous functional $f:E \to [0, 1]$ such that $F =f^{-1}(0)$. This implies $\mathcal A$ contains all closed and thus all open subsets of $E$. This completes the proof.

The Borel $\sigma$-algebra of a metric space is the smallest $\sigma$-algebra with respect to which all continuous functionals are measurable

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