Can Theorem 1.40 in Rudin's Real and Complex Analysis be strengthened when the $\sigma$-algebra is Borel?

Question

Let $(X, \mathcal F, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the Bochner integral. Then we have Theorem 1.40 in Rudin's Real and Complex Analysis, i.e.,

Theorem Let $f \in L_1(X, \mu, E)$ and $F$ be a non-empty closed subset of $E$. If $$ \varphi(A) :=\frac{1}{\mu(A)} \int_A f \mathrm d \mu \in F \quad \forall A \in \mathcal F \text{ s.t. } \mu(A) \in (0, \infty), $$ then $f(x) \in F$ for $\mu$-a.e. $x \in X$.

Now let $(X, \tau)$ be a topological space, $\mathcal F := \mathcal B(X)$ its Borel $\sigma$-algebra, and $\mathcal C$ the collection of all closed subsets of $X$.

If $X$ is a metric space, then $\mu$ is both inner and outer regular. This means we can approximate a Borel set $A \in \mathcal F$ with $\mu(A) < \infty$ from above by open sets and from below by closed sets. By dominated convergence theorem and by the fact that $F$ is closed, we can strengthen above theorem by restricting the test sets $A \in \mathcal F$ to those $A \in \tau$ or to those $A \in \mathcal C$.

If $X$ is not a metric space, then $\mu$ is not necessarily regular. Is there a counter-example where above theorem fails if $X$ is a topological space and if we replace $A \in \mathcal F$ with $A \in \tau$ (or with $A \in \mathcal C$)?

Answer 1

$\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}$A counterexample is as follows.

Let $X=(0,\infty)$ and $\tau:=\{\emptyset\}\cup\{(t,\infty)\colon t\in[0,\infty)\}$. Then $\F$ is the usual Borel $\sigma$-algebra over $X$. Let $\mu$ be the Lebesgue measure on $\F$.

Then there are no sets $A\in\tau$ with $\mu(A)\in(0,\infty)$. So, $$\frac1{\mu(A)} \int_A f\,d\mu\in F$$ for any Banach space $E$, any $f\in L^1(X,\mu,E)$, any nonempty closed $F\subseteq E$, and all sets $A\in\tau$ with $\mu(A)\in(0,\infty)$.

Taking now e.g. $E=\R$, $f(x):=e^{-x}$ for $x\in X$, and $F=[0,1/2]$, we see that it is not true that $f(x)\in F$ for $\mu$-almost all $x\in X$.

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To avoid having no open test sets $A$, with $\mu(A)\in(0,\infty)$, we can modify the above example by letting $X=(0,1)$, $\tau:=\{(t,1)\colon t\in[0,1]\}$, $E=\R$, $f(x):=e^{-x}$ for $x\in X$, and $F=[0,1-1/e]$.