Consider the space NBV$[0,1]$ of functions $f : [0,1] \to \mathbb{R}$ of bounded variation, normalized by the condition that $f(0) = 0$, $f$ is right-continuous on $(0, 1]$ and that $f$ has left-hand limits in $(0,1]$. Equip NBV$[0,1]$ with the weak($^*$) topology generated by Stieltjes-integration against continuous functions $C[0,1]$, that is by the mappings NBV$[0,1] \to \mathbb{R}$, $H \mapsto \int_0^1 f(t) H(dt)$ for $f \in C[0,1]$.
Now consider on NBV$[0,1]$ the Borel $\sigma$-algebra generated by this weak topology. We can also consider on NBV$[0,1]$ the $\sigma$-algebra generated by the evaluation mappings $H \mapsto H(t)$ for $t \in [0,1]$. Are these $\sigma$-algebras the equal?
Maybe this is helpful: I can show that on the subset of increasing functions these two $\sigma$-algebras indeed coincide. Moreover, the Borel $\sigma$-algebra on all of NBV$[0,1]$ is generated by the above integration mappings against continuous functions.
Yes, both $\sigma$-algebras are the same. It holds more:
Identify NBV$[0,1]$ with the space of signed Borel measures $ca([0,1])$:
The space $ca([0,1])$ is the dual of the Banach space $C[0,1]$ given by $\langle f, \mu \rangle := \int f \, d\mu$. Under the identification $\textrm{NBV}[0,1] = ca([0,1])$ the weak$^*$ topology on $ca([0,1])$ and the weak$^*$ topology on $\textrm{NBV[0,1]}$ coincide.
The following $\sigma$-algebras on $ca([0,1])$ coincide:
I omit the proof details. For the equivalences 1. $\Leftrightarrow$ 2. $\Leftrightarrow$ 3. see [Kallenberg, "Random Measures", Section 1], for 3. $\Leftrightarrow$ 4. see [Bogachev, "Measure Theory", Vol 2, Theorem 6.10.6] and for 4. $\Leftrightarrow$ 5. see [Bogachev, "Measure Theory", Vol 2, Theorem 6.3.4] (for this one has to show that the weak$^*$ topology on $ca[0,1]$ is perfectly normal).
From 2. it follows it follows that the Borel $\sigma$-algebra on $\textrm{NBV}[0,1]$ coincides with the $\sigma$-algebra generated by the evaluation mappings $H \mapsto H(t)$ for $t > 0$. Since $H(0) = 0$ for all $H \in \textrm{NBV[0,1]}$ it follows that the Borel $\sigma$-algebra is generated by the evaluation mappings for all $t \geq 0$.