You're right, I'll correct that part (I spend too much time working with ultrafilters, where the stronger claim I made would hold...). I used the fact that $R$ is a PID in defining $x$ and $B.$

@VivaanDaga Sorry, I hadn't seen the "sigma" in your question. A finitely additive measure $\mu$ with those properties is maximal iff its domain is $\mathcal{P}(\mathbb{R})^d,$ so there is such an extension iff $d \le 2.$ Section 2 of the Ciesielski/Pelc paper constructs a partition $\mathbb{R}^d=\bigcup_{k<\omega} N_k$ such that for any $k$ and isometries $\langle g_m \rangle,$ there are uncountably many disjoint isometric copies of $\bigcup_{m<\omega} g_m ``(N_k).$ Fix $S \subset \mathbb{R}^d.$ By maximality of $\mu,$ we have $\mu(S \cap N_k)=0,$ so $S$ is in the domain of $\mu.$

@VivaanDaga Yes, by Zorn's lemma.

I doubt any simple modification of this question will lead to a positive answer. E.g., neither $\bigcup [2^{-1} - 2^{-2n}, 2^{-1} - 2^{-2n-1}]$ nor its complement will be in the filter generated by $\mathcal{F}$ and $\mathcal{L}_{1/2}.$

Certainly one would need full $Z_2$ to prove the scheme assigning every definable class coding a pointwise convergent sequence of graphs of Baire-1 functions to a formula defining a double sequence of polynomials which effectively represents the limit Baire-2 function.

What I'm proposing might be doable in $\Pi^1_1-CA_0$ is showing that the formula defined in this construction sends each pointwise convergence sequence of continuous functions to the canonical sequence of rational polynomials with same pointwise limit. This is a single sentence, so it can only use a finite fragment of $Z_2.$ What would require full $Z_2$ is the theorem scheme which provides every formula which defines the graph of a Baire-1 function a formula defining its canonical rational polynomial sequence.

@SamSanders In the language of $Z_2,$ you can consider the equivalence relation of $r_1 \sim r_2$ if both reals code sequences of continuous functions which converge to the same output at each $x.$ Then the construction given provides canonical representatives of each equivalence class.

@PatrickLutz If $\mathbb{R}$ is a countable union of countable sets then every subset of $\mathbb{R}$ is also a countable union of countable sets. My point is that there is then $|\mathcal{P}(\mathbb{R})|$ many functions in Baire class 3 so they can't all be coded by a real.

@AsafKaragila It's easy to get a negative result for Baire 3. The indicator function of any countable union of countable sets is Baire 3, so if $\mathbb{R}$ is a countable union of countable sets, not all such functions will be codable by a real.