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Let $X$ denote the space of bounded Borel functions $f\colon [0,1] \to \mathbb{R}$. Let $M$ denote the space of finite Borel measures on $[0,1]$. What is the largest family $F \subset M$ such that for any $f\in X$ there exists a sequence $(f_n)_{n=1}^\infty \subset C[0,1]$ such that for any $\mu \in F$ $$ \int_{0}^{1} f_n(t) \, d\mu(t) \to \int_{0}^{1} f(t) \, d\mu(t) \quad as \quad n\to \infty, $$ i.e. $f_n \to f$ in the topology $\sigma(X,F)$.

In other words, what is the largest family $F \subset M$ that $C[0,1]$ is sequentially dense in $(X, \sigma(X, F)))$?

Clearly $F$ includes all absolutely continuous (with respect to Lebesgue) measures, because any bounded Borel function can be approximated with uniformly bounded continuous functions converging strongly in $L^1$. On the other hand $F\not = M$ as pointed out in a related post (which is actually a motivation for this question).

Update. If we asked $C[0,1]$ be dense (not sequentially dense) in $(X, \sigma(X,F))$, then one could have taken simply $F=M$. Indeed, for any finite set of measures $\mu_1$, ..., $\mu_n$ from $M$ we know that $C[0,1]$ is dense in $L^1([0,1], |\mu_1| + \ldots + |\mu_n|)$. But since we are asking $C[0,1]$ to be sequentially dense in $(X, \sigma(X,F))$, even existence of the largest family $F$ is not clear, as noted in the comment by @ChristianRemling . So existence of $F$ can be considered as a part of the question.

Another update. As pointed out in the comments by @ChristianRemling the term largest is ambiguous. So let me clarify this point. Let $\mathcal F$ denote the collection of all families $F \subset M$ such that $C[0,1]$ is sequentially dense in $(X, \sigma(X,F))$. By largest I meant maximal in the partially ordered set $(\mathcal F, \subset)$. Now a more precise version of my question has 3 parts:

  1. Does there exist a maximal element of $\mathcal F$?
  2. Does there exist a maximal element of $\mathcal F$ which is a superset of the family of all absolutely continuous measures?
  3. Can any of these maximal elements (if exists) be described explicitly?
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  • $\begingroup$ @ChristianRemling I agree, existence of $F$ is not clear to me at the moment. I thought it would follow from Zorn's lemma, but the point is that the question is about sequential density (added some details to the question). $\endgroup$
    – Skeeve
    Commented Mar 14, 2019 at 17:46
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    $\begingroup$ In fact, it's clear that there is no largest $F$ (in the natural sense that it contains all the others). For example, taking $F$ as any finite collection of Dirac measures works, but we can't take all of them: mathoverflow.net/questions/230028/… $\endgroup$
    – Christian Remling
    Commented Mar 14, 2019 at 21:53
  • $\begingroup$ @ChristianRemling I know that we can't take all Dirac measures. In fact this is exactly the way once proves $F\ne M$ (see also the linked post). But does this really prove non-existence of the largest family $F$? $\endgroup$
    – Skeeve
    Commented Mar 15, 2019 at 7:01
  • $\begingroup$ Say suppose $F\ne M$ is the largest family and we want to prove that it can be extended (to reach a contradiction with the maximality). I know that $C[0,1]$ is sequentially dense in $(X, \sigma(X,F))$ and also in $L^1(X,|\mu|)$, where $\mu \in M \setminus F$, but in general this gives two different approximating sequences. And it is not clear how to conclude from this that $C[0,1]$ is sequentially dense in $(X, \sigma(X, F \cup \{\mu\}))$. $\endgroup$
    – Skeeve
    Commented Mar 15, 2019 at 7:02
  • $\begingroup$ @ChristianRemling I agree that there is no $F_0$ such that $F \subset F_0$ for all $F$, thank you very much for pointing out that we are using different notions of "largest"! I have updated the question to avoid this ambiguity. $\endgroup$
    – Skeeve
    Commented Mar 15, 2019 at 19:06

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