Is every measurable set a measure-independent limit of open sets

Question

My main question is

Q1. Let $B$ be a Borel-measurable subset of $\mathbb R$. Is there a sequence of open sets $U_n$ independent of any measure such that for all Borel probability measures $\mu$ on $\mathbb R$, $\mu(B \Delta U_n) \to 0$?

Any open set is trivially a measure-independent limit of open sets.

Any closed set is an intersection of a nested sequence of open sets and hence is a measure-independent limit.

I wonder if Q1 can be answered negatively by showing that $\mathbb Q$ is a counter example.

Q1a. Is $\mathbb Q$ a counter example?

If $B$ is a measure-independent limit of $U_n$ then $1_B$ is a pointwise-limit of the sequence $1_{U_n}$ because of Dirac delta measures. Maybe this can be used to show that $\mathbb Q$ is a counter example, but I don't know how.

Motivation: last paragraph of the question in https://mathoverflow.net/questions/167823/the-borel-sigma-algebra-of-the-set-of-probability-measures?rq=1

Answer 1

I can sharpen PhoemueX's very nice answer a little bit. In particular, yes, $\mathbb{Q}$ is a counterexample.

You are quite right that, if a set $B$ is to be the measure-independent limit of a sequence of sets $A_n$, we must have $1_{A_n} \to 1_B$ pointwise. In particular, as PhoemueX says, we must have $B = \bigcup_{k \ge 1} \bigcap_{n \ge k} A_n$. So if $A_n = U_n$ are open then $B$ must be $G_{\delta,\sigma}$.

But we can say more: it must also be that $B = \bigcap_{k \ge 1} \bigcup_{n \ge 1} A_n$. In fact, it is a nice little exercise to show that for any sequence of sets $A_n$, we have $$\limsup_{n \to \infty} 1_{A_n} = 1_{\limsup A_n}, \qquad \liminf_{n \to \infty} 1_{A_n} = 1_{\liminf A_n}$$ where $$\limsup A_n := \bigcap_{k \ge 1} \bigcup_{n \ge k} A_n, \qquad \liminf A_n := \bigcup_{k \ge 1} \bigcap_{n \ge k} A_n.$$ So in particular if $1_B = \lim_{n \to \infty} 1_{A_n}$ pointwise, then we have $1_B = \limsup 1_{A_n} = \liminf 1_{A_n}$ and hence $B = \limsup A_n = \liminf A_n$.

And when $A_n = U_n$ are open, then $\limsup U_n$ is actually $G_\delta$. So in order for a set to be a measure-independent limit of open sets, it must be both $G_\delta$ and $G_{\delta,\sigma}$.

However, $\mathbb{Q}$ is not $G_\delta$. Since $\mathbb{Q}$ is dense, if it were $G_\delta$ it would be a dense $G_\delta$, or in other words comeager. But since $\mathbb{Q}$ is countable it is also meager. This contradicts the Baire category theorem: $\mathbb{R}$ is not meager, so no subset of $\mathbb{R}$ can be simultaneously meager and comeager.

Answer 2

This is not a complete answer, but maybe helpful nevertheless:

As you noted, it is a necessary condition that $\chi_{U_n}$ (the characteristic function/indicator function of $U_n$) fulfils $\chi_{U_n} \to \chi_B$ pointwise.

But conversely, if this is the case, then

$$ \mu(B \Delta U_n) = \int |\chi_B - \chi_{U_n} | \, d\mu \to 0 $$

by dominated convergence (the integrand is dominated by $2$, which is integrable, because $\mu$ is a finite measure).

Hence, what you are effectively asking is if for every Borel set $B$, there is some sequence $(U_n)_n$ of open sets with $\chi_{U_n} \to \chi_B$ pointwise.

Now note that every open set $U$ fulfils $\chi_U = \lim_n f_n$ pointwise for a suitable sequence $(f_n)_n$ of continuous functions (basically, take $U = \bigcup_n K_n$ with $K_n$ compact and $K_n \subset K_{n+1}$ and use (e.g.) Urysohns Lemma to construct $f_n \in C_c(U)$ with $f_n \equiv 1 $ on $K_n$).

Hence, every functio $\chi_U$ is of Baire class 1 (see http://en.wikipedia.org/wiki/Baire_function).

So if what you are asking is true, then every indicator function $\chi_B$ with $B$ Borel would be of Baire class (at most) two.

I highly doubt that this is true, but have not found an explicit counterexample/source where this is stated.

EDIT: Also, the pointwise limit $\chi_B = \chi_{U_n}$ implies that for $x \in B$, we have $\chi_{U_n}(x) \to 1$. Because of $\chi_{U_n}(x) \in \{0,1\}$, this yields some $n_x \in \Bbb{N}$ with $\chi_{U_n}(x) = 1$ for all $n \geq n_x$ and hence $x \in \bigcap_{n\geq n_x} U_n$. Hence,

$$ x \in \bigcup_{k \geq 1}\bigcap_{n \geq k} U_n. $$

A similar argument shows that if $x \notin \bigcup_{k} \bigcap_{n \geq k} U_n$, then $x \notin B$. Hence,

$$ B = \bigcup_{k \geq 1} \bigcap_{n \geq k}U_n, $$

which implies that $B$ is a $G_{\delta, \sigma}$ set, or (other notation, same statement) a $\Sigma_3^0$ set (see http://en.wikipedia.org/wiki/Borel_hierarchy and http://en.wikipedia.org/wiki/G%CE%B4_set for the notation used here).

The article http://en.wikipedia.org/wiki/Borel_hierarchy#Boldface_Borel_hierarchy claims (the last bullet point in that paragraph):

If $X$ is an uncountable Polish space, it can be shown that $\mathbf{\Sigma}^0_\alpha$ is not contained in $\mathbf{\Pi}^0_\alpha$ for any $\alpha < \omega_1$, and thus the hierarchy does not collapse.

This implies that not every Borel set is a $G_{\delta, \sigma}$ set. Hence, your claim is false.