If $\lim_{n\to \infty}\mu(B_n\setminus A_n )= 0 \& A_n\subset E\subset B_n$, then is $E$ measurable?

Question

I've asked the following question here:

Let $E,A_n,B_n$ be subsets of $\mathbb{R}^k$ with $A_n\subset E\subset B_n$. Suppose that $A_n,B_n$ are measurable sets and $$\lim_{n\to \infty} \mu(B_n \setminus A_n) = 0$$

Is $E$ also measurable?$\mu$ is a generic measure.

In the comments, I was given a hint to take $E$ as a Lebesgue measurable but not Borel measurable set and sandwich that between two Borel sets of same measure. One such $E$ that I could think of is the standard example given using Cantor functions:

Let $\psi (x)=x+\phi(x)$ where $\phi(x)$ is the Cantor function(continuous and non-decreasing). Thus $\psi:[0,1]\to [0,2]$ is invertible and $\psi,\psi^{-1}$ map Borel sets to Borel sets. Choose a non-Borel subset $S\subset \psi(C)$, then $E=\psi^{-1}(C)$ is Lebesgue measurable but not Borel measurable. What Borel sets can I use to sandwich $E$ which can be expressed as a limit of other Borel sets?

Answer 1

Take $E\subseteq \Bbb{R}$ a bounded measurable set that is not Borel.

Then by regularity of the Lebesgue measure there is a bounded $F_{\sigma}$ set $F$ and a bounded $G_{\delta}$ set $G$ such that $F\subseteq E \subseteq G$

and $F=\bigcup_nF_n$ and $G=\bigcap_n G_n$ and $G_1$ has finite measure.

Take $O_n=\bigcap_{k=1}^nG_n$ and $K_n=\bigcup_{k=1}^nF_n$

and $\mu(O_n \setminus K_n) \to \mu(G\setminus F)=0$

Answer 2

Let $\mathcal{M}$ be the $\sigma$-algebra under consideration. If $\mathcal{M}$ is $\mu$-complete (in the sense that if $N\in\mathcal{M}$, $\mu(N)=0$, and $A\subseteq N$, then $A\in\mathcal{M}$), then the answer is yes.

For your case, define $A=\cup_{n}A_{n}$ and $B=\cap_{n}B_{n}$, then $A$ and $B$ are measurable. Moreover, $B\setminus A\subseteq B_{n}\setminus A_{n}$, so $\mu(B\setminus A)\leq\mu(B_{n}\setminus A_{n})\rightarrow0$, so $\mu(B\setminus A)=0$.

Finally, $E\setminus A\subseteq B\setminus A$. Since the $\sigma$-algebra is $\mu$-complete, it follows that $E\setminus A$ is measurable and hence $E=A\cup\left(E\setminus A\right)$ is also measurable.

Note that the conclusion is false if $\mathcal{M}$ is not $\mu$-complete. For example, $\mathcal{B}(\mathbb{R})$, the Borel $\sigma$-algebra on $\mathbb{R}$ is not complete with respect to the Lebesgue measure $\lambda$. Then, there exists $N\in\mathcal{B}(\mathbb{R})$ and $E\subseteq N$ such that $\lambda(N)=0$ but $E\notin\mathcal{B}(\mathbb{R})$. In this case, take $A_{n}=\emptyset$, $B_{n}=N$...