Is the difference of a compact set and a measure zero set a Borel set?

Question

Let $E$ be a compact set. Is it true that for any measure zero set $O$, is the set $E\setminus O$ always a Borel set?

Answer 1

No. We will construct a counterexample in $\mathbb{R}$ with Lebesgue measure.

Let $\mathcal{C}$ denote the Cantor set and $c:[0,1]\to[0,1]$ the Cantor function. Recall that $c$ is continuous and monotonically increasing. Moreover, $c$ maps $\mathcal{C}$ to a set of unit measure. Define $f(x)=\frac12(x+c(x))$ and note that it maps $\mathcal{C}$ to the set $f[\mathcal{C}]$ whose measure is $\frac12$. Thus, $f[\mathcal{C}]$ has a non-measurable subset $V$. Moreover, $f$ is continuous and strictly increasing and thus possesses a continuous inverse $g(x)=f^{-1}(x)$. Therefore, $g[V]\subset\mathcal{C}$ is not Borel.

Now, set $E=\mathcal{C}$ and $O = g[V]$. Cantor set is compact and measure zero, so $E$ is compact and $O$ is measure zero. Moreover, since $\mathbb{R}$ is Hausdorff, $E$ is closed and therefore Borel. Now, assume $E\setminus O$ is Borel. Since $O\subset E$ we have $O=E\setminus (E\setminus O)$ contradicting the fact that $O$ is not Borel. Therefore, $E\setminus O$ is not Borel.

Answer 2

Alternate. $E = [0,1]$ is compact. There are $2^{\mathfrak c}$ measure zero sets $O$ contained in $E$, so there are $2^{\mathfrak c}$ sets $E \setminus O$. But there only $\mathfrak c$ Borel sets. Thus not every set $E \setminus O$ is Borel.