$A\subseteq \mathbb{R}$ s.t. $|G\smallsetminus A| = \infty$ for every open cover.

Question

I'm working through Axler's Measure and Integration and in exercise 2D number 3 he asks for an example $A\subseteq\mathbb{R}$ such that for every open set $A\subseteq G$ we have $|G\smallsetminus A| = \infty$. Here the vertical bars mean the outer measure.

I've tried many things with no success, such as the Cantor set, the complement of it, tiling the real line with translations of the Cantor set, the rationals, the irrationals.

Intuitively I can see that $A$ cannot be "too big" like all of $\mathbb{R}$ or all of the irrationals, since then we could take $G=\mathbb{R}$ and get a small measure difference. It can't be too small, because any interval can be arbitrarily approximated by open sets from above. So it somehow needs to be "diffuse everywhere and of a medium-ish size, I think.

Answer 1

I'll call $m$ the Lebesgue measure and $m^*$ the Lebesgue outer measure. It isn't as much a matter of "being big" as it is a matter of not being measurable. In fact, you should be aware of this fact:

Lemma 1: A subset $A\subseteq \Bbb R$ is Lebesgue measurable if and only if for all $\varepsilon>0$ there is an open set $G\supseteq A$ such that $m^*(G\setminus A)\le\varepsilon$.

The "only if" part is a standard result in most books. Anyways, if $A$ is measurable, then for all $n\ge 1$ call $A_n=A\cap B(0,n)$, and select $G_n\supseteq A$ an open set such that $m(G_n)=m^*(G_n)\le m^*(A_n)+2^{-n}\varepsilon=m(A_n)+2^{-n}\varepsilon$. Therefore $m(G_n\setminus A_n)\le 2^{-n}\varepsilon$ and thus \begin{align}m^*\left(\left(\bigcup_{n\ge 1} G_n\right)\setminus A\right)&=m\left(\left(\bigcup_{n\ge 1} G_n\right)\setminus \bigcup_{n\ge 1}A_n\right)\le m\left(\bigcup_{n\ge 1} (G_n\setminus A_n)\right)\le\\ &\le\sum_{n\ge 1}m(G_n\setminus A_n)\le\varepsilon\end{align}

For the "if" part, consider $G_n\supseteq A$ open such that $m^*(G_n\setminus A)\le \frac1n$. By monotonicity, $m^*\left(\left(\bigcap_{n\ge 1}G_n\right)\setminus A\right)\le\inf_{n\ge 1}\frac1n=0$, and therefore $A$ is difference of a $G_\delta$ set and a null set.

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Also, a useful lemma:

Lemma 2: Let $A\subseteq \Bbb R$ be a set and let $\{E_k\}_{k\in\Bbb N}$ be disjoint measurable sets such that $A\subseteq \bigcup_{k\in\Bbb N} E_k$. Then, $$m^*(A)=\sum_{k=0}^\infty m^*(A\cap E_k)$$

Only $\ge$ needs a proof. Notice that $A\supseteq \bigcup_{k=0}^n A\cap E_k$, and therefore, by monotonicity, only $m^*\left(\bigcup_{k=0}^n A\cap E_k\right)=\sum_{k=0}^n m^*(A\cap E_k)$ is needed. Let $n$ the first natural number that fails additivity. Evidently $n>0$. By the definitory property of measurable set, $$m^*\left(\bigcup_{k=0}^n A\cap E_k\right)=m^*\left(\left(\bigcup_{k=0}^n A\cap E_k\right)\setminus E_n\right)+m^*\left(\left(\bigcup_{k=0}^n A\cap E_k\right)\cap E_n\right)=\\=m^*\left(\bigcup_{k=0}^{n-1} A\cap E_k\right)+m^*(A\cap E_n)=\sum_{k=0}^n m^*(A\cap E_k)$$ Against the hypothesis on $n$.

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Now, let $X\subseteq (0,1)$ be a non-measurable subset, call $\varepsilon_X>0$ an epsilon coming from lemma 1, and call $n+X=\{n+x\,:\,x\in X\}$ its translation by $n$. It is clear that $\varepsilon_X$ is a good choice of $\varepsilon_{n+X}$ as well. Now, consider $Y=\bigcup_{n\in\Bbb N}(n+X)$ and let $G\supseteq Y$ open. We have that $$G\setminus Y\supseteq \bigcup_{n\in\Bbb N} G\cap(n,n+1)\setminus (n+X)$$

And by lemma 2: \begin{align}m^*(G\setminus Y)&\geq m^*\left(\bigcup_{n\in\Bbb N} G\cap(n,n+1)\setminus (n+X)\right)=\\&=\sum_{n\in\Bbb N}m^*(G\cap (n,n+1)\setminus (n+X))\ge\sum_{n\in\Bbb N}\varepsilon_X=\infty\end{align}

Answer 2

Let $A$ be any Bernstein set, i.e., a set $A\subseteq\mathbb R$ such that both $A$ and its complement meet every uncountable closed set. It easily follows that both $A$ and its complement meet every uncountable $F_\sigma$ set. (The same goes for uncountable Borel sets but that's more difficult and not needed here.) A Bernstein set is easily "constructed" by transfinite induction, using the facts that there are just $2^{\aleph_0}$ uncountable closed sets, and each one of them contains $2^{\aleph_0}$ points.

Consider any open set $G$ such that $A\subseteq G$; I claim that $|G\setminus A|=\infty$.

Let $H$ be an open set such that $G\setminus A\subseteq H$; I have to show that $|H|=\infty$. In fact, I will show that $\mathbb R\setminus H$ is countable.

Now $\mathbb R\setminus H\subseteq(\mathbb R\setminus G)\cup(G\setminus H)$, and $\mathbb R\setminus G$ is countable because it's a closed set which is disjoint from $A$, and $G\setminus H$ is countable because it's an $F_\sigma$ set which is disjoint from the complement of $A$.