Lebesgue measure of the open unit cube

Question

Let $U:=(0,1)\times…\times(0,1) \subset \mathbb{R^n}$ be the open unit cube and $\lambda_n$ the Lebesgue measure.

How to prove that

1) For all $\varepsilon \in (0,1)$ there is a compact set $C_\varepsilon \subset U$ without interior points such that $\lambda_n(C_\varepsilon)>1-\varepsilon$

2) There is no compact set $C \subset U$ with $\lambda_n(C)=\lambda_n(U)=1$

For 1) I tried this:

Since $U$ consists of open intervals and $\varepsilon >0$, there are two sets $C,O$ uch that $C$ is compact and $O$ open. So $C \subset U \subset O$ and $\lambda_n(O/C)<\varepsilon$. Here I don't know how to continue.

Is this way correct?

For 2) I don't see how it can be shown since on closed intervals this statement is true.

EDIT: 2) How to show it with additivity of measure?

Answer 1

Short answer to first question is by using inner regularity of Lebesgue measure. That’s what you have done.

Note that $U$\ $C$ $\subseteq$ $O$ \ $C$

Hence by monotonicity of measure $\lambda_n$($U$ \ $C$) $< \epsilon$ which gives $\lambda_n(C)> 1-\epsilon$.

Now for the second question,

Complement (wrt open unit cube) of any compact set in open unit cube is open and Lebesgue measure of open set is always non-zero. These observations prove second question using additivity of measure.

Answer 2

I think the they're looking for something more explicit with $1)$.

You can try the set $C = [\delta,1-\delta]\times [\delta,1-\delta]...\times [\delta,1-\delta]$ and choose $\delta$ wisely with respect to $\varepsilon$.

For the other part of the question I'll give you a hint that every compact set lies in some set of the form above.