# Lebesgue measure of the open unit cube

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?

• I have added some details to my answer. – Mayuresh L Oct 30 '18 at 13:18

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.

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.