Closed sets with empty interior measure zero

Question

Is the Lebesgue measure of a closed set with empty interior in $\mathbb{R}^{n}$ always zero?

Trying to understand something in the math notes that I don't understand, and if the above is true, it would make more sense. Not sure if true though!

Answer 1

You can just take

$$ U =\bigcup_n B_{1/2^n}(x_n), $$

where $(x_n)_n$ is an enumeration of $\Bbb{Q}$.

Then $U$ is open an of finite measure, so that $A=U^c$ is closed and of positive (infinite) measure.

But the interior of $A$ is void, because it contains no point of $\Bbb{Q}$.

This example avoids having to understand the construction of the Cantor set.

Answer 2

No. For example Cantor-like set, they are closed and nowhere dense, but they can have any measure in $(0,1)$.