How to characterize all sets with Lebesgue-Borel measure of zero?

Question

We know that the Lebesgue-Borel measure (L-B-measure) on $\mathcal B^d$ with $d \in \mathbb N$ of a point in $\mathbb R^d$ is $0$, also clear that the L-B-measure of a countable set of points is $0$, but there might be other more complex sets in $\mathcal B^d$ which have the L-B-measure of zero.

Is it possible in some general way to describe what sets have $0$ L-B-measure?

Answer 1

We could go back to the definition to describe sets with zero measure.

A set $A$ in $\Bbb R^d$ with $d\in\Bbb N$ has zero measure if and only if for every real $\epsilon>0$ there is a finite or countable open cover

$$A\subseteq \bigcup_{i=1}^{\infty} I_i$$

where each $I_k$ is a $d$-dimensional open interval in $\Bbb R^d$, and

$$\sum_{i=1}^{\infty} m(I_i)<\epsilon$$

where $m(I)$ is the usual simple definition of measure of an open interval. I.e.

$$m((a_1,b_1)\times(a_2,b_2)\times\cdots\times(a_d,b_d))=\prod_{i=1}^d(b_i-a_i)$$

More briefly, the infimum of all sums of the measures of the intervals in a finite or countable open cover of $A$ is zero.

Do you need something different from this?

Answer 2

The outer measure $m^o(S)$ of a set $S$ is the $GLB$ of $m(U)$ over all open $ U$ such that $S\subset U$ ....DEF'N : A set $S$ satisfies $m(S)=0$ iff $m^o(S)=0$ . Many uncountable sets have measure $0$. In one dimension the Cantor set $C$ has $m(C)=0$. I am talking about Lebesgue measure.