# Can anyone clarify the meaning of zero content?

I am having hard time understanding the definition of zero content. The following are the definitions of zero content in $\mathbb{R}$ and $\mathbb{R}^2.$

1. A set $Z \subset \mathbb{R}$ is said to have zero content if $\forall \epsilon > 0$ there is a collection of intervals $I_1, \ldots, I_L$ such that
(i) $Z \subset \bigcup_1^L I_l,$ and
(ii) the sum of the lengths of the $I_l$ is less than $\epsilon.$
2. A set $Z \subset \mathbb{R}^2$ is said to have zero content if $\forall \epsilon > 0$ there is a finite collection of rectangles $R_i$ such that
(i) $Z \subset \bigcup_1^M R_i$ and
(ii) the sum of areas of the $R_i$ is less than $\epsilon.$

Thanks!

• What do you mean by "clarify the meaning"? What kind of answer are you looking for here? – Omnomnomnom Nov 20 '14 at 16:47
• I am looking for an easier explanation of those definitions with examples of pictures if possible. – eChung00 Nov 20 '14 at 16:49
• There are some pretty pictures on the relevant wikipedia page – Omnomnomnom Nov 20 '14 at 16:53

Essentially, that means that Lebesgue measure of $Z$ (length in $\Bbb R$, area in $\Bbb R^2$ etc.) is zero. Measures are monotone, i.e. $\lambda(Z)\leq \lambda (\cup I_l)$ since $Z\subseteq \cup I_l$. Measures are additive, that is $\lambda (\cup I_l) = \sum \lambda(I_l) \leq \epsilon$. Hence, you get $\lambda (Z)\leq \epsilon$ for all $\epsilon>0$ and so $\lambda(Z) = 0$.
Measure theory is a bit abstract, and formal definition of $\lambda$ is rather technical. Yet, in early applications in analysis you often just need to know that something has measure of $0$, and you don't care much what is the measure whenever it is greater than $0$ - for example, you want boundaries of nice sets to have measure of $0$. For this reason, instead of giving the whole definition of measures, you simply define what does it mean to be of zero measure, or of zero content in your terminology.
• You should clarify what you mean by "essentially". Note, for example, that neither $\Bbb Q$ nor $\Bbb Q \times \Bbb Q$ have content zero by this definition. – Omnomnomnom Nov 20 '14 at 16:50
• @Omnomnomnom: I am not sure whether $M,L$ are assumed to be finite. – Ilya Nov 20 '14 at 16:53