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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!

Ben Grossmann
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eChung00
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1 Answers1

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Update: the answer below is not correct, see comments.

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.

SBF
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  • 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. – Ben Grossmann Nov 20 '14 at 16:50
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    It's not quite the same as zero Lebesgue measure, is it? It doesn't allow the cover to be infinite. So an unbounded set can't have zero content. – TonyK Nov 20 '14 at 16:51
  • That's what I thought too, but it looks like the concept in the OP wants a finite collection of intervals/boxes, so it doesn't get all of the Lebesgue null sets. – hmakholm left over Monica Nov 20 '14 at 16:53
  • @Omnomnomnom: I am not sure whether $M,L$ are assumed to be finite. – SBF Nov 20 '14 at 16:53
  • @HenningMakholm: perhaps, you are right. The link by Omnomnomnom made in the comment to OP looks more relevant. – SBF Nov 20 '14 at 16:54
  • The books says the subscripts $M$ and $L$ are finite. – eChung00 Nov 20 '14 at 16:55