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I'm having quite a bit of trouble understanding this intuitively.

From what I've learned, a subset $A$ of $R^n$ has ($n$-dimensional) measure zero if for every $\epsilon > 0$, there is a countable cover $\{U_1, U_2,...\}$ of A by closed rectangles such that $\sum_{i=1}^{\infty} v(U_i) < \epsilon$.

And for Jordan measure otherwise known as content zero, there's the definition; a subset $A$ of $R^n$ has ($n$-dimensional) content zero if for every $\epsilon > 0$, there is a finite cover $\{U_1,...U_n\}$ of A by closed rectangles such that $\sum_{i=1}^{n} v(U_i) < \epsilon$.

Now the idea of A having a cover, i.e A being contained in the union of all $U_i$ makes sense to me. Where I'm confused is the idea of the sum of all the volumes being $< \epsilon$ for any $\epsilon > 0$. If it's a cover, no matter how we choose the intervals, isn't the sum the same?

Not too sure where the flaw in my understanding is but consider $I_1=(a_1,b_1), I_2=(a_2,b_2)$ which has a given sum and is a cover of some set. If we refine this further to say $I_1=(a_1,b_1), I_2=(a_2,b_2), I_3=(a_3,b_3)$, isn't the sum still the same? So that we don't have the sum of the volumes less than any given $\epsilon$.

Some clarification so that this is more intuitive would be appreciated!

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Let's think the most trivial case, $A=\{0\}$ in $\mathbb R$. It's clear that $m(A)=0$. Why is that?

For example, $[-1,1]$ is a countable cover of $A$, so $m(A)\leq m[-1,1]=2$. But $[\frac{-1}{2},\frac12]$ is also a countable cover of $A$, which lead us to $m(A)\leq 1$. You can pick up a smaller cover for $A$, $[\frac{-1}{4},\frac14]$, so $m(A)\leq \frac12$.

This is (more or less) an intuitive approach. You can pick a countable cover whose measure is as close to zero you want (that is, smaller than any $\varepsilon$ you pick).

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  • $\begingroup$ Yeah, this example makes sense. But then when I consider the fact "that the measure of any finite set is also 0". Ex. {1,2} for example, I have trouble generalizing this idea to being able to pick countable covers whose measure is as close to zero as we want for sets past the trivial example. Is there not a limit to how small we can have our cover be for certain sets (i.e not less than any $\epsilon > 0$.) $\endgroup$ – SS' Feb 12 '17 at 0:45
  • $\begingroup$ All right. Then, you can use the same idea: $\{[\frac12, \frac32]\}$, $\{[\frac32, \frac52]\}$ is a countable covering of $A=\{1,2\}$ and its measure is $1$. But $\{[\frac34, \frac54]$, $[\frac74, \frac94]\}$ is also a countable covering of $A$, and its measure is $\frac12$. $\endgroup$ – A. Salguero-Alarcón Feb 12 '17 at 0:48
  • $\begingroup$ @SS' if you pick an $\varepsilon$, the cover $\{[1-\frac\varepsilon4, 1+\frac\varepsilon4], [2-\frac\varepsilon4, 2+\frac\varepsilon4]\}$ is a countable cover of $A=\{1,2\}$, and its measure is exactly $\varepsilon$. $\endgroup$ – A. Salguero-Alarcón Feb 12 '17 at 0:50
  • $\begingroup$ Yep this clears things up. Thanks ! : D $\endgroup$ – SS' Feb 12 '17 at 0:57
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It's not the same cover for each $\epsilon$!

For simplicity, ignore content and let $n=1$. A set $A$ has measure $0$ if for each $\epsilon>0$ there is some cover $\mathcal{C}_\epsilon=\{U_i^\epsilon: i\in\mathbb{N}\}$ of $A$ with the sum of the lengths of the intervals $U_i^\epsilon$ less than $\epsilon$. But that cover is allowed to change with $\epsilon$!

No single cover "witnesses" that $A$ has measure zero; instead, the existence of "arbitrarily small" covers does.

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