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What is the concept of set of measure zero? Please explain it in a easy language. Thank you.

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    $\begingroup$ It is important when asking that someone "explain it in... easy language" that you indicate the level of math education involved. Certainly a high school/gymnasium level of "easy" explanation will be more constrained than a beginning graduate math level. $\endgroup$
    – hardmath
    Dec 28, 2013 at 14:33

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From an application point of view, a set of measure zero has the property that you can change the value of the function at points in the set without affecting the value of the integral of the function.

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A set of measure zero, at least in terms of Lebesgue measure (as your tags suggest), is simply a set that's so small that we can contain it in countably many open balls with total volume being arbitrarily small.

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    $\begingroup$ This seems unclear, it sounds like you are only requiring the balls to arbitrarily small when we need the sum of the volumes to be arbitrarily small. Also, as a personal annoyance you really only need finitely many open balls. $\endgroup$ Dec 28, 2013 at 14:08
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    $\begingroup$ @PVAL You're right, I should have been clearer in my writing. However, I don't see how finitely many open balls suffice. For example, in $\mathbb{R}$, $m\left(\mathbb{Q}\right)=0$, but I can't cover $\mathbb{Q}$ in only finitely many open intervals of arbitrarily small length. $\endgroup$
    – Brian
    Dec 28, 2013 at 14:20
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    $\begingroup$ Yes, sorry you're right you need countably many. $\endgroup$ Dec 28, 2013 at 14:23
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Set of measure zero in terms of Lebesgue measure are sets that are "small" in some sense. e.g. Every k-dimensional subspace of $\mathbb{R}^n$ has measure zero if $k < n$. In other words, lines have no area, and planes have no volume.

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    $\begingroup$ It's really important to realize that some sets that don't seem "small" do indeed have measure $0$, and that measure isn't the only way to define "small" (i.e. meager sets). $\endgroup$
    – cderwin
    Dec 28, 2013 at 14:27
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A set of measure zero $E$ (under the Lebesgue measure in $\Bbb R^n$ has the following property, for any small $\epsilon$ I choose, there exists countably many cubes $I_k$ (intervals in one dimension) with $E\subseteq I_k$ and $\sum v(I_k)< \epsilon$ where $v(I_k)$ is just the volume of $I_k$.

To truly have any understanding of a measure zero set, you need to understand this definition and the crucial examples that come immediately afterwards. First, note any countable (and hence finite) subset of $\Bbb R^n$ is measure zero. The Cantor set shows there are uncountable measure zero sets. In higher dimensions, for any set $E\subseteq \Bbb R^{n-k}$ with measure $0$ , we have $E\times \Bbb R^k$ has measure zero in $\Bbb R^n$.

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The simplest explanation I could find to this is here, Link but the other answers here are good as well

Now perhaps what you really may be asking by "easy language" is what is the point of saying that a set has measure zero?

Well, in a way, a set of measure zero is supposed to convey that it is continuous. As a more accessible application, in advanced probability/statistics books you will see that they are talking about a sigma algebra on a Borel set (with measure zero). That is supposed to convey that the space you're working in is continuous and you won't have any "missing"/undefined probabilities. In other words, as far as you're concerned, it's the same as working in $\mathbb{R}^{n}$

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  • $\begingroup$ 'Well, in a way, a set of measure zero is supposed to convey that it is continuous.' What does it mean for a set to be continuous? $\endgroup$ Dec 28, 2013 at 15:30
  • $\begingroup$ Instead of working in R, suppose you were restricted to working in N. In other words, the only numbers you're allowed to work with are whole numbers. Suppose you want to calculate areas of geometric objects. Do you not see a problem with that? $\endgroup$
    – rocinante
    Dec 28, 2013 at 15:53
  • $\begingroup$ Yes there is a problem with that, that's why we have real analysis. How is that at all relevant to the question? $\endgroup$ Dec 28, 2013 at 15:56
  • $\begingroup$ You're asking what does it mean for a set to be continuous. I gave you an example of a set which is not continuous, N. N is not a set of measure zero. $\endgroup$
    – rocinante
    Dec 28, 2013 at 15:59
  • $\begingroup$ First of all, you still haven't said what it means for a set to be continuous. Secondly, $\mathbb{N}$ is a set of measure zero (with respect to Lebesgue measure). $\endgroup$ Dec 28, 2013 at 16:03

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