What is the concept of set of measure zero? Please explain it in a easy language. Thank you.
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 3 $\begingroup$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 $\Bbb R^n$.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$The simplest explanation I could find to this is here, 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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