Timeline for Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2023 at 0:19 | comment | added | Terry Tao | One reason why this question is non-trivial is that the analogous claim with ${\mathbb R}$ replaced by ${\mathbb Q}$ is of course false (${\mathbb Q}^n$ has measure zero), and so one must use a property of ${\mathbb R}$ (e.g., local compactness) that is not shared by ${\mathbb Q}$ in the proof. | |
| Apr 7, 2021 at 8:40 | answer | added | Pietro Majer | timeline score: 2 | |
| Apr 3, 2021 at 10:26 | comment | added | Dave L Renfro | (+1) Yesterday my first thought was this is trivial, but as I outlined an approach (comment above), I realized I needed to know $\mathbb R$ does not have measure zero. The problem seems to be that we need to prove that separately adding the lengths of a lot of tiny intervals whose union is an interval $I$ can't be super-efficient in the sense of giving a value less than $|I|$ (even showing the sum is greater than $r|I|$ for some possibly small positive constant $r$ would be enough). I thought someone would find a way of easily taking care of this, but in looking over the answers now . . . | |
| Apr 3, 2021 at 6:52 | vote | accept | Stefan Friedl | ||
| Apr 2, 2021 at 21:11 | answer | added | Pietro Majer | timeline score: 5 | |
| Apr 2, 2021 at 20:50 | answer | added | Mizar | timeline score: 1 | |
| Apr 2, 2021 at 20:38 | comment | added | Pietro Majer | The (exterior) measure of $A\subset \mathbb{R}^n$ is defined as the infimum of the measures of unions of cubes, over all countable coverings of $A$ by cubes . The well-known and now standard argument by Borel, to show that for a cube the infimum is not zero is, if I remember correctly, the birthplace of the notion of the Heine-Borel compactness. | |
| Apr 2, 2021 at 20:35 | answer | added | Abdelmalek Abdesselam | timeline score: 3 | |
| Apr 2, 2021 at 20:18 | answer | added | Will Sawin | timeline score: 16 | |
| Apr 2, 2021 at 20:12 | comment | added | Ben McKay | Look at Guillemin and Pollack, Differential Topology, Proposition on page 203, for a complete elementary geometric argument that open sets have positive measure. | |
| Apr 2, 2021 at 19:48 | answer | added | Iosif Pinelis | timeline score: 5 | |
| Apr 2, 2021 at 19:11 | comment | added | Willie Wong | But swatting this fly with Vitali seems an overkill. (Though if you want to prove "measure non zero" using strictly the definition of measure zero as you listed, you may have to do something like this.) | |
| Apr 2, 2021 at 19:11 | comment | added | Willie Wong | Depends on the level of precision you want: if $Q$ is a cube and $C_\alpha$ (with $\alpha\in A$) is a countable cover of $Q$ with cubes, then by Vitali's covering lemma, there is a countable $B\subseteq A$ such that $\{C_\beta: \beta \in B\}$ are pairwise disjoint, and if you dilate each $C_\beta$ by 5 you again get a cover. So this shows that any covering of a cube $Q$ with cubes will have $\sum |C_\alpha| \geq 5^{-d} |Q|$ where $d$ is the number of dimensions. This shows that cubes have non-zero measure. | |
| Apr 2, 2021 at 18:47 | comment | added | Stefan Friedl | Exactly. It's "obviously true", but not that obvious after all. | |
| Apr 2, 2021 at 18:32 | comment | added | Benjamin Steinberg | Basically you want to argue a cube can't have measure zero since every open subset contains a cube. | |
| Apr 2, 2021 at 17:57 | comment | added | Dave L Renfro | Off the top of my head for $n=1,$ if an open set has measure zero then, by monotonicity (easily proved), there exists an open interval having measure zero, and since each rational translate of this open interval has measure zero (easily proved), it follows that the union of the countably many rational translates of this interval has measure zero (countable union of measure zero sets has measure zero is easily proved). This reduces the result to proving that $\mathbb R$ does NOT have measure zero. Or am I missing something? | |
| Apr 2, 2021 at 17:22 | history | asked | Stefan Friedl | CC BY-SA 4.0 |