Timeline for Is arbitrary union of closed balls in $\mathbb{R}^n$ Lebesgue measurable?
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30 events
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| Nov 16 at 2:38 | comment | added | new account | @MiGang I think you are right but all we need is $A\subseteq B'$. | |
| May 8, 2021 at 10:09 | comment | added | Del | The fact that $\overline{B}\setminus B$ has zero Lebesgue measure when the radii are bounded below should follow from an old argument of Erdos that every point in the set has Lebesgue density at most $1/2$ (originally given for the levelsets of the distance function). With some more care it could be shown that it's actually $(n-1)$-rectifiable (covered by countably many $(n-1)$-dimensional Lipschitz graphs) since it has a one-sided cone property. A more recent result implies that the graphs can be taken of class $C^2$, since the set has a one-sided ball property, see arxiv.org/abs/1703.09561. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| S May 22, 2016 at 13:43 | history | suggested | Amir Sagiv | CC BY-SA 3.0 |
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| May 22, 2016 at 13:34 | review | Suggested edits | |||
| S May 22, 2016 at 13:43 | |||||
| May 22, 2016 at 11:17 | comment | added | Jeff Strom | In your step 1, it appears that every ball contains the origin? | |
| May 22, 2016 at 10:32 | comment | added | MiGang | I think your argument must be revised at the last step. That is, the closure need not be contained in $B'$. For example, consider an open cover of $\mathbb Q\cap (0,1)$ with total measure $0.5$. The closure is $[0,1]$. | |
| Feb 4, 2016 at 23:39 | comment | added | George Lowther | @Mizar: I fixed the answer (better late than never, I suppose), although it does make the answer rather more complicated. | |
| Feb 4, 2016 at 23:36 | history | edited | George Lowther | CC BY-SA 3.0 |
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| Dec 1, 2013 at 12:28 | comment | added | Mizar | Why is $t\to \mu(B_t)$ unimodal? I see this reduces immediately to a finite number of balls, but how do you reduce it to the case of two balls? (Sorry for necroposting) | |
| Oct 29, 2010 at 20:48 | vote | accept | CKD | ||
| Oct 29, 2010 at 2:35 | comment | added | George Lowther | Maybe the bit you were missing is that in a second countable space, the union of a collection of open sets equals the union of some countable subcollection. | |
| Oct 29, 2010 at 2:27 | comment | added | George Lowther | by step 1, we are only looking at a union of balls of radius at least r | |
| Oct 29, 2010 at 2:26 | comment | added | George Lowther | Intuitively the idea is that, as t increases, the distance between the balls increases, so the overlap decreases. The measure of their union must increase as t increases. | |
| Oct 29, 2010 at 2:25 | comment | added | CKD | @George: Thanks for your patience. My first question is that, in Step 2, you write B as a union of countably many open balls of radius at least r, but how can we control the radius of the balls in the countable union (to make it $\geq r$)? My second question is answered by the details you added. Thanks! | |
| Oct 29, 2010 at 2:17 | comment | added | George Lowther | Actually, differentiating that inequality wrt $\epsilon$ shows that the surface area of a union of unit balls is bounded by N times its volume. (Just seemed like an interesting point). | |
| Oct 29, 2010 at 2:12 | comment | added | George Lowther | @CKD: Added more detail. | |
| Oct 29, 2010 at 2:11 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Oct 29, 2010 at 1:03 | comment | added | CKD | George, I still have two questions concerning your sketch of proof. First, how can you guanrantee each of the open balls in the countable union has radius greater than or equal to 1? Second, I don't know how to use convexity to prove $\mu (B') \leq (1+\epsilon)^{N}\mu(B)$ | |
| Oct 27, 2010 at 1:27 | vote | accept | CKD | ||
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| Oct 27, 2010 at 0:20 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Oct 27, 2010 at 0:15 | comment | added | George Lowther | If S is a non-Lebesgue measurable subset of the positive reals, then the union set of circles centered at the origin with radius in S is not Lebesgue measurable. Neither is the union of unit circles centered at (x,0) for x in S (look at its intersection with Rx{y} for |y| < 1). | |
| Oct 26, 2010 at 23:58 | comment | added | Joel David Hamkins | I meant the boundary. In two dimensions, circles (without the interior). | |
| Oct 26, 2010 at 23:52 | comment | added | George Lowther | Actually, were you asking about my argument above or below the line? | |
| Oct 26, 2010 at 23:51 | comment | added | George Lowther | Circles? You mean closed 2-balls? It will handle arbitrary unions of convex sets with nonempty interior, which is precisely the result Faisal just quoted. | |
| Oct 26, 2010 at 23:44 | comment | added | Joel David Hamkins | Can your argument handle an arbitrary union of circles? | |
| Oct 26, 2010 at 23:37 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Oct 26, 2010 at 23:23 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Oct 26, 2010 at 23:16 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Oct 26, 2010 at 22:49 | history | answered | George Lowther | CC BY-SA 2.5 |