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I've encountered two statements regarding the Lebesgue measure that don't exactly contradict each other, but seem to me to be a little bit unintuitive when regarded with respect to one another. The statements are:

  1. For each interval $[a,b]$ there exists a set $K\subset[a,b]$ that is compact , totally disconnected and with positive measure (i.e. doesn't contain an interval).
  2. Let $A\subset\mathbb{R}$ be some measurable set, and $0\leq\alpha<1$ such that for every interval $I\subset\mathbb{R}$ it holds that $m(A\cap I)\leq\alpha m(I)$ then $m(A)=0$

The second statement seems to imply that any positive measure set must contain an interval up to a set of arbitrary measure $\epsilon>0$ (since a suitable $\alpha$ can't be found for a positive measure set).

On the other hand the first statement says that for any interval I can find a subset of any desired positive "not-full" measure. I realize these statements don't contradict, but they seem a little "contradictory in nature". Intuitively I would expect the first statement to somehow allow me to construct a positive measure set with the property desired in the second statement.

I don't really have a question here but was hoping maybe someone can shed their own perspective on how they see these two properties of the Lebesgue measure co-existing. Maybe you can give me some further intuition.

Thanks a lot!

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  • $\begingroup$ Certainly they can't mean to include $\alpha=1$ in the second statement, since every set $A$ satisfies $m(A\cap I) \le m(I)$. $\endgroup$ – Greg Martin Apr 6 '14 at 8:54
  • $\begingroup$ Of course. Thanks I corrected that. $\endgroup$ – user126743 Apr 6 '14 at 9:10
  • $\begingroup$ In part 2., the $\alpha$ is fixed. You can have a positive measure set $A$, such that $m(A \cap I)<1$ for all intervals, but as you take smaller and smaller intervals $I_n$ "zooming in" on a part of $A$, then $m(A \cap I_n)$ arbitrarily approaches 1. Part 2. works because the $\alpha$ bounds this measure away from 1 (by some tiny amount). $\endgroup$ – Travis Bemrose Dec 16 '15 at 4:15
  • $\begingroup$ I interpret (2) as saying a set that contains no interval and "contracts" the measure of all intervals by at least $\alpha$ upon intersection with them must be a set of measure zero. It does seem contradictory to (1) intuitively. But consider a fat Cantor set with positive measure. Maybe we can find a decreasing sequence of intervals that intersect this fat Cantor set, where no $\alpha<1$ will capture the "contraction" of all intersections. $\endgroup$ – jdods Jul 15 '16 at 23:19
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My reaction is just to trace through some examples. The set $K$ in part $1$ is something like a Cantor set-just fattened up to maintain positive measure, say for concreteness $1/2$. While we're simplifying, may as well take $[a,b]=[0,1]$ as well. $K$ needn't be uniformly distributed-indeed the point of your part 2 is that it can't be-so we may suppose we can remove some interval and increase the proportional measure of $K$. In the most usual construction, we might have $m(K\cap [0,3/8])=1/2$. Then if $K$ is self-symmetrical, as it may certainly be, we immediately get arbitrarily large proportions of $K$ in sufficiently small intervals.

This is all in contrast to the case of the usual Cantor set $C$: one way to prove $m(C)=0$ is to observe that for every interval $I$, $C$ is contained in a subset of $I$ of relative measure no more than $2/3$. For the construction of $C$ is uniform: at every stage we remove intervals of relative measure $2/3$. In contrast again this shows why we don't construct $K$ by uniform removal of even very small intervals.

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  • $\begingroup$ Thanks! I like your reply, and the point on uniformity is interesting as well and hadn't occurred to me. I guess I'll have to accept that these properties just don't sit together well in my mind lol. $\endgroup$ – user126743 Apr 6 '14 at 9:48
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    $\begingroup$ Glad you like it! Loosely quoted: "In mathematics, young man, you don't understand things-you get used to them." -von Neumann $\endgroup$ – Kevin Carlson Apr 6 '14 at 9:51
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    $\begingroup$ And today we would just as likely say "In mathematics, young woman...." $\endgroup$ – Greg Martin Apr 6 '14 at 20:40
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For 1: There is an open set $U$ containing all rationals such that $m(U)< (b-a).$ Thus $m([a,b]\setminus U) > 0.$ The set $[a,b]\setminus U$ is compact, is a subset of $[a,b],$ and since it contains no rational, has no interior.

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  • $\begingroup$ Note that this is a two year old Question, and it does not ask for a proof of 1. and 2., but rather for reconciling the "intuitions" OP has about them. $\endgroup$ – hardmath Jul 16 '16 at 2:36

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