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This is from problem $8$, Chapter II of Rudin's Real and Complex Analysis.

The problem asks for a Borel set $M$ on $R$, such that for any interval $I$, $M \cap I$ has measure greater than $0$ and less than $m(I)$.

I was thinking of taking the Cantor approach: taking $R$ to be the union of $[a,b]$ with $a$ and $b$ rationals, and for each $[a,b]$ we construct Cantor sets inside it. During theconstruction of each Cantor set, in order to have positive measure on it, we need to take off smaller and smaller intervals from it, namely the proportion goes to $0$. As a result, these Cantor sets are extremely "dense" on their ends. If for an interval $I$ it intersects with the Cantor set on $[a,b]$ while $b-a>>m(I)$, we shall expect the measure of intersection to be rather close to $m(I)$ and then we lose control on these cases.

Is there any way to fix this or shall I consider other approaches?

Thank you

Drarp
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VADupleix
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1 Answers1

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Instead of trying to understand your approach (which sounds complicated), let me tell you how I'd do it. I guess you know how to construct a closed nowhere dense set of positive measure inside a given interval, right? Enumerate all the rational intervals in a sequence $I_1,I_2,I_3,\dots$. Now construct an infinite sequence $M_1,N_1,M_2,N_2,M_3,N_3,\dots$ of pairwise disjoint closed nowhere dense sets of positive measure, with $M_k,N_k\subset I_k$. [1] The $F_\sigma$-set $M=M_1\cup M_2\cup M_3\cup\cdots$ does what you want. [2]

[1] Note that, at each step of the construction, you have constructed a finite number of closed nowhere dense sets, whose union is therefore a closed nowhere dense set. Hence the interval $I_k$ you are currently working in will contain a subinterval which is disjoint from that nowhere dense set. Construct the next closed nowhere dense set inside that subinterval.

[2] Any interval $I$ contains some rational interval $I_k$. Since $N_k\subseteq I_k\subseteq I$ and $N_k\cap M=\emptyset$, it follows that $M\cap I\subseteq I\setminus N_k$ and $m(M\cap I)\le m(I\setminus N_k)=m(I)-m(N_k)\lt m(I)$.

bof
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    Sorry could you be more specific on your M and N? I don't quite understand why you constructed two sets in each interval and also how can you make sure they can be disjoint? (I'm assuming you take I_i to be all possible rational intervals so they overlaps with each other). Thanks. – VADupleix Oct 07 '14 at 04:03
  • Thank you for the neat construction. It makes sense to me now. – VADupleix Oct 07 '14 at 18:49
  • I don't think we can actually say that $N_k \cap M = \emptyset$. I think the best we can say is that if $N = N_1 \cap N_2 \cap \cdots$, then $N \cap M = \emptyset$. However, we don't have any guarantee that $m(N) > 0$ -- in fact, if it were, we would arrive at a contradiction, since $N \subset I_k$ for all $k. – tyo Oct 30 '19 at 22:17
  • @tyo Actually $N_k\cap M=\emptyset$. This is because the sets $M_1,N_1,M_2,N_2,M_3,N_3,\dots$ are pairwise disjoint, and $M$ is defined as the union of the sets $M_1,M_2,M_3,\dots$. – bof Oct 30 '19 at 23:06
  • I must be confused then! When you say the sets are pairwise disjoint, do you mean $M_i \cap M_j = \emptyset$ when $i \neq j$? Or just that $M_i \cap N_i = \emptyset$? I don't know in either case how to guarantee that $M_i \cap N_j = \emptyset$ for all $i,j$, since $N_k \subset (I_k \setminus M_k) \iff N_k \subset \bigcap_\mathbb{N} I_k \setminus M_k$ just by normal DeMorgan's law stuff. – tyo Nov 03 '19 at 22:42
  • @tyo When I say that the sets $M_1,N_1,M_2,N_2,\dots$ are pairwise disjoint I mean that no two of them have a nonempty intersection. To put it more verbosely, I mean that $M_i\cap M_j=\emptyset$ for $I\ne j$, and $N_i\cap N_j=\emptyset$ for $i\ne j$, and $M_i\cap N_j=\emptyset$ for all $i$ and $j$. – bof Nov 04 '19 at 03:25
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    @tyo I construct $M_1\subset I_1$, then $N_1\subset I_1\setminus M_1$, then $M_2\subset I_2\setminus(M_1\cup N_1)$, then $N_2\subset I_2\setminus(M_1\cup N_1\cup M_2)$, and so on. Note that, since $M_1\cup N_1\cup M_2$ is nowhere dense, $I_2\setminus(M_1\cup N_1\cup M_2)$ contains an interval, call it $J_2$, So $N_2$ can be constructed as a subset of $J_2$. I don't know what your problem is. – bof Nov 04 '19 at 03:34
  • sorry, I wrote that wrong above -- I meant to say $N_k \subset (I_k \setminus M) \iff N_k \subset \bigcap_j I_k \setminus M_j$. But by your construction, that's just $\bigcap_{j > k} I_k \setminus M_j$, which absolutely includes $N_k$. My bad.

    My problem is that I didn't realize you were excluding $M_1 \cup N_1$ from $I_2$ before constructing $M_2$. I had tried a construction similar to yours, omitted this step, and ended up with something too dense to work -- I was having trouble seeing what we had done differently. Thank you for the step-by-step.

     – tyo Nov 05 '19 at 00:19
  • How do you ensure that $m(I \cap M) > 0$ for every interval $I$? And it seems also possible for $m(I \cap N_k) = 0$, which implies that $m(I \setminus N_k) = |I|$. – shark May 05 '23 at 01:10
  • @KKslider Like I said, $I$ contains a rational interval $I_k$, and $I\cap M\supseteq I_k\cap M_k=M_k$, so $m(I\cap M)\ge m(M_k)\gt0$. Likewise $m(I\cap N_k)\ge m(I_k\cap N_k)=m(N_k)\gt0$. – bof May 05 '23 at 07:06
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    @bof See it now, by far the most elegant construction I've seen. – shark May 05 '23 at 22:43