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This is Exercise $50$ from Tao's blog note: https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/#comments

$(1)$. Give an example of a compact set ${K \subset {\mathbb R}}$ of positive measure such that ${m(K \cap I) < |I|}$ for every interval ${I}$ of positive length. (Hint: first construct an open dense subset of ${[0,1]}$ of measure strictly less than ${1}$.)

$(2)$. Give an example of a measurable set ${E \subset {\mathbb R}}$ such that ${0 < m(E \cap I) < |I|}$ for every interval ${I}$ of positive length. (Hint: first work in a bounded interval, such as ${(-1,2)}$. The complement of the set ${K}$ in the first example is the union of at most countably many open intervals. Now fill in these open intervals and iterate.)

For part $(1)$, consider ${\mathbb Q} \cap (0,1) = \{r_1, r_2, \dots\}$. Let $0 < \varepsilon < 1$ be small enough that $\forall r_n \in {\mathbb Q} \cap (0,1)$, $s_n = (r_n - \frac{\varepsilon}{4^n}, r_n + \frac{\varepsilon}{4^n}) \subset (0,1)$. Let $D = \bigcup_n^\infty s_n$, then $D$ is an open dense subset of ${[0,1]}$ of measure strictly less than ${1}$. Now set $K = [0,1] \setminus D$. Then $K$ is compact, $m(K) > 0$, and contains no interval.

For part $(2)$, it seems that one can reduce to the case where $I$ is bounded, since any interval $I$ contains a bounded subinterval $I'$. Following the hint, any bounded interval like $(-1, 2)$ is contained in $\mathbb{R} \setminus K$, which is open and thus can be expressed as a countable union of disjoint open intervals.

From here, I didn't quite get what is suggested by "fill in these open intervals and iterate". Any further hint on the correct direction to proceed will be appreciated.

shark
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  • See Construction of a Borel set with positive but not full measure in each interval, especially (because of what your question specifically asks about) Nick Strehlke's answer there. – Dave L. Renfro Apr 27 '23 at 08:29
  • I am aware of that answer. Yet the hint in the second part of the question presented here contains a specific component (fill in the open intervals and iterate) that is not fully used in that answer, which exploited the denseness of $\mathbb{Q}$ in $\mathbb{R}$. – shark Apr 28 '23 at 15:33
  • @爸爸到底: Once you have constructed a set $A$ with the desired property in say the interval $[0,1]$, then $\bigcup_{n\in\mathbb{Z}} A+n$ will have the desired property on $\mathbb{R}$. – Mittens Apr 28 '23 at 19:10
  • @爸爸到底: In any open interval, say $I=(a, b)$, consider $I\setminus\mathbb{Q}$. As $m(I) = m(I\setminus \mathbb{Q})=\sup{m(K): K\subset I\setminus \mathbb{Q},, K\quad\text{compact}}$, there exists a compact set $K$ contained in $I$ whose measure is as close to $b-a$ as needed. As $K$ does not contain any rationals, $K$ is nowhere dense. Starting with $(0,1)$, you can then apply this to each of the connected components of $(0,1)\setminus K$, which happen to be open and there are a countable number go them. – Mittens Apr 28 '23 at 20:14
  • @OliverDíaz: In part $(1)$ we constructed a compact set that contains no interval, so any interval in $\mathbb{R}$ will be in $\mathbb{R} \setminus K$. What I think the hint in part $(2)$ is aiming at is to express $\mathbb{R} \setminus K$ as a countable union of disjoint open intervals $ \bigcup_{n=1}^\infty I_n$. Then as you mentioned, if one can find $A_n \subset I_n$ with the desired property in $I_n$, $A = \bigcup_{n=1}^\infty A_n$ will have the desired property in $\mathbb{R}$. Is the hint "fill in these open intervals and iterate" suggesting something like this? – shark Apr 28 '23 at 20:48
  • @OliverDíaz: Focusing on a bounded open interval $I = (a, b)$, we want to construct $A$ such that $A$ has positive but not full measure in any subinterval $I'$ of $I$. By inner regularity we get a compact $K \subset I \setminus \mathbb{Q}$ approximating $I$ from inside, what is the next move to look at? – shark Apr 28 '23 at 21:18
  • @KKslider: Start with finding $K_1\subset(0,1)$ with measure $>1/2$, then for each open component of the complement find nowhere dense compacts with measure $>1/2\times$ length of the component. Take the union $E_2$ and then define $K_2=K_1=K_1\cup E_2$. and continue now for $(0,1)\setminus K_2$ to build $K_3$ and so on. The construction is like the one presented here, but with a few adjustments. – Mittens Apr 28 '23 at 21:24
  • @OliverDíaz: So the desired set $A$ will be $(0,1) \setminus \bigcup_{n=1}^\infty K_n$ by this construction? – shark Apr 29 '23 at 00:39
  • Let us continue this discussion in chat. – shark Apr 29 '23 at 01:08
  • @OliverDíaz: For a follow up question: How do we show that $E = \bigcup_n E_n$ constructed in this way is such that $m(E \cap I') < |I'|$ for every subinterval $I'$? (I see it that $0 < m(E \cap I')$) – shark May 02 '23 at 07:38

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