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Let $\{q_k\}$ be an enumeration of the rational numbers in $[0,1]$, and, for $0 < \epsilon<\frac{1}{2}$, let $$A_\epsilon=[0,1]\bigcap\bigcup_{j=1}^\infty \left(q_j-\frac{\epsilon}{2^j}, q_j+\frac{\epsilon}{2^j}\right).$$ Show that $\partial A_\epsilon =[0,1]\setminus A_\epsilon$.

It looks to me that $A_\epsilon=[0,1]$. If we take the infinite union, $\bigcup_{j=1}^\infty \left(q_j-\frac{\epsilon}{2^j}, q_j+\frac{\epsilon}{2^j}\right)$, will we not get $[0,1]$?

How can I proceed?

user5826
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1 Answers1

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Let $x \notin A_\varepsilon$ and $r > 0$. Clearly $x \in B(x,r)$. On the other hand, we have that $$B(x, r)\cap \mathbb{Q} = \Big([0,1] \cap \langle x - r, x + r\rangle\Big) \cap \mathbb{Q} \ne \emptyset $$

because $\mathbb{Q}$ is dense in $\mathbb{R}$. In particular $B(x, r) \cap A_\varepsilon \ne \emptyset$ since $\mathbb{Q} \cap [0,1] \subseteq A_\varepsilon$.

Therefore $x \in \partial A_\varepsilon$ since an arbitrary ball around $x$ intersects both $A_\varepsilon$ and $A_\varepsilon^c$.

Now take $x \in A_\varepsilon$. Therefore, there exists $j \in \mathbb{N}$ such that $x \in \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle \cap [0,1]$ which is an open set in $[0,1]$ so there exists $r > 0$ such that $B(x,r) \subseteq \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle \cap [0,1]$. Therefore $B(x,r)$ does not intersect $A_\varepsilon^c$ so $x \notin \partial A_\varepsilon$.

We conclude $\partial A_\varepsilon = A_\varepsilon^c = [0,1] \setminus A_\varepsilon$.

Indeed $A_\varepsilon \ne [0,1]$ for $\varepsilon < \frac12$ because if $\lambda$ denotes the Lebesgue measure, we have: $$\lambda\left(A_\varepsilon\right) \le \lambda\left(\bigcup_{j\in\mathbb{N}} \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle\right) \le \sum_{j=1}^\infty \lambda\left(\left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle\right) = \sum_{j=1}^\infty \frac{\varepsilon}{2^{j-1}} = 2\varepsilon < 1$$

and $\lambda([0,1]) = 1$.

mechanodroid
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  • How is $\mathbb{Q} \cap [0,1] \subseteq A_\varepsilon$? You are saying all the rationals between $0$ and $1$ is a subset of $A_\epsilon$. If that were the case then $\mathbb{Q} \cap [0,1] = A_\varepsilon$ by how $A_\epsilon$ is defined. – user5826 Mar 20 '18 at 16:23
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    @AlJebr Yes, if $q_i \in \mathbb{Q} \cap [0,1]$ then $$q_i \in \left\langle q_i - \frac{\varepsilon}{2^i}, q_i+ \frac{\varepsilon}{2^i}\right\rangle \cap [0,1] \subseteq [0,1] \cap \bigcup_{j\in\mathbb{N}} \left\langle q_j - \frac{\varepsilon}{2^j}, q_j + \frac{\varepsilon}{2^j}\right\rangle = A_\varepsilon$$ – mechanodroid Mar 20 '18 at 16:24
  • $A_\epsilon$ contains irrationals, so how can $A_\epsilon =\mathbb Q\cap [0,1]$? – user5826 Mar 20 '18 at 16:36
  • @AlJebr They are not equal, we only have $\mathbb{Q} \cap [0,1] \subseteq A_{\varepsilon}$. Why do you think that it must imply that they are equal? $A_\varepsilon$ is a union of balls around all rationals in $[0,1]$. – mechanodroid Mar 20 '18 at 16:37
  • You have showed that the Lebesgue measure is $<1$, but how can I show that $A_\epsilon$ is not Riemann measurable? (I don't know Lebesgue theory yet) – user5826 Mar 20 '18 at 17:05
  • @AlJebr You mean that $\chi_{A_\varepsilon}$ is not Riemann integrable? Recall that a function is Riemann integrable if and only if it is continuous almost everywhere. That translates to characteristic functions of sets in the manner that $\chi_S$ is Riemann integrable if and only if the boundary $\partial S$ has measure zero. We have calculated the boundary of $A_\varepsilon$: $$\lambda(\partial A_\varepsilon) = \lambda([0,1] \setminus A_\varepsilon) = \lambda([0,1]) - \lambda(A_\varepsilon) = 1 - 2\varepsilon > 0$$ Therefore $A_\varepsilon$ is not Riemann measurable. – mechanodroid Mar 20 '18 at 17:09
  • Cool. But what is $\lambda$? – user5826 Mar 20 '18 at 17:12
  • I see you have used some Lebesgue theory. Would it be possible to see it without Lebesgue theory? – user5826 Mar 20 '18 at 17:13
  • @AlJebr $\lambda$ is the Lebesgue measure. I don't know, see perhaps here. – mechanodroid Mar 20 '18 at 17:22