Problem. For each $\alpha \in \mathbb{R}$, let $S_{\alpha}=\{(x,y,z)\in \mathbb{R}^3~|~ x^2+y^2+z^2=\alpha^2\}$.
Let $E=\bigcup_{\alpha\in \mathbb{R\setminus \mathbb{Q}}}S_\alpha$. Which of the followings are true?
- The Lebesgue measure of $E$ is infinite?
- E contains an nonempty open set.
- E is path-connected .
- Every open set containing $E^c$ has infinite Lebesgue measure.
My Solution.
True. Since $E^c=\bigcup_{\alpha\in \mathbb{Q}}S_\alpha$ and $\mu(S_\alpha)=0$ So by countable additivity of $\mu$, $E^c$ has measure zero.
False. Since $E$ and $E^c$ both are Dense in $\mathbb{R^3}$.
False. Since any two sphere of irrational radius always there exists an intermediate sphere of rational radius between them.
True. I think there is only one open set containing $E^c$ namely $\mathbb{R^3}$. And $\mu(\mathbb{R^3})=\infty.$
But the answer key indicates the options: 1 is only True. Then what is wrong with my conclusion about option 4. Please let me know where I made mistake. Thank You..
As you pointed out, we know $E^c = \bigcup_{\alpha \in \mathbb{Q}} S_{\alpha}$. Let $(a_n)_n$ be an enumeration of the rationals. Then for each $n \in \mathbb{N}$, define: $$ r_n = \min \left\{ \frac{1}{2^n}, \frac{1}{{a_n}^4 2^n} \right\} $$ where $r_n = 1/2^n$ if $a_n = 0$, and $$ R_n = \{(x,y,z) \in \mathbb{R}^3 \, | \, {a_n}^2 - r_n < x^2 + y^2 + z^2 < {a_n}^2 + r_n \} $$ Remark that each $R_n$ is open, and that the size of $R_n$ is given by the difference of two spheres, i.e.: \begin{align*} \mu(R_n) &= \mu(B(0,{a_n}^2+r_n)) - \mu(B({a_n}^2-r_n)) \\ &= \frac{4}{3} \pi ({a_n}^2 + r_n)^3 - \frac{4}{3} \pi ({a_n}^2 - r_n)^3 \\ &= \frac{4}{3} \pi \left( 6 {a_n}^4 r_n + 2 {r_n}^3 \right) \\ &\leq \frac{4}{3} \pi \left( \frac{6}{2^n} + \frac{2}{2^{3n}} \right) \\ &\leq \frac{48 \pi}{3} \cdot \frac{1}{2^n} \end{align*}
Since each $R_n$ is open, their union is open. It is also clear that the union of the $R_n$ contains $E^c$, and finally we have: $$ \mu \left( \bigcup_{n \in \mathbb{N}} R_n \right) \leq \sum \limits_{n=1}^{\infty} \frac{48 \pi}{3} \cdot \frac{1}{2^n} = \frac{48 \pi}{3} $$ This gives an example of an open set that contains $E^c$ which has finite measure.