I'm looking through past papers, and I'm stuck on this particular question, there's no answers available. I have ideas but I'm not sure how to write them formally without losing marks.
$\mathbb Q \cap [0,1] $ as {$ x_1,x_2,... $}.
For each $n \in \mathbb N $ define
$I_n := (x_n - 2^{-n-2}, x_n + 2^{-n-2})$
Also define;
$U:= \bigcup\limits_{n=1}^{\infty} I_{n}$
i) Show $U$ is a bounded open set
ii) Show $ [0,1] \subseteq \overline U $
iii) Show if $I$ is a union of finitely many closed intervals contained in $U$, then $v(I)< \frac{1}{2} $
iv) Show $U$ is not measurable.
- For (i) would I use the fact every open set in $\mathbb R$ is a union of disjoint open intervals?
And that since $ x_n - 2^{-n-2}$ and $x_n + 2^{-n-2}$ are bounded so must be the union of these bounded sets?
(ii) the smallest closed set containing $U$, I can see the logic behind it but I'm unsure how to prove this formally
For (iii) would I show $I$ is compact and use that fact to get a more tractable subset of $v$ containing $I$?
For (iv) I considered using the fact the set $\mathbb Q \cap [0,1]$ is not Riemann measurable.
Apologies for the length.If anyone could supply formal answers for me to analyse or hints, I'd be very grateful.
For (i), that's really the opposite direction. Just furnish an explicit bound on all the elements to get the boundedness. Then use the known fact that any union of open sets is open to get the openness.
For (ii), notice that $\mathbb{Q} \cap [0,1] \subset U$. What's the closure of $\mathbb{Q} \cap [0,1]$?
For (iii), the point is that the infinite sum of the lengths of all the intervals is $1/2$. So if you subtract a positive number from that, you get something less than $1/2$.
For (iv), here is the idea. A set $A$ is Riemann measurable if and only if $\partial A$ has measure zero. In this case the measure of $\overline{U}$ is at least $1$ by part (ii). But by (iii), how could the measure of $U$, if it were defined, be any bigger than $1/2$?