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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.

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    Some thoughts on (i): an arbitrary union of open sets is open, a countable union of bounded sets is not necessarily bounded (in this case, it is). – 2017-01-30

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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 $1/2$?

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    Thank you for your answer. For (i) Would the bound be [-1,1]? (ii) the closure is [0,1] right? Is that answer too short? Sorry, I'm still a little confused. Is there anyway you could provide step-by-step answers for me to follow and examine? Especially for (iii) and (iv) Thank you again! – 2017-01-30
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    @smallscot For (i) notice that everything in $U$ is within $1/8$ of one of the $x_k$, so a bound is $[-1/8,9/8]$. For (ii), yes, the closure is $[0,1]$, so monotonicity of the closure gives you the desired result. For (iii) you pretty much just translate what I said: given a finite union of closed intervals contained in $U$, it is contained in a finite number of the intervals in the union (by compactness), so its measure is less than the sum of the lengths of that finite set of intervals. But that sum, call it $\sum_{i \in A} m(I_i)$, must be less than $\sum_{i=1}^\infty m(I_i)=1/2$. – 2017-01-30
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    @smallscot For (iv) you really should try to spend some time thinking, because that's where the point of this exercise lies. (i)-(iii) are just lemmas to get you to (iv). – 2017-01-30