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Proof that the rationals have a measure zero

A question on a problem set that I've been looking at introduces the idea of a set having a measure zero, which is that for any ϵ>0\epsilon > 0 there is a sequence of (finite or countable) balls (Bn)(B_n) with radii rnr_n so that N⊆⋃nBnN \subseteq \bigcup_n B_n and ∑nrn<ϵ\sum_n r_n < \epsilon.

Now in one of the parts of the question was whether a dense set can have a measure of zero. Now I've heard (or read) that the rationals have a measure of zero so I tried showing that.

My proof basically involves ordering Q\mathbb{Q} such that Q={qn∣n∈N}\mathbb{Q} = \{ q_n | n \in \mathbb{N} \}, and create a subset QN={qn∣n≤N}\mathbb{Q}_N = \{ q_n | n \leq N \}, then for each qn∈QNq_n \in \mathbb{Q}_N surrounding it with a neighborhood with centre qnq_n and radius rn<ϵ/Nr_n < \epsilon/N. Now clearly the sum of radii for this subset is ∑nrn=Nrn<N(ϵ/N)=ϵ\sum_n r_n = Nr_n < N(\epsilon/N) = \epsilon. Finally, we take the limit as N→∞N \rightarrow \infty to establish that ∑Qrn<ϵ\sum_{\mathbb{Q}} r_n < \epsilon and that Q\mathbb{Q} does indeed have measure zero.

But is this proof sound? Does taking the limit to infinity cause any problems here?
Original post by 0x2a

But is this proof sound? Does taking the limit to infinity cause any problems here?


rN r_N would be a better notation, as it's clear from your working that you consider it a function of NN, rather than varying for each element of the set QN\mathbb{Q}_N.

What would rnr_n be when you consider the whole of Q\mathbb{Q}?

Spoiler



Edit: It would be easier, IMO, to consider Q as a whole to start with, and use the fact that you can order the elements of Q.
(edited 8 years ago)
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Original post by ghostwalker
rN r_N would be a better notation, as it's clear from your working that you consider it a function of NN, rather than varying for each element of the set QN\mathbb{Q}_N.

What would rnr_n be when you consider the whole of Q\mathbb{Q}?

Spoiler


Edit: It would be easier, IMO, to consider Q as a whole to start with, and use the fact that you can order the elements of Q.

I am not sure I understand the stuff in the spoilers. Are you saying that ∑nrn\sum_n r_n cannot stay constant as n→∞n \rightarrow \infty?

A thought of a method which was ordering Q\mathbb{Q} as before and sending each qn↦1/nq_n \mapsto 1/n and I already know that the set E={1/n ∣ n∈N}E = \{ 1/n \ | \ n \in \mathbb{N} \} is of measure zero. But obviously the measure of a set is not preserved under bijections (i.e R→C\mathbb{R} \rightarrow C where CC is the Cantor set). So I don't really know what to do.
Original post by 0x2a
I am not sure I understand the stuff in the spoilers. Are you saying that ∑nrn\sum_n r_n cannot stay constant as n→∞n \rightarrow \infty?


You're choosing a constant r_N for a given N. Which works when N is finite.

But this does not extend to there being a constant r_N when dealing with Q\mathbb{Q}.


A thought of a method which was ordering Q\mathbb{Q} as before and sending each qn↦1/nq_n \mapsto 1/n and I already know that the set E={1/n ∣ n∈N}E = \{ 1/n \ | \ n \in \mathbb{N} \} is of measure zero. But obviously the measure of a set is not preserved under bijections (i.e R→C\mathbb{R} \rightarrow C where CC is the Cantor set). So I don't really know what to do.


Ordering Q as before you just need to find rir_i as radii of balls BiB_i centred on qiq_i such that ∑i=1∞ri\displaystyle\sum_{i=1}^{\infty} r_i is finite. If you can do that, then you can scale those rir_i to have any sum you want, in particular ∑ri<ϵ\sum r_i < \epsilon

Can't comment on bijections as my measure theory is too rusty.
(edited 8 years ago)
It is a fact that measure is sigma-additive: that is, that the measure of a countable union of sets is the sum of the measures of the sets. In particular, the measure of a countable union of sets of measure zero has measure zero (you should prove this!). Can you find countably many sets that can easily be proven to have measure zero whose union is Q?
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Original post by ghostwalker
Ordering Q as before you just need to find rir_i as radii of balls BiB_i centred on qiq_i such that ∑i=1∞ri\displaystyle\sum_{i=1}^{\infty} r_i is finite. If you can do that, then you can scale those rir_i to have any sum you want, in particular ∑ri<ϵ\sum r_i < \epsilon

Can't comment on bijections as my measure theory is too rusty.

Ah of course. Thank you :smile: Now it's so obvious and I didn't get it :angry:

Original post by BlueSam3
It is a fact that measure is sigma-additive: that is, that the measure of a countable union of sets is the sum of the measures of the sets. In particular, the measure of a countable union of sets of measure zero has measure zero (you should prove this!). Can you find countably many sets that can easily be proven to have measure zero whose union is Q?

I haven't learnt any measure theory yet, but you get that Q\mathbb{Q} has measure zero from the union of all unique singletons of rationals I'm guessing?
Original post by 0x2a
Ah of course. Thank you :smile: Now it's so obvious and I didn't get it :angry:


I haven't learnt any measure theory yet, but you get that Q\mathbb{Q} has measure zero from the union of all unique singletons of rationals I'm guessing?


Yup (though you'd be amazed how many people overcomplicate that and take it as things like the union of the sets
Unparseable latex formula:

A_n = \left{\frac{k}{n} |k \in \mathbb{N}\right}

). When I've got time, I'll take a look at how to prove the countable unions thing from your definition. Your definition definitely implies the measure-theoretic definition of having measure zero, but it looks like it might get a touch fiddly to prove the closure under countable unions without resorting to measure theory.
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Original post by BlueSam3
Yup (though you'd be amazed how many people overcomplicate that and take it as things like the union of the sets
Unparseable latex formula:

A_n = \left{\frac{k}{n} |k \in \mathbb{N}\right}

). When I've got time, I'll take a look at how to prove the countable unions thing from your definition. Your definition definitely implies the measure-theoretic definition of having measure zero, but it looks like it might get a touch fiddly to prove the closure under countable unions without resorting to measure theory.

I think ordering all the sets you are looking at as {En ∣ n∈N}\{ E_n \ | \ n \in \mathbb{N} \}, and then letting the sum of radii of the open balls that cover each EnE_n to be ϵ/2n\epsilon/2^n and then summing the sum of the radii themselves gives you ϵ\epsilon, and then ϵ\epsilon is arbitrary so the countable union of sets of measure zero is measure zero. Does this work?

This is how I proved that Q\mathbb{Q} is of measure zero by given each singleton {qn}\{q_n\} an open ball of radius ϵ/2n\epsilon/2^n with centre qnq_n.
Original post by 0x2a
I think ordering all the sets you are looking at as {En ∣ n∈N}\{ E_n \ | \ n \in \mathbb{N} \}, and then letting the sum of radii of the open balls that cover each EnE_n to be ϵ/2n\epsilon/2^n and then summing the sum of the radii themselves gives you ϵ\epsilon, and then ϵ\epsilon is arbitrary so the countable union of sets of measure zero is measure zero. Does this work?

This is how I proved that Q\mathbb{Q} is of measure zero by given each singleton {qn}\{q_n\} an open ball of radius ϵ/2n\epsilon/2^n with centre qnq_n.


Yeah, looks like it could be made to work. Probably easier to think in terms of unions, though: If A=⋃n∈NAnA = \bigcup_{n\in\mathbb{N}} A_n, and each AnA_n has measure zero then with BN=⋃n≤NAnB_N = \bigcup_{n\leq N} A_n each BNB_N can very easily be shown to have measure zero, and the BNB_N are nested, so given any ε>0\varepsilon > 0, we can pick our balls (for each NN such that the radii sum to something less than (say) ε2\frac{\varepsilon}{2}, and in particular, we can pick our balls such that each extends the previous set (since the An=Bn∖Bn−1A_n = B_n \setminus B_{n-1} have measure zero), so the sum of the radii of all of the balls (which cover AA) is the limit of the sum of the radii of the balls covering each BnB_n, so is at most ε2\frac{\varepsilon}{2}, so is certainly less than ε\varepsilon.
This is a useful technique for lots of things in measure theory.

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