Proving Rational Difference Exists in Finite Measure Subset [0,1]

[AxiomOfChoice]

This is a practice final exam problem that has been giving me fits: Let E be a Lebesgue measurable subset of the interval [0,1] that has finite measure. Show that there exist two points x,y \in E such that x-y is rational.

 

[berkeman]

This is a practice final exam problem that has been giving me fits: Let E be a Lebesgue measurable subset of the interval [0,1] that has finite measure. Show that there exist two points x,y \in E such that x-y is rational.

Moved to HH. Could you please post your work so far? Thanks.

 

[AxiomOfChoice]

Well, I haven't done much, as I'm rather stumped here. My professor gave me the following hint: "countable unions." He also mentioned that examining the non-measurable set constructed with the axiom of choice might help. But this sort of has me even more baffled.

 

[jbunniii]

Why don't you start by summarizing for us the construction of the non-measurable set you're referring to? If it's the usual construction in most textbooks, it should start by defining an equivalence relation on the unit interval, such that two points x and y are equivalent if and only if x - y is rational. That seems like a promising start.

 

[AxiomOfChoice]

Why don't you start by summarizing for us the construction of the non-measurable set you're referring to? If it's the usual construction in most textbooks, it should start by defining an equivalence relation on the unit interval, such that two points x and y are equivalent if and only if x - y is rational. That seems like a promising start.

That is exactly it. We consider an equivalence relation on [0,1] defined as follows: x\sim y if x - y \in \mathbb Q. We then employ the Axiom of Choice and choose one member x_\alpha from each equivalence class E_\alpha, and define the set N = \{x_\alpha \}. The uncountability of N follows from the fact that N must satisfy 1 \leq \sum_{k=1}^\infty m(N) \leq 3 (where m denotes Lebesgue measure).

 

[jbunniii]

OK, now let's consider your set E. Suppose the assertion were not true. Then there do not exist two distinct points x,y \in E such that x - y \in \mathbb{Q}.

Now consider the implication in terms of the equivalence classes E_\alpha. What is the maximum number of elements of E that can be contained in each class E_\alpha?

 

[jbunniii]

P.S. I am assuming that in the problem statement, "finite measure" should really be "finite NON-ZERO measure." Otherwise the statement is false, as the example E = \{0, 1/\sqrt{2}\} demonstrates.

 

[AxiomOfChoice]

P.S. I am assuming that in the problem statement, "finite measure" should really be "finite NON-ZERO measure." Otherwise the statement is false, as the example E = \{0, 1/\sqrt{2}\} demonstrates.

You're right. Actually, instead of finite, I should have said "positive" (which I'm assuming means 0 < mE < \infty).

 

[AxiomOfChoice]

OK, now let's consider your set E. Suppose the assertion were not true. Then there do not exist two distinct points x,y \in E such that x - y \in \mathbb{Q}.

Now consider the implication in terms of the equivalence classes E_\alpha. What is the maximum number of elements of E that can be contained in each class E_\alpha?

(1) You're very kind to help me out with this. Thanks. I'd buy you a beer if I could. (Or a glass of wine; whichever floats your boat.)

(2) Ah, I think I see...so each equivalence class consists of a single point: Given x,y\in E, x \sim y iff x = y. I guess this means there must be uncountably many equivalence classes, then, since mE > 0 implies it's necessarily uncountable. More importantly, don't we have N = \{ x_\alpha \} = E? I guess we can derive a contradiction from this?

 

[jbunniii]

Yes, "positive" makes more sense than "finite." In fact, m(E) \leq m([0,1]) = 1 so finiteness is automatic.

OK, so far we have established that each equivalence class contains at most one point of E.

Now we can use the axiom of choice to construct a set S containing one point of E \cap E_\alpha for each equivalence class E_\alpha for which E \cap E_\alpha is non-empty. Since each E_\alpha contains at most one point of E, what can you say about S?

 

[jbunniii]

P.S. I love Belgian ales.

 

[AxiomOfChoice]

Yes, "positive" makes more sense than "finite." In fact, m(E) \leq m([0,1]) = 1 so finiteness is automatic.

OK, so far we have established that each equivalence class contains at most one point of E.

Now we can use the axiom of choice to construct a set S containing one point of E \cap E_\alpha for each equivalence class E_\alpha for which E \cap E_\alpha is non-empty. Since each E_\alpha contains at most one point of E, what can you say about S?

Ok. I think my argument above has shown that each equivalence class consists of *exactly* one point, and that if we apply the Axiom of Choice (hehe) and pluck one point out of each equivalence class, at the end of the day we get N = E. So, if we define the translates N_k = N + r_k = E + r_k for an enumeration of the rationals in [-1,1], we find that

<br /> mE \leq \sum_{k = 1}^\infty mN_k \leq 3,<br />

since it can be shown that the sets N_k are disjoint, E \subset \bigcup_{k = 1}^\infty N_k (just take r_k = 0) and \bigcup_{k = 1}^\infty N_k \subset [-1,2]. But this implies, by translation invariance of m, that

<br /> \sum_{k=1}^\infty mE \leq 3.<br />

This is absurd, since we assumed mE &gt; 0 \Rightarrow \sum_1^\infty mE = \infty. Hence E is not measurable. Have I missed anything?

 

[AxiomOfChoice]

P.S. I love Belgian ales.

La Chouffe, maybe? Chimay?

 

[jbunniii]

Replying to your updated message:

(2) Ah, I think I see...so each equivalence class consists of a single point: Given x,y\in E, x \sim y iff x = y. I guess this means there must be uncountably many equivalence classes, then, since mE &gt; 0 implies it's necessarily uncountable. More importantly, don't we have N = \{ x_\alpha \} = E? I guess we can derive a contradiction from this?

Let's be sure we agree about everything so far. Using your notation:

The equivalence classes are called E_\alpha.

Now each E_\alpha \cap E contains either zero or one element. For convenience, define

A = \{\alpha : E_\alpha \cap E \neq \emptyset\}

Using the axiom of choice, for each \alpha \in A we choose x_\alpha \in E_\alpha \cap E. We thereby construct this set:

N = \bigcup_{\alpha \in A} \{x_\alpha\}.

As you pointed out, we actually have N = E.

Now let

\{r_i\}_{i=1}^{\infty}

be an enumeration of the rationals in [0,1]

And define the following sets:

N_i = N + r_i

where the addition is performed modulo 1.

Then, exactly as in the construction of the standard non-measurable set, we have

N_i \cap N_j = \emptyset

whenever i \neq j, i.e., the sets are disjoint.

Then consider

\bigcup_{i=1}^{\infty} N_i

What can you say about the measure of this set, and why is that a contradiction?

P.S. For some reason the preview function is behaving strangely - it puts the wrong stuff in each TeX section. So I will have to fix any typos after I post the message.

 

[jbunniii]

Ok. I think my argument above has shown that each equivalence class consists of *exactly* one point, and that if we apply the Axiom of Choice (hehe) and pluck one point out of each equivalence class, at the end of the day we get N = E. So, if we define the translates N_k = N + r_k = E + r_k for an enumeration of the rationals in [-1,1], we find that

<br /> mE \leq \bigcup_{k = 1}^\infty mN_k \leq 3,<br />

I assume you mean

\sum_{k=1}^\infty

not

\bigcup_{k = 1}^\infty

Yes, now you use the fact (just as in the standard construction of the nonmeasurable set) that the N_k are pairwise disjoint, and also the translation invariance of Lebesgue measure, to obtain

\bigcup_{k=1}^\infty N_k \subset [-1,2]

and thus

\sum_{k=1}^\infty m(E) \leq 3

That can't be true if m(E) &gt; 0. (Note that it COULD be true if m(E) = 0, which is why the positive hypothesis is important.)

Thus we have a contradiction. Therefore our assumption was wrong, and E must indeed contain distinct x,y such that x - y \in \mathbb{Q}.

 

[AxiomOfChoice]

I assume you mean

\sum_{k=1}^\infty

not

\bigcup_{k = 1}^\infty

Yes, now you use the fact (just as in the standard construction of the nonmeasurable set) that the N_k are pairwise disjoint, and also the translation invariance of Lebesgue measure, to obtain

\bigcup_{k=1}^\infty N_k \subset [-1,2]

and thus

\sum_{k=1}^\infty m(E) \leq 3

That can't be true if 0 &lt; m(E) &lt; \infty. (Note that it COULD be true if m(E) = 0, which is why the positive hypothesis is important.)

Thus we have a contradiction. Therefore our assumption was wrong, and E must indeed contain distinct x,y such that x - y \in \mathbb{Q}.

jbunniii,

Yes, I meant \sum instead of \bigcup. You've also pointed out to me that there is actually no contradiction present in

<br /> mE \leq \sum_{k=1}^\infty mE.<br />

The contradiction comes from the *other* inequality, which would force mE = 0, which is impossible. I have made the appropriate corrections to my posts above.

Thanks again!

 

[jbunniii]

Looks good! By the way, notice that we didn't actually need the axiom of choice in this case to construct the set N. We could have simply said "let N = E" and shown directly that the N_k's are pairwise disjoint.

 

[jbunniii]

La Chouffe, maybe? Chimay?

I'm considering attending this event this afternoon:

http://www.artisanwinedepot.com/ProductDetails.asp?ProductCode=EVENT-CRAFTBEERBATTLE

Needless to say, I think the Belgians will thrash the Germans.