# Homework Help: Measure theory question

1. May 7, 2010

### 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.

2. May 8, 2010

### Staff: Mentor

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

3. May 8, 2010

### 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.

4. May 8, 2010

### 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.

5. May 8, 2010

### AxiomOfChoice

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).

Last edited: May 8, 2010
6. May 8, 2010

### 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 $E_\alpha$?

7. May 8, 2010

### 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.

8. May 8, 2010

### AxiomOfChoice

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

9. May 8, 2010

### AxiomOfChoice

(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?

Last edited: May 8, 2010
10. May 8, 2010

### 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 $S$?

11. May 8, 2010

### jbunniii

P.S. I love Belgian ales.

12. May 8, 2010

### AxiomOfChoice

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

$$mE \leq \sum_{k = 1}^\infty mN_k \leq 3,$$

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

$$\sum_{k=1}^\infty mE \leq 3.$$

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

Last edited: May 8, 2010
13. May 8, 2010

### AxiomOfChoice

La Chouffe, maybe? Chimay?

14. May 8, 2010

### jbunniii

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.

15. May 8, 2010

### jbunniii

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 $m(E) > 0$. (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 $x - y \in \mathbb{Q}$.

Last edited: May 8, 2010
16. May 8, 2010

### AxiomOfChoice

jbunniii,

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

$$mE \leq \sum_{k=1}^\infty mE.$$

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!

17. May 8, 2010

### 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.

18. May 8, 2010