# Real analysis exam problem

Hi,

I was leafing through some old exams of our Real analysis course, and I found this puzzling problem:

"Let A⊂ℝ be Lebesgue-measurable so that for all a∈A, i = 1,2, ...

(1) m1( {x∈ℝ | a+(3/4)i-2 < x < a + i-2} ) < i-3

Claim: m1(A) = 0."

Initially I thought this may have something to do with the Lebesgue density theorem that has been used a lot during the course. However, to me it looks like condition (1) doesn't really set any boundaries to what kind of set A could be. (1) only tells us that :

(2) m1({x∈ℝ | a+(3/4)i-2 < x < a + i-2}) = lenght(a+(3/4)i-2, a + i-2) = |a + i-2 - a+(3/4)i-2| = 1/4 i-2.

Now, 1/4 i-2 < i-3 is true for any i =1,2,3. Condition (2) seems to be true for ANY set E⊂ℝ, if i = 1,2, or 3, even those for which m1(E) >0 (such as an interval). The only condition we get for the set A is that it has to be Lebesgue-measurable.

Any help would be appreciated.

Now, 1/4 i-2 < i-3 is true for any i =1,2,3.

This is incorrect. For i>4, i^(-3) < 1/4 i^(-2)

Edit: Ah, I think I see where you're confused. You seem to have missed the ellipses. The condition must hold for all natural numbers, not just for the first three.

For i>4, i^(-3) < 1/4 i^(-2)

Yeah, but what is demanded is the opposite: 1/4 i^(-2) < i^(-3), which is true for i < 4, namely 1, 2 and 3.
You are correct to say that for i >4, i^(-3) < 1/4 i^(-2), but that is not demanded.

So, to me it seems like condition (1) in my initial post only holds if i equals to 1, 2 or 3 - and for any set of the real numbers, not only sets A that have m(A) = 0, which is the claim (and my problem).

If your are sure that ##m_1## is known (and not arbitrary), maybe it is enough to prove that A is equal to the empty set.

If your are sure that ##m_1## is known (and not arbitrary), maybe it is enough to prove that A is equal to the empty set.

##m_1## i here means Lebesgue-measure in the first dimension of ℝ (ℝ1 = ℝ). ##m_n## would be Lebesgue-measure in the nth dimension of ℝ, ℝn.

It assumed that all elements of A have a certain property. In the light of this premise, would it not be contradictory to say that A has no elemets at all?

No, for example the set ##\{x \in \mathbb{R}|x^2<0\}## is the empty set.

This is nitpicking, but as I understand it, " for all a∈A" means "for all elements a that belong to A", which already states that A has elements. To leave an opening for the possibility of an empty set, it should say: "if a∈A (, condition (1) holds)". To say that all a∈∅ have a certain (albeit perhaps impossible) property, would not be correct since you are stating that there are elements in ∅ and attaching properties to them.

But if I'm wrong (or the author of the problem forgot to leave the opening for an empty set) how would I proceed in showing that A is empty?

As I see it, noone says there are elements in A. You just say if a is an element ##a \in A##, the property ... must hold. If no ##a \in \mathbb{R}## fulfills that property, A must be empty.

Hi,

I was leafing through some old exams of our Real analysis course, and I found this puzzling problem:

"Let A⊂ℝ be Lebesgue-measurable so that for all a∈A, i = 1,2, ...

(1) m1( {x∈ℝ | a+(3/4)i-2 < x < a + i-2} ) < i-3

Claim: m1(A) = 0."

Initially I thought this may have something to do with the Lebesgue density theorem that has been used a lot during the course. However, to me it looks like condition (1) doesn't really set any boundaries to what kind of set A could be. (1) only tells us that :

(2) m1({x∈ℝ | a+(3/4)i-2 < x < a + i-2}) = lenght(a+(3/4)i-2, a + i-2) = |a + i-2 - a+(3/4)i-2| = 1/4 i-2.

Now, 1/4 i-2 < i-3 is true for any i =1,2,3. Condition (2) seems to be true for ANY set E⊂ℝ, if i = 1,2, or 3, even those for which m1(E) >0 (such as an interval). The only condition we get for the set A is that it has to be Lebesgue-measurable.

Any help would be appreciated.

Assume ##b\in A, b \in \mathbb{R}##. Then, following from your calculation ##\frac{1}{4}i^{-2}<i^{-3}## for all ##i \in \mathbb{N}##. But this is wrong, since for ##i=4## insertion yield ##\frac{1}{64}=\frac{1}{64}##. Contradiction. So there is no ##b \in A##, and A is empty.

Then, following from your calculation ##\frac{1}{4}i^{-2}<i^{-3}## for all ##i \in \mathbb{N}##. But this is wrong, since for ##i=4## insertion yield ##\frac{1}{64}=\frac{1}{64}##. Contradiction. So there is no ##b \in A##, and A is empty.

This solution would indeed make sense if the condition has to hold for all i∈ℕ. Actually, how I initially read the problem was "for all a∈A and i must be an integer" but now I realize that it probably was meant "for all a∈A and for all i = 1,2,...".

If the solution really is that A is empty, it's surprisingly simple - this problem was the last one on the exam, which is usually the most difficult one. But then again, if condition (1) is demanded for all i∈ℕ, it would seem that there are no real numbers a∈A that would satisfy it.

EDIT: "difficult" is a wrong word. I guess the problem was difficult as I didn't come up with the empty set explanation in the first place, lol. But I'd expect a solution that requires more advanced methods, such as the Lebesgue density theorem, which was part of the more "advaced stuff" that was studied during the course.

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Hi,

sometimes such problems are meant to be relatively easy, but sometimes they just are a victim of ambigious notation. Maybe you can ask somebody who is closer involved to find out what the expected solution was. :)

mathwonk