Real analysis exam problem

  • #1
36
0
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.
 

Answers and Replies

  • #2
299
20
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.
 
  • #3
36
0
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).
 
  • #4
94
10
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.
 
  • #5
36
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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?
 
  • #6
94
10
No, for example the set ##\{x \in \mathbb{R}|x^2<0\}## is the empty set.
 
  • #7
36
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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?
 
  • #8
94
10
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.
 
  • #9
36
0
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.
 
Last edited:
  • #10
94
10
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. :)
 
  • #11
mathwonk
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In my opinion darthmatter is correct. the condition essentially says that if an element of A exists then 1/4n^2 < 1/n^3 for all n. Since this is false as soon as n >4, there are no elements of A.

By the way I hate smart alecky problems like this. This does not test any knowledge whatsoever of real analysis except the simplest limit facts and the meaning of logical implications. It is just confusing. One should always ask oneself when writing a test, what am i testing? does passing this test mean someone knows what i want them to about the subject, or am i just trying to trick people?
 
  • #12
94
10
Yeah I also don't think it is a good problem then. But it sounds like that kind of problem rarely occurs in the exam anyway.
 

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