Jordan Content/Measure Problem

[asif zaidi]

Hello

We are doing Jordan content/measure and I think I understand it but I seem to be having problem the following h/w assignment.

Problem Statement:

Let I be a generalized interval and let f be the constant function 1. Find a subset D of I such that f (restricted to region D) is not integrable

Problem Solution

A function is integrable iff the boundary has Jordan content 0. I can prove that the boundary of a constant function has measure 0.

So I cannot see where this function is non-integrable.

What am I missing?

Thanks

Asif

The Attempt at a Solution

 

[HallsofIvy]

Hello

We are doing Jordan content/measure and I think I understand it but I seem to be having problem the following h/w assignment.

Problem Statement:

Let I be a generalized interval and let f be the constant function 1. Find a subset D of I such that f (restricted to region D) is not integrable

Problem Solution

A function is integrable iff the boundary has Jordan content 0. I can prove that the boundary of a constant function has measure 0.

So I cannot see where this function is non-integrable.

What am I missing?

Thanks

Asif

The Attempt at a Solution

Try some really complicated subset: like, for example, the set of all irrational numbers in that interval. What is its boundary? What is the measure of that boundary?

 

[asif zaidi]

For irrational sets: the Jordan measure is 0. What I cannot tell is the boundary.

That is my problem with the assignment also - I thought that for an integral to exist it had to be bounded. But the problem doesn't specify it. It says simply over the generalized interval.

Does this mean that since bounds are not specified then the constant function cannot be integrable even if the Jordan measure is 0.

Thanks

Asif

 

[matt grime]

The jordan measure of the irrationals inside the interval [0,1] is not 0. In fact it doesn't exist, that being what you're supposed to prove.

What is you're definition of boundary?

The closure less the interior?
THe closure intersection with the closure of the complement?
Something else?

 

[asif zaidi]

By closure I have understood to mean the boundary point where the function values cannot be made arbitrarily close. It is at these points that I looked at the constant function 1 and showed the Jordan content to be 0.

But I cannot find where the subsets are not integrable.

 

[matt grime]

I suggest you review all of the definitions of all of the terms involved; I don't understand what you've claimed is your definition of the boundary. Your definition seems to imply that the boundary of any set is a single point.

It is immediate from the definitions that the boundary of the irrationals inside the interval [0,1] is not of jordan content 0.

 

[matt grime]

Aside: there's no need to send me PMs about questions that are being dealt with in threads.

If you don't know the definition of boundary then you can't do any of these questions.

The boundary of a set S is defined to be (amongst other equivalent things), the closure of S less the interior of S.

If S is Qn[0,1], ie the rational numbers in the interval [0,1], what is the closure of S? What is the interior of S?

The aim is *not* to show that S hasn't got Jordan measure 0 (in fact, it doesn't have a Jordan measure - that is the aim), but to show that the boundary of S doesn't have Jordan measure 0.