# Homework Help: Measurable Sets/ Proofs: Apostol

1. May 19, 2006

Prove that each of the following sets is measurable, and has zero area: (a) a set consisting of a single point (b) a set consisting of a finite number of points in a plane (c) the union of a finite collection of line segments in a plane

(a) To prove that a set is measurable you have to say: Let Q be a region that can be enclosed between two step regions S and T. S is a subset of Q which is a subset of T. If there is only one number c that satisfies $$a(S) \leq c \leq a(T)$$ then Q is measurable, and a(Q) = c. So a set cant be a subset of a point, so S = Q. But Q can be a subset of T. But T has to equal Q, so c = 0 and Q is measurable.

Would you do this for parts b and c?

Last edited: May 19, 2006
2. May 19, 2006

### HallsofIvy

But you haven't shown that a set consisting of a single point is a measurable set so that you don't know that reasoning applies! I recommend you look at the definition of "measurable set" in the plane and show that that applies to a singleton set. After you have show that that is a measurable set, then your reasoning shows that the measure is 0. For (b) do you have a theorem about the measure of the union of two (or more) measurable sets? For (c), once again, go back to the definition of "measurable set" and apply it to a single line in the plane.

3. May 19, 2006

(a)All it says in the book is that a measurable set is a set in a plane in which an area can be assigned. A point is an set in a plane to which an area can be assigned ($$a(S) \geq 0$$.

(b) A finite number of points is a set in a plane to which an area can be applied. So it is measurable. $$a(S U T) = a(S) + a(T) - a(S & T) = 0 + 0 - 0$$

(c) a line segment is a set in a plane to which an area can be assigned. Thus it is a measurable set, with a(S) = 0. Thus $$a(S U T) = a(S) + a(T) - a(S & T) = 0 +0 - 0$$

Is this correct?

4. May 20, 2006

1 a. A measurable set is a set in the plane to which an area can be assigned. For each set S in M, a(S) . So a point is a set S in the plane to which an area can be assigned. Thus, $$a(S) \geq 0$$ . But how do we know that a(S) = 0?

b. A set consisting of a finite number of points in a plane is a set S in the plane to which an area can be assigned. Using the additive property, a(S U T ) = a(S) + a(T) – a(S & T) = 0 + 0 – 0.
c. A line segment is a set S in the plane to which an area can be assigned. So $$a(S) \geq 0$$ . But how do we know that the area is 0? So a(S U T ) = a(S) + a(T) – a(S & T) = 0 + 0 – 0.

Is this correct?

5. May 20, 2006

### matt grime

how do you know this can be done?

It is a bit more complicated than you are giving credit here. In particular you are vastly simplifying the real definition of a measurable set and confusing it with the explanation of what it means.

First you need to take R, then define a set of measurable sets, B with measure u:R-->R, where B and u satisfy certain axioms.

Presumably you are letting (B,u) be the Borel structure. Ie B is the sigma algebra generated by sets [a,b) and u assigns the measure b-a to [a,b).

assuming that you can prove a one point set is indeed an element of B then it must have measure smaller than e for any e>0, ie measure zero.

6. May 20, 2006

matt_grime, this is from an introductory calculus book (Apostol). It neverl explained anything about Borel structures or sigma algebra. The directions say to use the axioms of area to prove the statements.

Thanks

7. May 20, 2006

### matt grime

Well you'd better write out the axioms of area because if you're going to label your thread as a measure theory question then I'm going to assume you are using measure theory. Not having a copy of Apostol to hand, nor having any interest in getting hold of a copy, I think you might want to help out those who are trying to help you.

8. May 20, 2006

### HallsofIvy

Okay, so this is "measure" in the Riemann sense rather than the Lebesque sense. If a measurable set is a set to which an "area" can be assigned, consistent with the axioms of area, then you will need to show that assigning area 0 to any singleton set is consistent with those axioms.