Outer Measure is monotonic

[mat]

Assume A and B are subsets of [0, 1] and that A is a subset of B. Prove m*(A)<=m*(B).

Here is my proof, but I am not sure if the italicized statement is true:

Since A is a subset of B, Let B =(B\A) U A.

Then m*(B) = m*(B\A) + m*(A) and since m*(B\A)>= 0 (by definition)

we get that m*(B)>=m*(A).

Thanks for any insight!

 

[daon2]

It is true. Can you prove it?

Try a more general statement: If A,B are disjoint and measurable, then m*(A u B)=m*(A)+m*(B).

 

[mat]

okay, so it seems that I have been getting my "m*" and "mu*" confused. So mu* of a bounded set A is the outer (Lesbegue) measure defined as glb{m(G): G is open and bounded and A is contained in G}.
I have to show that mu*(A)<=mu*(B) (with the same assumptions on A and B as my first post).

So, the statement that you asked me to prove is not necessarily true for mu*.

Please help

 

[daon2]

okay, so it seems that I have been getting my "m*" and "mu*" confused. So mu* of a bounded set A is the outer (Lesbegue) measure defined as glb{m(G): G is open and bounded and A is contained in G}.
I have to show that mu*(A)<=mu*(B) (with the same assumptions on A and B as my first post).

So, the statement that you asked me to prove is not necessarily true for mu*.

Please help

It will all depend on your definitions. For instance, Royden defines measurable the following way: A set E is measurable if and only if m*(A n E)+m*(A n E') = m*(A) for any set A. If you can use this and know A,B are measurable with A a subset of B, m*B = m*(B n A) + m*(B n A') = m*(A)+m*(B\A).

So if you know A and B are measurable, that's all you need. Otherwise, a contradiction would work, try to formalize the following: If m*A > m*B, I can draw a region around B whose volume (length) is less then than the greatest lower bound of all possible regions one can draw around A. Now think: contradiction, right?