Proving Finite Lebesgue Measure of A When B Is Contained Within

[mikesmith00]

Is there anyway to prove that if I have (m is the lebesgue measure) m(A\B) is finite, then m(A) is finite? It seems intuitive to me, but I'm having trouble coming up with rigorous mathematical reasoning for it. B is completely contained within A. If there's anything else that might clarify this, let me know, because this is bugging me. Thanks for any help that you can offer.

 

[rochfor1]

The reason you're having trouble proving it is that it's not true (at least when you consider the Lebesgue measure on the whole real line). Consider the trivial example of A=all reals and $$B = (-\infty,0)\cup(0,\infty)$$. For which m(A\B)=m({0})=0, but A has infinite measure. For a somewhat less trivial example take A=all reals and B=all irrationals. It is very important to note that although the "identity" m(A\B)=m(A)-m(B) seems plausible and intuitive, it fails in many infinite measure spaces.

 

[mikesmith00]

Thank you. I'm working on getting ready for my real analysis comps, and trying to go through all of the books that were listed on the syllabus for the exam. This was part of a question in one of the books that I'm working on (Stein and Shakarachi vol 3) and I'm trying to show that if f is integrable, then $$\sum m(E_{2^{k}}) < \infty$$ summed from $$k= -\infty$$ to $$\infty$$ with $$E_{2^{k}}=\left\{x:f(x)>2^{k}\right\}$$. My first attempt was that I was able to show that if I found disjoint sets $$F_{k} = E_{2^{k}}{\setminus}E_{2^{k+1}}$$ then $$\sum m(F_{k}) < \infty$$ summed from $$k= -\infty$$ to $$\infty$$. I was trying to use this to show that if we know that sum F is finite, then sum E is finite. Looks like I might have to start on a different approach. Thanks for your help again.

 

[mikesmith00]

I believe I have it solved. If we're taking the measure of the union, and I have two unions that are equal to each other, then the measures are equal, so then if the measure of one is finite, then the other must be finite as well. If you see any flaws in this logic, any contributions to this are more than welcome.