# Lebesgue measurability proof - check my proof?

1. Dec 1, 2011

### jinsing

1. The problem statement, all variables and given/known data

If f is bounded on the measurable set S, and the measure of S is finite, and P,Q are partitions of S, then L(f,P) \leq U(f,Q)

2. Relevant equations

Lebesgue measurability/integrability, refinements

3. The attempt at a solution

Not sure if this is totally right, but:

Assume the hypotheses. Let P = {E1, E2,...,En} and Q={F1, F2,...,Fm}. Let R = {A_ji}, where $$E_j \cap F_i = \bigcup_{i=0}^{k_j} A_{ji}.$$ Then R is a refinement of P \cap Q. Since S has finite measure and is measurable on a bounded function f, then it follows that

$$m \mu(S) \leq L(f,P\cap Q) \leq L(f,R) \leq U(f,R) \leq U(f, P \cap Q) \leq M \mu(S).$$

But if R is a refinement of P \cap Q, then clearly R is a refinement of P and R is a refinment of Q. Then it follows that L(f,P) \leq U(f,Q).

You might be able to tell, but I was sort of grasping for straws towards the end. Any useful hints?

Thanks!

2. Dec 1, 2011

### jinsing

Okay, so I've looked at this a bit more, and I have a more specific question.

I've figured out P \cap Q isn't necessarily a refinement of S. I think the basic idea of my proof is correct, but how would you go about defining a refinement so that for every A in R, there is an E in P and a F in Q such that A is a subset of E \cap F?

3. Dec 1, 2011

### jgens

How about this: R is a refinement of P and R is a refinement of Q. This means L(f,P) < L(f,R) < U(f,R) < U(f,Q). All that there is to the argument then is checking that L(f,P) < L(f,R) holds and that U(f,R) < U(f,Q) holds, and you can do this by creating your refinement one step at a time.

Am I missing something here?