Lebesgue measurability proof - check my proof?

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The discussion revolves around proving that if a function f is bounded on a measurable set S with finite measure, then the lower sum L(f,P) is less than or equal to the upper sum U(f,Q) for partitions P and Q of S. The proof attempts to establish this by defining a refinement R of the intersection of P and Q, demonstrating that L(f,P) is less than L(f,R) and U(f,R) is less than U(f,Q). It acknowledges the challenge of ensuring that R is a proper refinement of both P and Q. The user seeks clarification on how to define this refinement effectively and whether their approach to proving the inequalities is sound. The discussion highlights the complexities involved in establishing the relationships between lower and upper sums through refinements.
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Homework Statement



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)

Homework Equations



Lebesgue measurability/integrability, refinements

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!
 
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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?
 
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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