Let (X, M, u) be a measure space. The definition of the Lebesgue integral over X for a function defined on X and with values in [0, oo] is given in terms of simple functions, that is,
Int f du = supremum {Int phi du | phi is a measurable simple function satisfying 0 <= phi <= f}. Some authors require phi to take on only finite values, but others allow phi to take on oo. It's not hard to show that in both cases the value of Int f du is the same. But if we allow simple functions to take on oo, then the standard proof of the Monotone Convergence Theorem, the one due to H. Levy, is not valid, Because, for c in (0, 1) and phi a measurable simple function such that 0 <= phi <= f, we define X_n = {x in X | f_n(x) >= c phi(x)}, n=1,2 ,3 ...f_n being a monotone increasing sequence of measurable functions on X. If we require phi to have only finite values, then X = Union X_n. But if we allow phi to take on oo, then this is not true. That's the reason for my question.
2007-07-19
09:50:36
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2 answers
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asked by
Steiner
7
in
Mathematics