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contradiction to the condition

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Let (X, E, u) be a measure space with u a non-negative
measure. Suppose that
1. f : X -> R is measurable
2. f (x) >= 0 a.e. with respect to u.
3. integral (over X) f du = 0
Prove that f (x) = 0 a.e. with respect to u.

note: E = Capital Sigma
u = lowercase mu

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Solution Summary

A contradiction to the condition is embedded.

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Proof:
is a measurable space with a non-negative measure.
First, a.e., with respect to . It means that except a ...

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