Information about contradiction to condition

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Reference no: EM13129235

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

Reference no: EM13129235

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