integral is 0 if and only if the function is 0 almost everywhere

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Proof

(⟸)

If $f = 0$ almost everywhere

0 \leq \int_{E} f \leq \int_{E} 0 = 0

(⟹)

Let

F_{n} = {x \in E : f (x) > \frac{1}{n}}, F = {x \in E : f (x) > 0}

Then $F = ⋃_{n} F_{n}$ and $F_{1} \subset F_{2} \subset \dots$ and for all $n$

0 \leq \frac{1}{n} m (F_{n}) = \int_{F_{n}} \frac{1}{n} \leq \int_{F_{n}} f \leq \int_{E} f = 0

Then $\frac{1}{n} m (F_{n}) = 0 ⟹ m (F_{n}) = 0$ for all $n$ . But then by continuity of measure, we have

m (F) = m (⋃_{n = 1}^{\infty} F_{n}) = lim_{n \to \infty} m (F_{n}) = 0

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Functional Analysis Lecture 11	2025-07-15

integral is 0 if and only if the function is 0 almost everywhere

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