function relations almost everywhere hold in the integral

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Theorem

If $f, g \in L^{+} (E)$ and $f \leq g$ almost everywhere on $E$ then

\int_{E} f \leq \int_{E} g

Proof

Let $F = {x \in E : f (x) \leq g (x)}$ .

This is measurable since $g - f$ is measurable (sums of measurable functions are measurable)
And $m (F^{c}) = 0$ . Then

\begin{aligned} \int_{E} f & = \int_{F} f + {\int_{F^{c}} f}^{0} \\ = \int_{F} f \\ \leq \int_{F} g + \int_{F^{c}} g \\ = \int_{E} g \end{aligned}

\begin{matrix} ◼ \end{matrix}

File	Last Modified
Functional Analysis Lecture 11	2025-07-15
Monotone Convergence Theorem	2025-07-14
sets of measure zero do not affect the integral	2025-07-14

function relations almost everywhere hold in the integral

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