sets of measure zero do not affect the integral

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Theorem

If ${f_{n}}_{n}$ is a sequence in $L^{+} (E)$ such that $f_{1} (x) \leq f_{2} (x) \leq \dots$ for almost all $x \in E$ and

(*) {\begin{cases} f_{1} (x) \leq f_{2} (x) \leq f_{3} (x) \dots \\ lim_{n \to \infty} f_{n} (x) = f (x) \end{cases}

Then

\int_{E} f = lim_{n \to \infty} \int_{E} f_{n}

Proof

Let

F = {x \in E | (*) holds}

Then $m (E ∖ F) = 0$ by assumption. Thus $f - χ_{F} f = 0$ a.e. and $f_{n} - χ_{F} f_{n} = 0$ a.e. also. Then the Monotone Convergence Theorem says

\int_{E} f = \int_{E} f χ_{F} = \int_{F} f = lim_{n \to \infty} \int_{F} f_{n}

Where the first equality holds since function relations almost everywhere hold in the integral and the last holds by the Monotone Convergence Theorem. This then becomes

lim_{n \to \infty} \int_{F} f_{n} = lim_{n \to \infty} \int_{E} f_{n}

Because $m (E ∖ F) = 0$ and so any integral over the region is 0.

ie, sets of measure zero do not affect the Lebesgue Integral.

File	Last Modified
Functional Analysis Lecture 11	2025-07-15

File

Last Modified

Functional Analysis Lecture 11

2025-07-15

sets of measure zero do not affect the integral

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