sets of measure zero do not affect the integral

[[concept]]

Topics

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Theorem

If $\{f_n{\}}_n$ is a sequence in $L^{+}(E)$ such that $f_1(x)\le f_2(x)\le \dots$ for almost all $x\in E$ and

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 $F=\{x\in E|(\ast )\text{ holds}\}$ Then $m(E\setminus F)=0$ by assumption. Thus $f-{\chi }_{F}f=0$ a.e. and $f_n-{\chi }_F{f}_n=0$ a.e. also. Then the Monotone Convergence Theorem says ${\int }_{E}f={\int }_{E}f{\chi }_F={\int }_{F}f={\lim }_{n\to \infty }{\int }_F{f}_n$ 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 $m(E\setminus F)=0$ and so any integral over the region is 0.

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

References

References

See Also

Mentions

Mentions

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