Integral of Characteristic Function/Corollary

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $E \in \Sigma$ be a measurable set, and let $\chi_E: X \to \R$ be its characteristic function.

Then:

$\displaystyle \int \chi_E \, \mathrm d\mu = \mu \left({E}\right)$

where the integral sign denotes the $\mu$-integral of $\chi_E$.


Proof

By Integral of Characteristic Function, have:

$I_\mu \left({\chi_E}\right) = \mu \left({E}\right)$

where $I_\mu \left({\chi_E}\right)$ is the $\mu$-integral of $\chi_E$.

From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, it also holds that:

$\displaystyle \int \chi_E \, \mathrm d\mu = I_\mu \left({\chi_E}\right)$


Combining these equalities gives the result.

$\blacksquare$


Sources