Integral of Characteristic Function

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


Corollary

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

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


Proof

Let $a_1 = 1$ and $E_1 = E$.

As in the definition of standard representation, denote $a_0 = 0$ and $E_0 = X \setminus E_1$.


Then for $x \in X$, we have:

$\chi_E \left({x}\right) = 0 \cdot \chi_{E_0} \left({x}\right) + 1 \cdot \chi_{E_1} \left({x}\right)$

since $E_1 = E$.

Hence $\chi_E = a_0 \chi_{E_0} + a_1 \chi_{E_1}$ is a standard representation for $\chi_E$.


Thus, by definition of $\mu$-integral:

$\displaystyle \int \chi_E \, \mathrm d \mu = a_0 \mu \left({E_0}\right) + a_1 \mu \left({E_1}\right) = \mu \left({E}\right)$

as desired.

$\blacksquare$


Sources