Characteristic Function Measurable iff Set Measurable

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Theorem

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

Let $E \subseteq X$.


Then the following are equivalent:

$(1): \quad E \in \Sigma$; i.e., $E$ is a $\Sigma$-measurable set
$(2): \quad \chi_E: X \to \left\{{0, 1}\right\}$, the characteristic function of $E$, is $\Sigma$-measurable


Proof

$(1)$ implies $(2)$

Assume that $E \in \Sigma$.


It is clear that $x \notin \left\{{0, 1}\right\}$ implies $\chi_E^{-1} \left({x}\right) = \varnothing$.

Hence Preimage of Union under Mapping and Characteristic Function Determined by 1-Fiber yield, for any $\alpha \in \R$:

$\left\{{x \in X: \chi_E \left({x}\right) \ge \alpha}\right\} = \begin{cases}\varnothing & \text{if $1 < \alpha$}\\ E & \text{if $0 < \alpha \le 1$}\\ X & \text{if $\alpha \le 0$}\end{cases}$

Since $\Sigma$ is a $\sigma$-algebra, $X \in \Sigma$.

By assumption, also $E \in \Sigma$.

Lastly, Sigma-Algebra Contains Empty Set ensures $\varnothing \in \Sigma$.


This establishes $(4)$ of Characterization of Measurable Functions.

Hence $\chi_E$ is $\Sigma$-measurable.

$\Box$


$(2)$ implies $(1)$

Assume that $\chi_E$ is $\Sigma$-measurable.


Since for all $x \in X$, it holds that:

$\chi_E \left({x}\right) > 0 \iff x \in E$

it follows that:

$E = \chi^{-1} \left({\left({0 \,.\,.\, +\infty}\right)}\right)$

and thus $E \in \Sigma$ as $\chi_E$ is $\Sigma$-measurable.

$\blacksquare$


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