Characteristic Function Measurable iff Set Measurable

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Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $E \subseteq X$.


Then the following are equivalent:

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


Proof

$(1)$ implies $(2)$

Assume that $E \in \Sigma$.


It is clear that $x \notin \set {0, 1}$ implies $\map {\chi_E^{-1} } x = \O$.

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

$\set {x \in X: \map {\chi_E} x \ge \alpha} = \begin{cases} \O & \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 $\O \in \Sigma$.


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:

$\map {\chi_E} x > 0 \iff x \in E$

it follows that:

$E = \map {\chi^{-1} } {\openint 0 {+\infty} }$

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

$\blacksquare$


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