Characteristic Function of Union/Variant 3

Theorem

Let $A, B \subseteq S$.

Let $\chi_{A \mathop \cup B}$ be the characteristic function of their union $A \cup B$.

Then:

$\chi_{A \mathop \cup B} = \max \set {\chi_A, \chi_B}$

where $\max$ denotes the max operation.

Proof

Suppose $\map {\chi_{A \mathop \cup B} } s = 0$.

Then $s \notin A \cup B$, so $s \notin A$ and $s \notin B$.

Hence:

$\map {\chi_A} s = \map {\chi_B} s = 0$

and by definition of max operation:

$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$

Conversely, suppose:

$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$

Then it follows that:

$\map {\chi_A} s = \map {\chi_B} s = 0$

because characteristic functions are $0$ or $1$.

Hence $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.

That is:

$\map {\chi_{A \mathop \cup B} } s = 0$

Above considerations give:

$\map {\chi_{A \mathop \cup B} } s = 0 \iff \max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$

and applying Characteristic Function Determined by 0-Fiber, the result follows.

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $5 \ \text{(v)}$