Characteristic Function of Intersection/Variant 2

Theorem

Let $A, B \subseteq S$.

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

Then:

$\chi_{A \mathop \cap B} = \min \set {\chi_A, \chi_B}$

Proof

By Characteristic Function Determined by 1-Fiber, it suffices to show that:

$\min \set {\map {\chi_A} s, \map {\chi_B} s} = 1 \iff s \in A \cap B$

By definition of characteristic function, we have:

$\min \set {\map {\chi_A} s, \map {\chi_B} s} = 1$

if and only if:

$\map {\chi_A} s = \map {\chi_B} s = 1$

because $\chi_A, \chi_B$ take values in $\set {0, 1}$.

By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.

That is:

$s \in A \cap B$

by definition of set intersection.

$\blacksquare$

Sources

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