Characteristic Function of Set Difference
Theorem
Let $A, B \subseteq S$.
Then:
- $\chi_{A \mathop \setminus B} = \chi_A - \chi_{A \cap B}$
where:
- $A \setminus B$ denotes set difference
- $\chi$ denotes characteristic function.
Proof
Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$.
Then by definition of characteristic function:
- $s \in A \setminus B$
That is, by definition of set difference, $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$.
Hence:
- $\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$
Conclude that:
- $\map {\chi_{A \mathop \setminus B} } s = 1 \implies \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
Now conversely suppose $\map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$.
Considering that $\map {\chi_{A \cap B} } s \ge 0$, it must be that $\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$.
Hence, $s \in A$, and $s \notin A \cap B$.
By definition of set intersection, this means $s \notin B$.
Hence $s \in A \setminus B$ by definition of set difference.
Thus:
- $\map {\chi_{A \mathop \setminus B} } s = 1$
Hence, it has been established that:
- $\map {\chi_{A \mathop \setminus B} } s = 1 \iff \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
which, by application of Characteristic Function Determined by 1-Fiber, gives the result.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $5 \ \text{(iii)}$