Additive Function is Strongly Additive

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Theorem

Let $\mathcal S$ be an algebra of sets.

Let $f: \mathcal S \to \overline {\R}$ be an additive function on $\mathcal S$.


Then $f$ is also strongly additive.

That is:

$\forall A, B \in \mathcal S: \map f {A \cup B} + \map f {A \cap B} = \map f A + \map f B$


Proof

From Set Difference and Intersection form Partition, we have that:

$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
$B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.


So, by the definition of additive function:

$\map f A = \map f {A \setminus B} + \map f {A \cap B}$
$\map f B = \map f {B \setminus A} + \map f {A \cap B}$


We also have from Set Difference Disjoint with Reverse that:

$\paren {A \setminus B} \cap \paren {B \setminus A} = \O$


Hence:

\(\displaystyle \map f A + \map f B\) \(=\) \(\displaystyle \map f {A \setminus B} + 2 \map f {A \cap B} + \map f {B \setminus A}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\paren {A \setminus B} \cup \paren {A \cap B} \cup \paren {B \setminus A} } + \map f {A \cap B}\) Set Difference and Intersection form Partition: Corollary 1
\(\displaystyle \) \(=\) \(\displaystyle \map f {A \cup B} + \map f {A \cap B}\)

Hence the result.

$\blacksquare$