Additive Function is Strongly Additive/Proof 2

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Theorem

Let $\SS$ be an algebra of sets.

Let $f: \SS \to \overline \R$ be an additive function on $\SS$.


Then $f$ is also strongly additive.

That is:

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


Proof

Recall that $\sqcup$ denotes the disjoint union.


First, since:

\(\ds A \cup B\) \(=\) \(\ds A \cup \paren {B \setminus A}\) Set Difference Union Second Set is Union
\(\ds \) \(=\) \(\ds A \sqcup \paren {B \setminus A}\) in view of Definition of Set Difference

we have:

$\paren 1 : \quad \map f {A \cup B} = \map f A + \map f {B \setminus A}$


Secondly, since

\(\ds B\) \(=\) \(\ds \paren {B \setminus A} \cup \paren {A \cap B}\) Set Difference Union Intersection
\(\ds \) \(=\) \(\ds \paren {B \setminus A} \sqcup \paren {A \cap B}\) Set Difference and Intersection are Disjoint

we have:

$\paren 2 : \quad \map f B = \map f {B \setminus A} + \map f {A \cap B}$


On the other hand:

\(\ds A \cup B\) \(=\) \(\ds \paren {A \cap B} \cup \paren {A \cup B}\) as $A \cap B \subseteq A \cup B$
\(\ds \) \(=\) \(\ds \paren {A \cap B} \cup \paren { \paren {A \cup B} \setminus \paren {A \cap B } }\) Set Difference Union Second Set is Union
\(\ds \) \(=\) \(\ds \paren {A \cap B} \sqcup \paren { \paren {A \cup B} \setminus \paren {A \cap B } }\) in view of Definition of Set Difference

so that:

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

In particular, we have neither:

$\map f {A \cup B} = + \infty \land \map f {A \cap B} = - \infty$

nor:

$\map f {A \cup B} = - \infty \land \map f {A \cap B} = + \infty$

Thus the sum in $\overline {\R}$:

$\map f {A \cup B} + \map f {A \cap B}$

is well-defined.


Finally:

\(\ds \map f {A \cup B} + \map f {A \cap B}\) \(=\) \(\ds \map f A + \map f {B \setminus A} + \map f {A \cap B}\) by $\paren 1$
\(\ds \) \(=\) \(\ds \map f A + \map f B\) by $\paren 2$

$\blacksquare$