Additive Function is Strongly Additive
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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 1
From Set Difference and Intersection form Partition:
- $A$ 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 is Disjoint with Reverse that:
- $\paren {A \setminus B} \cap \paren {B \setminus A} = \O$
From Sum of Additive Function Values is Well-Defined, it follows that $\map f A + \map f B$ is well-defined.
Hence:
| \(\ds \map f A + \map f B\) | \(=\) | \(\ds \map f {A \setminus B} + 2 \map f {A \cap B} + \map f {B \setminus A}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f {\paren {A \setminus B} \cup \paren {A \cap B} \cup \paren {B \setminus A} } + \map f {A \cap B}\) | Set Differences and Intersection form Partition of Union | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map f {A \cup B} + \map f {A \cap B}\) |
Hence the result.
$\blacksquare$
Proof 2
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 $\paren 2$ |
$\blacksquare$