Additive Function on Empty Set is Zero
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Theorem
Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be an additive function on $\AA$.
Then $\map f \O = 0$.
Proof
From Properties of Algebras of Sets:
- $\O \in \AA$
Let $X \in \AA$.
Then:
\(\ds X \cap \O\) | \(=\) | \(\ds \O\) | Intersection with Empty Set | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f X + \map f \O\) | \(=\) | \(\ds \map f {X \cup \O}\) | Definition of Additive Function (Measure Theory) | ||||||||||
\(\ds \) | \(=\) | \(\ds \map f X\) | Union with Empty Set | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f \O\) | \(=\) | \(\ds 0\) |
$\blacksquare$