Countably Additive Function of Null Set
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Theorem
Let $\AA$ be a $\sigma$-algebra.
Let $\overline \R$ denote the extended set of real numbers.
Let $f: \AA \to \overline \R$ be a function be a countably additive function.
Suppose that there exists at least one $A \in \AA$ where $\map f A$ is a finite number.
Then:
- $\map f \O = 0$
Proof
Suppose that $A \in \AA$ such that $\map f A$ is a finite number.
So, let $\map f A = x$.
Consider the sequence $\sequence {S_i} \subseteq \AA$ defined as:
- $\forall i \in \N: S_i = \begin{cases}
A & : i = 0 \\ \O & : i > 0 \end{cases}$
Then:
- $\ds \bigcup_{i \mathop \ge 0} S_i = A$
Hence:
\(\ds \map f A\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f {\bigcup_{i \mathop \ge 0} S_i}\) | \(=\) | \(\ds x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{i \mathop \ge 0} \map f {S_i}\) | \(=\) | \(\ds x\) | Definition of Countably Additive Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f A + \sum_{i \mathop > 0} \map f \O\) | \(=\) | \(\ds x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{i \mathop > 0} \map f \O\) | \(=\) | \(\ds 0\) |
It follows directly that:
- $\map f \O = 0$
$\blacksquare$
Proof 2
This follows immediately from Countably Additive Function Dichotomy by Empty Set.
$\blacksquare$