Measure is Finitely Additive Function
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Theorem
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\mu: \Sigma \to \overline {\R}$ be a measure on $\Sigma$.
Then $\mu$ is finitely additive.
Proof
From the definition of a measure, $\mu$ is countably additive.
From Countably Additive Function also Finitely Additive, $\mu$ is finitely additive.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $4.3 \ \text{(i)}$
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $1.2$: Measures