Measure is Finitely Additive Function

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