Null Sets Closed under Subset

Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.

Then $M$ is also a $\mu$-null set.

Proof

As $\mu$ is a measure, $\mu \left({M}\right) \ge 0$.

Also, by Measure is Monotone, $\mu \left({M}\right) \le \mu \left({N}\right) = 0$.

Hence $\mu \left({M}\right) = 0$, and $M$ is a $\mu$-null set.

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 4$: Problem $10 \ \text{(ii)}$