ThmDex – An index of mathematical definitions, results, and conjectures.
Result R3513 on D3802: Subnull set
Measurable subnull set has measure zero
Formulation 0
Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space such that
(i) $\mu(F) = 0$ for $F \in \mathcal{F}$
Then \begin{equation} \forall \, E \subseteq F \, (E \in \mathcal{F} \quad \Rightarrow \quad \mu(E) = 0) \end{equation}
Proofs
Proof 0
Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space such that
(i) $\mu(F) = 0$ for $F \in \mathcal{F}$
Let $E \subseteq F$ such that $E \in \mathcal{F}$. Denote $E^{\complement} : = F \setminus E$. The sets $E$ and $E^{\complement}$ are disjoint with $E \cup E^{\complement} = F$, so additivity of measure implies \begin{equation} 0 = \mu(F) = \mu(E \cup E^{\complement}) = \mu(E) + \mu(E^{\complement}) \geq \mu(E) \geq 0 \end{equation} Hence, $\mu(E) = 0$. $\square$