ThmDex – An index of mathematical definitions, results, and conjectures.
Result R975 on D1158: Measure space
Isotonicity of unsigned basic measure

Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.
Then $$\forall \, E, F \in \mathcal{F} \, (E \subseteq F \quad \implies \quad \mu(E) \leq \mu(F))$$

Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.
Let $E, F \in \mathcal{F}$ each be a D1109: Measurable set in $M$ such that
 (i) $$E \subseteq F$$
Then $$\mu(E) \leq \mu(F)$$
Subresults
 ▶ R2090: Isotonicity of probability measure
Proofs
Proof 0
Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.
Let $E, F \in \mathcal{F}$ such that $E \subseteq F$. Result R977: Ambient set is union of subset and complement of subset yields the decomposition $F = E \cup (F \setminus E)$. Applying R976: Finite disjoint additivity of unsigned basic measure to this union we then obtain $$\mu(F) = \mu(E \cup (F \setminus E)) = \mu(E) + \mu(F \setminus E)$$ Since $\mu \geq 0$, we conclude that $$\mu(F) = \mu(E) + \mu(F \setminus E) \geq \mu(E)$$ $\square$