ThmDex – An index of mathematical definitions, results, and conjectures.

Isotonicity of unsigned basic measure

Formulation 0

Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.

Then \begin{equation} \forall \, E, F \in \mathcal{F} \, (E \subseteq F \quad \implies \quad \mu(E) \leq \mu(F)) \end{equation}

Formulation 1

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)	\begin{equation} E \subseteq F \end{equation}

Then \begin{equation} \mu(E) \leq \mu(F) \end{equation}

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 \begin{equation} \mu(F) = \mu(E \cup (F \setminus E)) = \mu(E) + \mu(F \setminus E) \end{equation} Since $\mu \geq 0$, we conclude that \begin{equation} \mu(F) = \mu(E) + \mu(F \setminus E) \geq \mu(E) \end{equation} $\square$