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

Upper and lower bounds for codomain set of finite unsigned basic measure

Formulation 0

Let $M = (X, \mathcal{F})$ be a D1108: Measurable space.
Let $\mu : \mathcal{F} \to [0, \infty]$ be a D85: Unsigned basic measure on $M$ such that

(i)	$\mu(X) < \infty$

Then \begin{equation} \mu(\mathcal{F}) \subseteq [0, \mu(X)] \end{equation}

Formulation 1

Let $M = (X, \mathcal{F})$ be a D1108: Measurable space.
Let $\mu : \mathcal{F} \to [0, \infty]$ be a D85: Unsigned basic measure on $M$ such that

(i)	$\mu(X) < \infty$

Then \begin{equation} 0 \leq \mu \leq \mu(X) \end{equation}

Proofs

Proof 0

Let $M = (X, \mathcal{F})$ be a D1108: Measurable space.
Let $\mu : \mathcal{F} \to [0, \infty]$ be a D85: Unsigned basic measure on $M$ such that

(i)	$\mu(X) < \infty$

Fix $E \in \mathcal{F}$. Now result R7: Empty set is subset of every set and the definition of a D84: Sigma-algebra guarantee the inclusions $\emptyset \subseteq E \subseteq X$. Thus R975: Isotonicity of unsigned basic measure and again the definition of a measure imply \begin{equation} 0 = \mu(\emptyset) \leq \mu(E) \leq \mu(X) \end{equation} Since $E \in \mathcal{F}$ was arbitrary, this implies that $\mu(F) \in [0, \mu(X)]$ for all $F \in \mathcal{F}$, which finishes the proof. $\square$