Pre-measure/New page
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Grade: C
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The message provided is:
I've sort of forged my own path with regards to what we call a pre-measure. This is because I have found it easier to consider a pre-measure as a very similar thing to a measure, and then shown what other authors call pre-measures extend uniquely to my kind of pre-measure. They also call my kind of pre-measure a pre-measure.
Definition
Given a ring of sets, [ilmath]\mathcal{R} [/ilmath], a (non-negative) pre-measure is an extended real valued countably additive set function, [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\overline{\mathbb{R}_{\ge 0} } [/ilmath] with the following properties:- [ilmath]\bar{\mu}(\emptyset)=0[/ilmath]
- [ilmath]\forall(A_n)_{n=1}^\infty\subseteq\mathcal{R}[/ilmath][ilmath]\text{ pairwise disjoint} [/ilmath][math]\left[\left(\bigudot_{n=1}^\infty A_n\right)\in\mathcal{R}\implies\left(\bar{\mu}\left(\bigudot_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\bar{\mu}(A_n)\right)\right][/math][Note 1][Note 2]
Caution:Be aware that this convention differs from what many authors define as a pre-measure (although it meets their definition) see conventions and terminology below
Conventions and terminology
See also
- Pre-measurable space
- Measure - which we obtain by:
- Semi-ring of sets
- Types of set algebras
Notes
- ↑ There is a slight abuse of notation here, by the nature of implies if the LHS is false, we do not care if the RHS is true or false! However if the LHS is false here, the RHS doesn't even make sense (as [ilmath]\bar{\mu}(A)[/ilmath] makes no sense when [ilmath]A[/ilmath] is not in the domain of [ilmath]\bar{\mu} [/ilmath])
- ↑ By using [ilmath]\bigudot[/ilmath] we are making it clear that the sequence is that of pairwise disjoint sets and going forward we shall not write "pairwise disjoint" when the union symbol implies it. For clarity the full statement is thus:
- [math]\forall(A_n)_{n=1}^\infty\subseteq\mathcal{R}\left[\underbrace{\left(\forall i,j\in\mathbb{N}[i\ne j\implies A_i\cap A_j=\emptyset]\right)}_{\text{the }S_n\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(\left(\bigudot_{n=1}^\infty A_n\right)\in\mathcal{R}\implies\left(\bar{\mu}\left(\bigudot_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\bar{\mu}(A_n)\right)\right)\right][/math]
References
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