Integral of a positive function (measure theory)/Definition

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Definition

Let [ilmath](X,\mathcal{A},\mu)[/ilmath] be a measure space, the [ilmath]\mu[/ilmath]-integral of a positive numerical function, [ilmath]f\in\mathcal{M}^+_{\bar{\mathbb{R} } }(\mathcal{A}) [/ilmath][Note 1][Note 2] is[1]:

  • [math]\int f\mathrm{d}\mu:=\text{Sup}\left\{I_\mu(g)\ \Big\vert\ g\le f, g\in\mathcal{E}^+(\mathcal{A})\right\}[/math][Note 3]

Recall that:



TODO: Link to [ilmath]\mathcal{E} [/ilmath] somewhere, are they numeric or real valued?



TODO: Can every simple function be made into a standard representation, thus what is [ilmath]\mathcal{E} [/ilmath] exactly and what is the domain of [ilmath]I_\mu[/ilmath] exactly?


Notes

  1. So [ilmath]f:X\rightarrow\bar{\mathbb{R} }^+[/ilmath]
  2. Notice that [ilmath]f[/ilmath] is [ilmath]\mathcal{A}/\bar{\mathcal{B} } [/ilmath]-measurable by definition, as [ilmath]\mathcal{M}_\mathcal{Z}(\mathcal{A})[/ilmath] denotes all the measurable functions that are [ilmath]\mathcal{A}/\mathcal{Z} [/ilmath]-measurable, we just use the [ilmath]+[/ilmath] as a slight abuse of notation to denote all the positive ones (with respect to the standard order on [ilmath]\bar{\mathbb{R} } [/ilmath] - the extended reals)
  3. The [ilmath]g\le f[/ilmath] is an abuse of notation for saying that [ilmath]g[/ilmath] is everywhere less than [ilmath]f[/ilmath], we could have written:
    • [math]\int f\mathrm{d}\mu=\text{Sup}\left\{I_\mu(g)\ \Big\vert\ g\le f, g\in\mathcal{E}^+\right\}=\text{Sup}\left\{I_\mu(g)\ \Big\vert\ g\in\left\{h\in\mathcal{E}^+(\mathcal{A})\ \big\vert\ \forall x\in X\left(h(x)\le f(x)\right)\right\}\right\}[/math] instead.
    Inline with: Notation for dealing with (extended) real-valued measurable maps

References

  1. Measures, Integrals and Martingales - René L. Schilling