Monotonicity of the integral of non-negative extended-real-valued measurable functions with respect to a measure
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Statement
Let [ilmath](X,\mathcal{A},\mu)[/ilmath] be a measure space and let [ilmath]f,g\in[/ilmath][ilmath]\mathcal{M}_{\bar{\mathbb{R} }_{\ge 0} }(\mathcal{A})[/ilmath]^{[Note 1]}, then^{[1]}:
- if [ilmath]f\le g[/ilmath] - i.e.: [ilmath]\forall x\in X[f(x)\le g(x)][/ilmath] - then:
- [ilmath]\int f\ \mathrm{d}\mu\le\int g\ \mathrm{d}\mu[/ilmath]
Proof
Recall the definition of the integral of a non-negative numerical function^{Author's note:}^{[Note 2]}:
- [math]\int f\ \mathrm{d}\mu:\eq\text{Sup}\left(\left\{I_\mu(s)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \Big\vert\ s\in\mathcal{E}_{\ge 0}(\mathcal{A})\wedge s\le f\right\} \right)[/math], where:
- [ilmath]s\le f[/ilmath] is an abuse of notation for [ilmath]\forall x\in X[s(x)\le f(x)][/ilmath] and
- [ilmath]s\in\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath] means [ilmath]s\in[/ilmath][[Set of all simple functions over a set|[ilmath]\mathcal{E}(\mathcal{A})[/ilmath] and [ilmath]\forall x\in X[s(x)\ge 0][/ilmath]
- Let [ilmath]f,g\in\mathcal{M}_{\bar{\mathbb{R} }_{\ge 0} }(\mathcal{A}) [/ilmath] be given and suppose [ilmath]f\le g[/ilmath]
- Define [ilmath]A:\eq\left\{r\in\mathcal{E}_{\ge 0}(\mathcal{A})\ \Big\vert\ r\le f\right\} [/ilmath]
- Define [ilmath]B:\eq\left\{s\in\mathcal{E}_{\ge 0}(\mathcal{A})\ \Big\vert\ s\le g\right\} [/ilmath]
- Lemma: we claim that [ilmath]A\subseteq B[/ilmath], using the implies-subset relation we see [ilmath]\big(A\subseteq B\big)\iff\big(\forall a\in A[a\in B]\big)[/ilmath]
- Let [ilmath]a\in A[/ilmath] be given, notice that by definition [ilmath]A\subseteq\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath], so [ilmath]a\in\mathcal{E}_{\ge 0}(\mathcal{A}) [/ilmath]
- Then [ilmath]a\le f[/ilmath], or [ilmath]\forall x\in X[a(x)\le f(x)][/ilmath]
- But [ilmath]f\le g[/ilmath] by hypothesis, i.e. [ilmath]\forall x\in X[f(x)\le g(x)][/ilmath], so we see [ilmath]\forall x\in X[a(x)\le g(x)][/ilmath] of [ilmath]a\le g[/ilmath]
- We see that both:
- [ilmath]a\in\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath] - of which all elements of [ilmath]B[/ilmath] are too
- [ilmath]a\le g[/ilmath] - the definition of a member of [ilmath]\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath] to be in [ilmath]B[/ilmath]
- So [ilmath]a\in B[/ilmath]
- We see that both:
- Since [ilmath]a\in A[/ilmath] was arbitrary we have shown this for all [ilmath]a\in A[/ilmath] - as required to prove the lemma
- Let [ilmath]a\in A[/ilmath] be given, notice that by definition [ilmath]A\subseteq\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath], so [ilmath]a\in\mathcal{E}_{\ge 0}(\mathcal{A}) [/ilmath]
- Define [ilmath]C:\eq\{I_\mu(r)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \big\vert\ r\in A\} [/ilmath] and
- Define [ilmath]D:\eq\{I_\mu(s)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \big\vert\ s\in B\} [/ilmath]
- As [ilmath]I_\mu(\cdot)[/ilmath] is a mapping (See: integral of a simple function (measure theory)) we see [ilmath]C\subseteq D[/ilmath]
- Thus [ilmath]\text{Sup}(C)\le\text{Sup}(D)[/ilmath]
- i.e.: [ilmath]\int f\ \mathrm{d}\mu\le\int g\ \mathrm{d}\mu[/ilmath] - as required.
- Thus [ilmath]\text{Sup}(C)\le\text{Sup}(D)[/ilmath]
- As [ilmath]I_\mu(\cdot)[/ilmath] is a mapping (See: integral of a simple function (measure theory)) we see [ilmath]C\subseteq D[/ilmath]
- Define [ilmath]D:\eq\{I_\mu(s)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \big\vert\ s\in B\} [/ilmath]
- Lemma: we claim that [ilmath]A\subseteq B[/ilmath], using the implies-subset relation we see [ilmath]\big(A\subseteq B\big)\iff\big(\forall a\in A[a\in B]\big)[/ilmath]
- Define [ilmath]B:\eq\left\{s\in\mathcal{E}_{\ge 0}(\mathcal{A})\ \Big\vert\ s\le g\right\} [/ilmath]
- Define [ilmath]A:\eq\left\{r\in\mathcal{E}_{\ge 0}(\mathcal{A})\ \Big\vert\ r\le f\right\} [/ilmath]
Grade: C
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Notes
- ↑ functions which map to the extended real values are sometimes called numerical functions
- ↑ Right now this is at integral of a positive function (measure theory), this is technically incorrect, but the information can be found here.
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