Method of exhaustion #
If μ, ν are two measures with ν s-finite, then there exists a set s such that
μ is sigma-finite on s, and for all sets t ⊆ sᶜ, either ν t = 0 or μ t = ∞.
Main definitions #
MeasureTheory.Measure.sigmaFiniteSetWRT: if such a set exists,μ.sigmaFiniteSetWRT νis a measurable set such thatμ.restrict (μ.sigmaFiniteSetWRT ν)is sigma-finite and for all setst ⊆ (μ.sigmaFiniteSetWRT ν)ᶜ, eitherν t = 0orμ t = ∞. If no such set exists (which is only possible ifνis not s-finite), we defineμ.sigmaFiniteSetWRT ν = ∅.MeasureTheory.Measure.sigmaFiniteSet: for an s-finite measureμ, a measurable set such thatμ.restrict μ.sigmaFiniteSetis sigma-finite, and for all setss ⊆ μ.sigmaFiniteSetᶜ, eitherμ s = 0orμ s = ∞. Defined asμ.sigmaFiniteSetWRT μ.
Main statements #
measure_eq_top_of_subset_compl_sigmaFiniteSetWRT: for s-finiteν, for all setssin(sigmaFiniteSetWRT μ ν)ᶜ, ifν s ≠ 0thenμ s = ∞.An instance showing that
μ.restrict (sigmaFiniteSetWRT μ ν)is sigma-finite.restrict_compl_sigmaFiniteSetWRT: ifμ ≪ νandνis s-finite, thenμ.restrict (μ.sigmaFiniteSetWRT ν)ᶜ = ∞ • ν.restrict (μ.sigmaFiniteSetWRT ν)ᶜ. As a consequence, that restriction is s-finite.An instance showing that
μ.restrict μ.sigmaFiniteSetis sigma-finite.restrict_compl_sigmaFiniteSet_eq_zero_or_top: the measureμ.restrict μ.sigmaFiniteSetᶜtakes only two values: 0 and ∞ .measure_compl_sigmaFiniteSet_eq_zero_iff_sigmaFinite: a measureμis sigma-finite iffμ μ.sigmaFiniteSetᶜ = 0.
References #
- [P. R. Halmos, Measure theory, 17.3 and 30.11][halmos1950measure]
A measurable set such that μ.restrict (μ.sigmaFiniteSetWRT ν) is sigma-finite and for all
measurable sets t ⊆ sᶜ, either ν t = 0 or μ t = ∞.
Equations
- μ.sigmaFiniteSetWRT ν = if h : ∃ (s : Set α), MeasurableSet s ∧ MeasureTheory.SigmaFinite (μ.restrict s) ∧ ∀ t ⊆ sᶜ, ν t ≠ 0 → μ t = ⊤ then h.choose else ∅
Instances For
We prove that the condition in the definition of sigmaFiniteSetWRT is true for finite
measures. Since every s-finite measure is absolutely continuous with respect to a finite measure,
the condition will then also be true for s-finite measures.
Let C be the supremum of ν s over all measurable sets s such that μ.restrict s is
sigma-finite. C is finite since ν is a finite measure. Then there exists a measurable set t
with μ.restrict t sigma-finite such that ν t ≥ C - 1/n.
A measurable set such that μ.restrict (μ.sigmaFiniteSetGE ν n) is sigma-finite and
for C the supremum of ν s over all measurable sets s with μ.restrict s sigma-finite,
ν (μ.sigmaFiniteSetGE ν n) ≥ C - 1/n.
Equations
- μ.sigmaFiniteSetGE ν n = ⋯.choose
Instances For
A measurable set such that μ.restrict (μ.sigmaFiniteSetWRT' ν) is sigma-finite and
ν (μ.sigmaFiniteSetWRT' ν) has maximal measure among such sets.
Equations
- μ.sigmaFiniteSetWRT' ν = ⋃ (n : ℕ), μ.sigmaFiniteSetGE ν n
Instances For
μ.sigmaFiniteSetWRT' ν has maximal ν-measure among all measurable sets s with sigma-finite
μ.restrict s.
Auxiliary lemma for measure_eq_top_of_subset_compl_sigmaFiniteSetWRT'.
For all sets s in (μ.sigmaFiniteSetWRT ν)ᶜ, if ν s ≠ 0 then μ s = ∞.
For all sets s in (μ.sigmaFiniteSetWRT ν)ᶜ, if ν s ≠ 0 then μ s = ∞.
A measurable set such that μ.restrict μ.sigmaFiniteSet is sigma-finite,
and for all measurable sets s ⊆ μ.sigmaFiniteSetᶜ, either μ s = 0 or μ s = ∞.
Equations
- μ.sigmaFiniteSet = μ.sigmaFiniteSetWRT μ
Instances For
The measure μ.restrict μ.sigmaFiniteSetᶜ takes only two values: 0 and ∞ .
The restriction of an s-finite measure μ to μ.sigmaFiniteSet is sigma-finite.
An s-finite measure μ is sigma-finite iff μ μ.sigmaFiniteSetᶜ = 0.