LeanPool.SardMoreira.LebesgueDensity #
If a finite measure μ is absolutely continuous with respect to a σ-finite measure ν,
then μ s → 0 as ν s → 0. More precisely, for any ε ≠ 0 there exists δ > 0
such that all sets of ν measure less than δ have a μ measure less than ε.
The previous lemma is not true unless we assume that ν is a finite measure.
Indeed, for ν = volume on ℝ, μ = ν.withDensity (Real.nnabs ·),
we can choose s = Set.Icc a (a + δ / 2) for a large a,
and get an arbitriraly large value of μ s.
If a finite measure μ is absolutely continuous with respect to a σ-finite measure ν,
then μ s → 0 as ν s → 0. More precisely, for any ε ≠ 0 there exists δ > 0
such that all sets of ν measure less than δ have a μ measure less than ε.
If $s_b$ is a family of sets such that $\{(a, b) \mid a \in s_b\}$ is a measurable set, then for any s-finite measure $\mu$, the function $b \mapsto \mu(s_b)$ is measurable.
This is a version of measurable_measure_prod_mk_right.
Let e : X → Y be a dense topological embedding, let Z be a regular space.
For each y : Y, let U y be an open set such that y ∈ closure (U y).
Let s be an open set, let x : X be a point.
Suppose that f : Y → Z is continuous within U y at all y ∈ s close to e x.
Then f (e x') tends to z as x' tends to x, e x' ∈ s,
if and only if f y tends to z as y ∈ s tends to e x.
If X = ℚ, Y = ℝ, and e = Rat.cast,
then this lemma can be used to restate convergence of a function defined on real numbers
in terms of convergence of a function on rational numbers,
which is more convenient for measure theory, because there are only countably many rational numbers.
Parametrized version of ae_tendsto_measure_inter_div_of_measurableSet.
Parametrized version of ae_tendsto_measure_inter_div_of_measurableSet.