Lebesgue integral over finite and countable types, sets and measures #
The lemmas in this file require at least one of the following of the Lebesgue integral:
- The type of the set of integration is finite or countable
- The set of integration is finite or countable
- The measure is finite, s-finite or sigma-finite
Markov's inequality for the counting measure with hypothesis using tsum
in ℝ≥0∞
.
Markov's inequality for the counting measure with hypothesis using tsum
in ℝ≥0
.
If μ
is an s-finite measure, then for any function f
there exists a measurable function g ≤ f
that has the same Lebesgue integral over every set.
For the integral over the whole space, the statement is true without extra assumptions,
see exists_measurable_le_lintegral_eq
.
See also MeasureTheory.Measure.restrict_toMeasurable_of_sFinite
for a similar result.
In a sigma-finite measure space, there exists an integrable function which is positive everywhere (and with an arbitrarily small integral).
If the Lebesgue integral of a function is bounded by some constant on all sets with finite measure in a sub-σ-algebra and the measure is σ-finite on that sub-σ-algebra, then the integral over the whole space is bounded by that same constant.
Alias of MeasureTheory.lintegral_le_of_forall_fin_meas_trim_le
.
If the Lebesgue integral of a function is bounded by some constant on all sets with finite measure in a sub-σ-algebra and the measure is σ-finite on that sub-σ-algebra, then the integral over the whole space is bounded by that same constant.
If the Lebesgue integral of a function is bounded by some constant on all sets with finite measure and the measure is σ-finite, then the integral over the whole space is bounded by that same constant.