Hahn decomposition #
This file proves the Hahn decomposition theorem (signed version). The Hahn decomposition theorem
states that, given a signed measure s
, there exist complementary, measurable sets i
and j
,
such that i
is positive and j
is negative with respect to s
; that is, s
restricted on i
is non-negative and s
restricted on j
is non-positive.
The Hahn decomposition theorem leads to many other results in measure theory, most notably, the Jordan decomposition theorem, the Lebesgue decomposition theorem and the Radon-Nikodym theorem.
Main results #
MeasureTheory.SignedMeasure.exists_isCompl_positive_negative
: the Hahn decomposition theorem.MeasureTheory.SignedMeasure.exists_subset_restrict_nonpos
: A measurable set of negative measure contains a negative subset.
Notation #
We use the notations 0 ≤[i] s
and s ≤[i] 0
to denote the usual definitions of a set i
being positive/negative with respect to the signed measure s
.
Tags #
Hahn decomposition theorem
exists_subset_restrict_nonpos #
In this section we will prove that a set i
whose measure is negative contains a negative subset
j
with respect to the signed measure s
(i.e. s ≤[j] 0
), whose measure is negative. This lemma
is used to prove the Hahn decomposition theorem.
To prove this lemma, we will construct a sequence of measurable sets $(A_n){n \in \mathbb{N}}$, such that, for all $n$, $s(A{n + 1})$ is close to maximal among subsets of $i \setminus \bigcup_{k \le n} A_k$.
This sequence of sets does not necessarily exist. However, if this sequence terminates; that is, there does not exists any sets satisfying the property, the last $A_n$ will be a negative subset of negative measure, hence proving our claim.
In the case that the sequence does not terminate, it is easy to see that $i \setminus \bigcup_{k = 0}^\infty A_k$ is the required negative set.
To implement this in Lean, we define several auxiliary definitions.
- given the sets
i
and the natural numbern
,ExistsOneDivLT s i n
is the property that there exists a measurable setk ⊆ i
such that1 / (n + 1) < s k
. - given the sets
i
and thati
is not negative,findExistsOneDivLT s i
is the least natural numbern
such thatExistsOneDivLT s i n
. - given the sets
i
and thati
is not negative,someExistsOneDivLT
chooses the setk
fromExistsOneDivLT s i (findExistsOneDivLT s i)
. - lastly, given the set
i
,restrictNonposSeq s i
is the sequence of sets defined inductively whererestrictNonposSeq s i 0 = someExistsOneDivLT s (i \ ∅)
andrestrictNonposSeq s i (n + 1) = someExistsOneDivLT s (i \ ⋃ k ≤ n, restrictNonposSeq k)
. This definition represents the sequence $(A_n)$ in the proof as described above.
With these definitions, we are able consider the case where the sequence terminates separately,
allowing us to prove exists_subset_restrict_nonpos
.
A measurable set of negative measure has a negative subset of negative measure.
The set of measures of the set of measurable negative sets.
Equations
Instances For
Alternative formulation of measure_theory.signed_measure.exists_is_compl_positive_negative
(the Hahn decomposition theorem) using set complements.
The Hahn decomposition theorem: Given a signed measure s
, there exist
complement measurable sets i
and j
such that i
is positive, j
is negative.
The symmetric difference of two Hahn decompositions has measure zero.