Theorems on the convergence of bounded monotonic sequences
In the mathematical field of real analysis, the monotone convergence theorem is any of a number of related theorems proving the convergence of monotonic sequences (sequences that are decreasing or increasing) that are also bounded. Informally, the theorems state that if a sequence is increasing and bounded above, it will converge to its smallest upper bound, its supremum; in the same way, if a sequence is decreasing and is bounded below it will convergence to its largest lower bound, its infimum.
Convergence of a monotone sequence of real numbers[edit]
Let be the set of terms of . By assumption, is non-empty and bounded above by . By the least-upper-bound property of real numbers, exists and . Now, for every , there exists such that , since otherwise is a strictly smaller upper bound of , contradicting the definition of the supremum . Then since is non decreasing, and is an upper bound, for every , we have
Hence, by definition .
The proof of lemma 2 is analogous or follows from lemma 1 by considering .
There is a variant of Lemma 1 and 2 where we allow unbounded sequences in the extended real numbers, the real numbers with and added.
In the extended real numbers every set has a supremum (resp. infimum) which of course may be (resp. ) if the set is unbounded. An important use of the extended reals is that any set of non negative numbers has a well defined sum
where are the upper extended non negative real numbers.
If for all natural numbers and , is a non-negative real number and , then[2]: 168
In particular, the left hand side is finite if and only if the right hand side is.
The theorem states e.g. that if you have an infinite matrix of non-negative real numbers such that
the rows are weakly increasing and bounded with
then
for each column, the column sum is bounded hence convergent, and the limit of the column sums is equal to the sum of the "limit column" which element wise is the supremum over the row.
The following result is a generalisation of the monotone convergence of non negative sums theorem above. It is due to Beppo Levi, who proved a slight generalization in 1906 of an earlier result by Henri Lebesgue.[3] In what follows, denotes the -algebra of Borel sets on the upper extended non negative real numbers . By definition, contains the set and all Borel subsets of
Let be a measure space, and . Consider a pointwise non-decreasing sequence of -measurable non-negative functions , i.e., for every and every ,
Set the pointwise supremum of the sequence to be . That is, for every ,
Then is -measurable and
Remark 1. The integrals and the supremum may be finite or infinite, but the left hand side is finite if and only if the right hand side is.
Remark 2. Under the assumptions of the theorem,
(Note that the second chain of equalities follows from monoticity of the integral (lemma 2 below).
Remark 3. The theorem remains true if its assumptions hold -almost everywhere. In other words, it is enough that there is a null set such that the sequence non-decreases for every To see why this is true, we start with an observation that allowing the sequence to pointwise non-decrease almost everywhere causes its pointwise limit to be undefined on some null set . On that null set, may then be defined arbitrarily, e.g. as zero, or in any other way that preserves measurability. To see why this will not affect the outcome of the theorem, note that since we have, for every
and
provided that is -measurable.[4]: section 21.38 (These equalities follow directly from the definition of the Lebesgue integral for a non-negative function).
Remark 4. The proof below does not use any properties of the Lebesgue integral except those established here. The theorem, thus, can be used to prove other basic properties, such as linearity, pertaining to Lebesgue integration.
This proof does not rely on Fatou's lemma; however, we do explain how that lemma might be used. Those not interested in this independency of the proof may skip the intermediate results below.
We need two basic lemmas
In the proof below, we apply the monotonic property of the Lebesgue integral to non-negative functions only. Specifically (see Remark 4),
lemma 1 (monotonicity of the Lebesgue integral). let the functions be -measurable.
If everywhere on then
If and then
Proof. Denote by the set of simple-measurable functions such that
everywhere on
1. Since we have
By definition of the Lebesgue integral and the properties of supremum,
2. Let be the indicator function of the set It can be deduced from the definition of the Lebesgue integral that
if we notice that, for every outside of Combined with the previous property, the inequality implies
Lemma 1. Let be a measurable space. Consider a simple -measurable non-negative function . For a subset , define
Then is a measure on .
Monotonicity follows from lemma 1. Here, we will only prove countable additivity, leaving the rest up to the reader. Let , be a decomposition of as a countable pairwise disjoint union of measurable subsets . Write
for a finite number of non-negative constants and measurable sets . By definition of the Lebesgue integral
as required.
Here we used that for fixed , all the sets are pairwise disjoint, the countable additivity of and the fact that since all the summands are non-negative, the sum of the series, be it finite or infinite, is independent of the order of the summation and so summations can be interchanged.
The proof can be based on Fatou's lemma instead of a direct proof, because Fatou's lemma can be proved independent of the monotone convergence theorem. The measurability follows directly and the crucial "inverse Inequality" (step 3) is a direct consequence of the inequality
However the monotone convergence theorem is some ways more primitive than Fatou's lemma and the following is a direct proof.
Denote by the set of simple -measurable functions ( nor included!) such that
on .
Step 1. The function is –measurable, and the integral is well-defined (albeit possibly infinite)[4]: section 21.3
Since from we get , so it suffices to show that is measurable. To see this it suffices to prove that the inverse image of an interval under is an element of the sigma-algebra on . This is because intervals with generate the Borel sigma algebra on the extended non negative reals by countable intersection and union. Since the is a non decreasing sequence
Therefore since and
Hence is a measurable set,
being the countable intersection of measurable sets, each the inverse image of a Borel set under a -measurable function . This shows that is -measurable.
Since the integral is defined as a sup
over simple functions and is thus well-defined but the sup may possibly be infinite.
Step 2. We have the inequality
This follows directly from the monotonicity of the integral and the definition of supremum: since we have , hence .
step 3 We have the reverse inequality
.
We need some extra machinery.
Given a simple function and a positive real number , define
Then because the sequence is non decreasing. Moreover because either and for all , or and for .
step 3(a).
The set is measurable.
To see this from first principles, write , for some finite collection of non-negative constants , and measurable sets which we may assume are pairwise disjoint and with union (here denotes the indicator function of the set ). Then if we have if and only if . Therefore
which since the are measurable, is a union of (disjoint) measurable sets
Step 3(b). For every simple -measurable non-negative function ,
To prove this, define . By Lemma 1, is a measure on . By "continuity from below" (Lemma 2),
as required.
Step 3(c). We now prove that, for every ,
Indeed, using the definition of , the monotonicity of the Lebesgue integral and the non-negativity of we have
for every .
Hence by step 3(b):
Since the left hand side is just a finite sum we get
Under similar hypotheses to Beppo Levi's theorem, it is possible to relax the hypothesis of monotonicity.[5] As before, let be a measure space and . Again, will be a sequence of -measurable non-negative functions . However, we do not assume they are pointwise non-decreasing. Instead, we assume that converges for almost every , we define to be the pointwise limit of , and we assume additionally that pointwise almost everywhere for all . Then is -measurable, and exists, and
As before, measurability follows from the fact that almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has
by Fatou's lemma, and then, by standard properties of limits and monotonicity,
Therefore , and both are equal to . It follows that exists and equals .