Monotone convergence theorem


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 that are also bounded. Informally, the theorems state that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum; in the same way, if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum.

Convergence of a monotone sequence of real numbers

Lemma 1

If a sequence of real numbers is increasing and bounded above, then its supremum is the limit.

Proof

Let be such a sequence, and let be the set of terms of. By assumption, is non-empty and bounded above. By the least-upper-bound property of real numbers, exists and is finite. Now, for every, there exists such that, since otherwise is an upper bound of, which contradicts to the definition of. Then since is increasing, and is its upper bound, for every, we have. Hence, by definition, the limit of is

Lemma 2

If a sequence of real numbers is decreasing and bounded below, then its infimum is the limit.

Proof

The proof is similar to the proof for the case when the sequence is increasing and bounded above,

Theorem

If is a monotone sequence of real numbers, then this sequence has a finite limit if and only if the sequence is bounded.

Proof

Theorem

If for all natural numbers j and k, aj,k is a non-negative real number and aj,k ≤ aj+1,k, then
The theorem states that if you have an infinite matrix of non-negative real numbers such that
  1. the columns are weakly increasing and bounded, and
  2. for each row, the series whose terms are given by this row has a convergent sum,
then the limit of the sums of the rows is equal to the sum of the series whose term k is given by the limit of column k. The series has a convergent sum if and only if the sequence of row sums is bounded and therefore convergent.
As an example, consider the infinite series of rows
where n approaches infinity. Here the matrix entry in row n and column k is
the columns are indeed weakly increasing with n and bounded, while the rows only have finitely many nonzero terms, so condition 2 is satisfied; the theorem now says that you can compute the limit of the row sums by taking the sum of the column limits, namely .

Beppo Levi's monotone convergence theorem for Lebesgue integral

The following result is due to Beppo Levi and Henri Lebesgue. In what follows, denotes the -algebra of Borel sets on. By definition, contains the set and all Borel subsets of

Theorem

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 limit of the sequence to be. That is, for every ,
Then is -measurable and
Remark 1. The integrals may be finite or infinite.
Remark 2. 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
provided that is -measurable..
Remark 3. Under assumptions of the theorem,
.
Remark 4. The proof below does not use any properties of Lebesgue integral except those established here. The theorem, thus, can be used to prove other basic properties, such as linearity, pertaining to Lebesgue integration.
Remark 5. In the proof below, we apply the monotonic property of Lebesgue integral to non-negative functions only. Specifically, let the functions be -measurable.
Proof. Denote the set of simple -measurable functions such that
everywhere on
1. Since we have
By definition of Lebesgue integral and the properties of supremum,
2. Let be the indicator function of the set It can be deduced from the definition of Lebesgue integral that
if we notice that, for every outside of Combined with the previous property, the inequality implies

Proof

This proof does not rely on Fatou's lemma. However, we do explain how that lemma might be used.
For those not interested in independent proof, the intermediate results below may be skipped.

Intermediate results

Lebesgue integral as measure
Lemma 1. Let be a measurable space. Consider a simple -measurable non-negative function. For a subset, define
Then is a measure on.
Proof
Monotonicity follows from Remark 5. Here, we will only prove countable additivity, leaving the rest up to the reader. Let, where all the sets are pairwise disjoint. Due to simplicity,
for some finite non-negative constants and pairwise disjoint sets such that. By definition of Lebesgue integral,
Since all the sets are pairwise disjoint, the countable additivity of
gives us
Since all the summands are non-negative, the sum of the series, whether this sum is finite or infinite, cannot change if summation order does. For that reason,
as required.
"Continuity from below"
The following property is a direct consequence of the definition of measure.
Lemma 2. Let be a measure, and, where
is a non-decreasing chain with all its sets -measurable. Then

Proof of theorem

Step 1. We begin by showing that is –measurable.
Note. If we were using Fatou's lemma, the measurability would follow easily from Remark 3.
To do this without using Fatou's lemma, it is sufficient to show that the inverse image of an interval under is an element of the sigma-algebra on, because intervals generate the Borel sigma algebra on the reals. Since is a closed interval, and, for every,,
Thus,
Being the inverse image of a Borel set under a -measurable function, each set in the countable intersection is an element of. Since -algebras are, by definition, closed under countable intersections, this shows that is -measurable, and the integral is well defined.
Step 2. We will first show that
The definition of and monotonicity of imply that, for every and every. By monotonicity of Lebesgue integral,
and
Note that the limit on the right exists because, due to monotonicity, the sequence is non-decreasing.
End of Step 2.
We now prove the reverse inequality. We seek to show that
Proof using Fatou's lemma. Per Remark 3, the inequality we want to prove is equivalent to
But the latter follows immediately from Fatou's lemma, and the proof is complete.
Independent proof. To prove the inequality without using Fatou's lemma, we need some extra machinery. Denote the set of simple -measurable functions such that
on.
Step 3. Given a simple function and a real number, define
Then,, and.
Step 3a. To prove the first claim, let, for some finite collection of pairwise disjoint measurable sets such that, some non-negative constants, and denoting the indicator function of the set.
For every holds if and only if Given that the sets are pairwise disjoint,
Since the pre-image of the Borel set
under the measurable function is measurable, and -algebras, by definition, are closed under finite intersection and unions, the first claim follows.
Step 3b. To prove the second claim, note that, for each and every,
Step 3c. To prove the third claim, we show that.
Indeed, if, to the contrary,, then an element
exists such that, for every. Taking the limit as, we get
But by initial assumption,. This is a contradiction.
Step 4. For every simple -measurable non-negative function,
To prove this, define. By Lemma 1, is a measure on. By "continuity from below",
as required.
Step 5. We now prove that, for every,
Indeed, using the definition of, the non-negativity of, and the monotonicity of Lebesgue integral, we have
for every. In accordance with Step 4, as, the inequality becomes
Taking the limit as yields
as required.
Step 6. We are now able to prove the reverse inequality, i.e.
Indeed, by non-negativity, and For the calculation below, the non-negativity of is essential. Applying the definition of Lebesgue integral and the inequality established in Step 5, we have
The proof is complete.