In the mathematical field of real analysis, the **monotone convergence theorem** is any of a number of related theorems proving the good convergence behaviour of monotonic sequences, i.e. sequences that are non-increasing, or non-decreasing. In its simplest form, it says that a non-decreasing bounded-above sequence of real numbers converges to its smallest upper bound, its supremum. Likewise, a non-increasing bounded-below sequence converges to its largest lower bound, its infimum. In particular, infinite sums of non-negative numbers converge to the supremum of the partial sums if and only if the partial sums are bounded.

For sums of non-negative increasing sequences , it says that taking the sum and the supremum can be interchanged.

In more advanced mathematics the monotone convergence theorem usually refers to a fundamental result in measure theory due to Lebesgue and Beppo Levi that says that for sequences of non-negative pointwise-increasing measurable functions , taking the integral and the supremum can be interchanged with the result being finite if either one is finite.

For a non-decreasing and bounded-above sequence of real numbers

the limit exists and equals its supremum:

For a non-increasing and bounded-below sequence of real numbers

the limit exists and equals its infimum:

- .

Let be the set of values 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 .

If is a monotone sequence of real numbers, i.e., if for every or for every , then this sequence has a finite limit if and only if the sequence is bounded.^{[1]}

- "If"-direction: The proof follows directly from the lemmas.
- "Only If"-direction: By (ε, δ)-definition of limit, every sequence with a finite limit is necessarily bounded.

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 summation order independent sum

where are the upper extended non negative real numbers. For a series of non negative numbers

so this sum coincides with the sum of a series if both are defined. In particular the sum of a series of non negative numbers does not depend on the order of summation.

Let be a sequence of non-negative real numbers indexed by natural numbers and . Suppose that for all . Then^{[2]}^{: 168 }

**Remark**
The suprema and the sums may be finite or infinite but the left hand side is finite if and only if the right hand side is.

**proof**: Since we have so . Conversely, we can interchange sup and sum for finite sums so
hence .

The theorem states that if you have an infinite matrix of non-negative real numbers such that

- the rows are weakly increasing and each is bounded where the bounds are summable

then

- for each column, the non decreasing column sums are 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.

As an example, consider the expansion

Now set

for and for , then with and

- .

The right hand side is a non decreasing sequence in , therefore

- .

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 a measurable set. Let be a pointwise non-decreasing sequence of -measurable non-negative functions, i.e. each function is -measurable and for every and every ,

Then the pointwise supremum

is a -measurable function and

**Remark 1.** The integrals and the suprema 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
hence

**2.** The functions where is the indicator function of , are easily seen to be measurable and . Now apply **1**.

**Lemma 1.** Let be a measurable space. Consider a simple -measurable non-negative function . For a measurable subset , define

Then is a measure on .

Write with and measurable sets . Then

Since finite positive linear combinations of measures are countably additive, to prove countable additivity of it suffices to prove that, the set function defined by is countably additive for all . But this follows directly from the countable additivity of .

**Lemma 2.** Let be a measure, and , where

is a non-decreasing chain with all its sets -measurable. Then

Set , then we decompose and likewise as a union of disjoint measurable sets. Therefore , and so .

Set . 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 }

From we get . Hence we have to show that is -measurable. To see this, it suffices to prove that is -measurable for all , because the intervals generate the Borel sigma algebra on the extended non negative reals by complementing and taking countable intersections, unions.

Now since the is a non decreasing sequence, if and only if for all . Since we already know that and we conclude that

Hence is a measurable set, being the countable intersection of the measurable sets .

Since the integral is well defined (but possibly infinite) as

- .

**Step 2.** We have the inequality

This is equivalent to for all which follows directly from and the monotonicity of the integral.

**step 3** We have the reverse inequality

- .

By the definition of integral as a supremum step 3 is equivalent to

for every . It is tempting to prove for sufficiently large, but this does not work e.g. if is simple and the . We need an "epsilon of room" to manoeuvre.

Given a simple function and an , define

**step 3(a).** We have

- is -measurable.

Ad 1: Write , for non-negative constants , and measurable sets , which we may assume are pairwise disjoint and with union . Then for we have if and only if so

which is measurable because the are measurable.

Ad 2: the sequence is non decreasing.

Ad 3: Fix . Either so hence , or so for sufficiently large hence .

Now by the definition of and the monotonicity of the Lebesgue integral we have

Hence by lemma 2, "continuity from below" and (3(a).3):

The left hand side is a finite sum and the inequality can be rewritten as

which gives step 3 by first taking the supremum over and then the supremum over .

The proof of Beppo Levi's theorem is complete.

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

The proof can also be based on Fatou's lemma instead of a direct proof as above, because Fatou's lemma can be proved independent of the monotone convergence theorem. However the monotone convergence theorem is in some ways more primitive than Fatou's lemma and the following is a direct proof.

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 .