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.

Convergence of a monotone sequence of real numbers

Every bounded-above monotonically nondecreasing sequence of real numbers is convergent in the real numbers because the supremum exists and is a real number. The proposition does not apply to rational numbers because the supremum of a sequence of rational numbers may be irrational.

Proposition

(A) For a non-decreasing and bounded-above sequence of real numbers the limit exists and equals its supremum: (B) For a non-increasing and bounded-below sequence of real numbers the limit exists and equals its infimum:

Proof

Let be the set of values of. By assumption, { a_n } is non-empty and bounded above by K. By the least-upper-bound property of real numbers, exists and c \le K. Now, for every, there exists N such that , since otherwise is a strictly smaller upper bound of { a_n }, contradicting the definition of the supremum c. Then since is non decreasing, and c is an upper bound, for every n > N, we have Hence, by definition. The proof of the (B) part is analogous or follows from (A) by considering.

Theorem

If is a monotone sequence of real numbers, i.e., if for every n \ge 1 or for every n \ge 1, then this sequence has a finite limit if and only if the sequence is bounded.

Proof

Convergence of a monotone series

There is a variant of the proposition above where we allow unbounded sequences in the extended real numbers, the real numbers with \infty and -\infty added. In the extended real numbers every set has a supremum (resp. infimum) which of course may be \infty (resp. -\infty) 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.

Monotone convergence of non negative sums

Let be a sequence of non-negative real numbers indexed by natural numbers i and k. Suppose that for all i, k. Then

Proof

Since we have so. Conversely, we can interchange sup and sum for finite sums by reverting to the limit definition, so hence.

Examples

Matrices

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.

e

Consider the expansion Now set for i \le k and a_{i,k} = 0 for i > k, then with and The right hand side is a non decreasing sequence in k, therefore

Beppo Levi's lemma

The following result is a generalisation of the monotone convergence of non negative sums theorem above to the measure theoretic setting. It is a cornerstone of measure and integration theory with many applications and has Fatou's lemma and the dominated convergence theorem as direct consequence. It is due to Beppo Levi, who proved a slight generalization in 1906 of an earlier result by Henri Lebesgue. Let denotes the \sigma-algebra of Borel sets on the upper extended non negative real numbers [0,+\infty]. By definition, contains the set {+\infty} and all Borel subsets of

Theorem (monotone convergence theorem for non-negative measurable functions)

Let be a measure space, and X\in\Sigma a measurable set. Let be a pointwise non-decreasing sequence of -measurable non-negative functions, i.e. each function is -measurable and for every {k\geq 1} and every {x\in X}, 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, 1. Note that the second chain of equalities follows from monoticity of the integral (lemma 2 below). Thus we can also write the conclusion of the theorem as with the tacit understanding that the limits are allowed to be infinite. Remark 3. The theorem remains true if its assumptions hold \mu-almost everywhere. In other words, it is enough that there is a null set N such that the sequence {f_n(x)} non-decreases for every To see why this is true, we start with an observation that allowing the sequence { f_n } to pointwise non-decrease almost everywhere causes its pointwise limit f to be undefined on some null set N. On that null set, f 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 {\mu(N)=0}, we have, for every k, provided that f is -measurable. (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.

Proof

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.

Intermediate results

We need three basic lemmas. In the proof below, we apply the monotonic property of the Lebesgue integral to non-negative functions only. Specifically (see Remark 4),

Monotonicity of the Lebesgue integral

lemma 1 . let the functions be -measurable. Proof. Denote by the set of simple -measurable functions such that everywhere on X. 1. Since f \leq g, we have hence 2. The functions where is the indicator function of X_i, are easily seen to be measurable and. Now apply 1.

Lebesgue integral as measure

Lemma 2. Let be a measurable space. Consider a simple -measurable non-negative function. For a measurable subset, define Then \nu_s is a measure on.

proof (lemma 2)

Write with and measurable sets. Then Since finite positive linear combinations of countably additive set functions are countably additive, to prove countable additivity of \nu_s it suffices to prove that, the set function defined by is countably additive for all. But this follows directly from the countable additivity of \mu.

Continuity from below

Lemma 3. Let \mu be a measure, and, where is a non-decreasing chain with all its sets \mu-measurable. Then

proof (lemma 3)

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

Proof of theorem

Set. Denote by the set of simple -measurable functions such that on X. Step 1. The function f is –measurable, and the integral is well-defined (albeit possibly infinite) From we get. Hence we have to show that f is -measurable. To see this, it suffices to prove that is \Sigma-measurable for all, because the intervals [0,t] generate the Borel sigma algebra on the extended non negative reals [0,\infty] by complementing and taking countable intersections, complements and countable unions. Now since the f_k(x) is a non decreasing sequence, if and only if for all k. Since we already know that f\ge 0 and f_k\ge 0 we conclude that Hence is a measurable set, being the countable intersection of the measurable sets. Since f \ge 0 the integral is well defined (but possibly infinite) as Step 2. We have the inequality This is equivalent to for all k which follows directly from and "monotonicity of the integral" (lemma 1). 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 k >K_s sufficiently large, but this does not work e.g. if f is itself simple and the f_k < f. However, we can get ourself an "epsilon of room" to manoeuvre and avoid this problem. Step 3 is also equivalent to for every simple function and every where for the equality we used that the left hand side of the inequality is a finite sum. This we will prove. Given and, define We claim the sets have the following properties: Assuming the claim, by the definition of and "monotonicity of the Lebesgue integral" (lemma 1) we have Hence by "Lebesgue integral of a simple function as measure" (lemma 2), and "continuity from below" (lemma 3) we get: which we set out to prove. Thus it remains to prove the claim. Ad 1: Write, for non-negative constants , and measurable sets , which we may assume are pairwise disjoint and with union. Then for x\in A_i we have if and only if so which is measurable since the f_k are measurable. Ad 2: For we have so Ad 3: Fix x \in X. If s(x) = 0 then, hence. Otherwise, s(x) > 0 and so for N_x sufficiently large, hence. The proof of the monotone convergence theorem is complete.

Relaxing the monotonicity assumption

Under similar hypotheses to Beppo Levi's theorem, it is possible to relax the hypothesis of monotonicity. 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 x, we define f to be the pointwise limit of, and we assume additionally that f_k \le f pointwise almost everywhere for all k. Then f is -measurable, and exists, and

Proof based on Fatou's lemma

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. It easily follows from the monotone convergence theorem and proof of Fatou's lemma is similar and arguably slightly less natural than the proof above. 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, since (monotonicity), Therefore