Monotone convergence theorem

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 $$a_1 \le a_2 \le a_3 \le ...\le K$$  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 $$0 \le a_{i,1} \le a_{i,2} \le \cdots $$, 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 $$0 \le f_1(x) \le f_2(x) \le \cdots$$, 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
Bounded above monotone non decreasing sequences are convergent in the real numbers because suprema exist there. Note that the proposition is wrong when considered in the rational numbers.

Proposition
(A) For a non-decreasing and bounded-above sequence of real numbers
 * $$a_1 \le a_2 \le a_3 \le...\le K < \infty,$$

the limit $$\lim_{n \to \infty} a_n$$ exists and equals its supremum:
 * $$\lim_{n \to \infty} a_n = \sup_n a_n \le K.$$

(B) For a non-increasing and bounded-below sequence of real numbers
 * $$a_1 \ge a_2 \ge a_3 \ge \cdots \ge L > -\infty,$$

the limit $$ \lim_{n \to \infty} a_n$$ exists and equals its infimum:
 * $$\lim_{n \to \infty} a_n = \inf_n a_n \ge L$$.

Proof
Let $$\{ a_n \}_{n\in\mathbb{N}}$$ be the set of values of $$ (a_n)_{n\in\mathbb{N}} $$. By assumption, $$\{ a_n \}$$ is non-empty and bounded above by $$K$$. By the least-upper-bound property of real numbers, $c = \sup_n \{a_n\}$ exists and $$ c \le K$$. Now, for every $$\varepsilon > 0$$, there exists $$N$$ such that $$c\ge a_N > c - \varepsilon $$, since otherwise $$c - \varepsilon $$ is a strictly smaller upper bound of $$\{ a_n \}$$, contradicting the definition of the supremum $$c$$. Then since $$(a_n)_{n\in\mathbb{N}}$$ is non decreasing, and $$c$$ is an upper bound, for every $$n > N$$, we have
 * $$|c - a_n| = c -a_n \leq c - a_N = |c -a_N|< \varepsilon. $$

Hence, by definition $$ \lim_{n \to \infty} a_n = c =\sup_n a_n$$.

The proof of the (B) part is analogous or follows from (A) by considering $$\{-a_n\}_{n \in \N}$$.

Theorem
If $$(a_n)_{n\in\mathbb{N}}$$ is a monotone sequence of real numbers, i.e., if $$a_n \le a_{n+1}$$ for every $$n \ge 1$$ or $$a_n \ge a_{n+1}$$ for every $$n \ge 1$$, then this sequence has a finite limit if and only if the sequence is bounded.

Proof

 * "If"-direction: The proof follows directly from the proposition.
 * "Only If"-direction: By (ε, δ)-definition of limit, every sequence $$(a_n)_{n\in\mathbb{N}}$$ with a finite limit $$L$$ is necessarily bounded.

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.
 * $$ \bar\R = \R \cup \{-\infty, \infty\}$$

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 $$ a_i \ge 0, i \in I $$ has a well defined summation order independent sum
 * $$ \sum_{i \in I} a_i = \sup_{J \subset I,\ |J|< \infty} \sum_{j \in J} a_j \in \bar \R_{\ge 0}$$

where $$\bar\R_{\ge 0} = [0, \infty] \subset \bar \R$$ are the upper extended non negative real numbers. For a series of non negative numbers
 * $$\sum_{i = 1}^\infty a_i = \lim_{k \to \infty} \sum_{i = 1}^k a_i = \sup_k \sum_{i =1}^k a_i = \sup_{J \subset \N, |J| < \infty} \sum_{j \in J} a_j = \sum_{i \in \N} a_i,$$

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.

Theorem (monotone convergence of non negative sums)
Let $$a_{i,k} \ge 0 $$ be a sequence of non-negative real numbers indexed by natural numbers $$i$$ and $$k$$. Suppose that $$a_{i,k} \le a_{i,k+1}$$ for all $$i, k$$. Then
 * $$\sup_k \sum_i a_{i,k} = \sum_i \sup_k a_{i,k} \in \bar\R_{\ge 0}.$$

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 $$a_{i,k} \le \sup_k a_{i,k}$$ we have $$\sum_i a_{i,k} \le \sum_i \sup_k a_{i,k}$$ so $$\sup_k \sum_i a_{i,k} \le \sum_i \sup_k a_{i,k} $$. Conversely, we can interchange sup and sum for finite sums so $$\sum_{i = 1}^N \sup_k a_{i,k} = \sup_k \sum_{i =1}^N a_{i,k} \le \sup_k \sum_{i =1}^\infty a_{i,k}$$ hence $$\sum_{i = 1}^\infty \sup_k a_{i,k} \le \sup_k \sum_{i =1}^\infty a_{i,k}$$.

The theorem states that if you have an infinite matrix of non-negative real numbers $$ a_{i,k} \ge 0$$ such that then
 * the rows are weakly increasing and each is bounded $$a_{i,k} \le K_i$$ where the bounds are summable $$\sum_i K_i <\infty$$
 * for each column, the non decreasing column sums $$\sum_i a_{i,k} \le \sum K_i $$ are bounded hence convergent, and the limit of the column sums is equal to the sum of the "limit column" $$ \sup_k a_{i,k}$$ which element wise is the supremum over the row.

As an example, consider the expansion
 * $$ \left( 1+ \frac1k\right)^k = \sum_{i=0}^k \binom ki \frac1{k^i}

$$ Now set
 * $$ a_{i,k} = \binom ki \frac1{k^i} = \frac1{i!} \cdot \frac kk \cdot \frac{k-1}k\cdot \cdots \frac{k-i+1}k.

$$ for $$ i \le k $$ and $$ a_{i,k} = 0$$ for $$ i > k $$, then $$0\le a_{i,k} \le a_{i,k+1}$$ with $$\sup_k a_{i,k} = \frac 1{i!}<\infty $$ and
 * $$\left( 1+ \frac1k\right)^k = \sum_{i =0}^\infty a_{i,k}$$.

The right hand side is a non decreasing sequence in $$k$$, therefore
 * $$ \lim_{k \to \infty}

\left( 1+ \frac1k\right)^k = \sup_k \sum_{i=0}^\infty a_{i,k} = \sum_{i = 0}^\infty \sup_k a_{i,k} = \sum_{i = 0}^\infty \frac1{i!} = e$$.

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 $$\operatorname{\mathcal B}_{\bar\R_{\geq 0}}$$ denotes the $$\sigma$$-algebra of Borel sets on the upper extended non negative real numbers $$[0,+\infty]$$. By definition, $$\operatorname{\mathcal B}_{\bar\R_{\geq 0}}$$ contains the set $$\{+\infty\}$$ and all Borel subsets of $$\R_{\geq 0}.$$

Theorem (monotone convergence theorem for non-negative measurable functions)
Let $$(\Omega,\Sigma,\mu)$$ be a measure space, and $$X\in\Sigma$$ a measurable set. Let $$\{f_k\}^\infty_{k=1}$$ be a pointwise non-decreasing sequence of $$(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})$$-measurable non-negative functions, i.e. each function $$f_k:X\to [0,+\infty]$$ is $$(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})$$-measurable and for every $${k\geq 1}$$ and every $${x\in X}$$,


 * $$ 0 \leq f_k(x) \leq f_{k+1}(x)\leq\infty. $$

Then the pointwise supremum
 * $$ \sup_k f_k : x \mapsto \sup_k f_k(x) $$

is a $$(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})$$-measurable function and


 * $$\sup_k \int_X f_k \,d\mu = \int_X \sup_k f_k \,d\mu.$$

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. $\textstyle \lim_{k \to \infty} f_k(x) = \sup_k f_k(x) = \limsup_{k \to \infty} f_k(x) = \liminf_{k \to \infty} f_k(x) $

2. $\textstyle \lim_{k \to \infty} \int_X f_k \,d\mu = \sup_k \int_X f_k \,d\mu = \liminf_{k \to \infty} \int_X f_k \,d\mu = \limsup_{k \to \infty} \int_X f_k \,d\mu. $

(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 $$\mu$$-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 $${x\in X\setminus N}.$$ 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 $$k,$$


 * $$ \int_X f_k \,d\mu = \int_{X \setminus N} f_k \,d\mu$$ and $$\int_X f \,d\mu = \int_{X \setminus N} f \,d\mu, $$

provided that $$f$$ is $$(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})$$-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 $$f,g : X \to [0,+\infty]$$ be $$(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})$$-measurable.


 * If $$f \leq g$$ everywhere on $$X,$$ then


 * $$\int_X f\,d\mu \leq \int_X g\,d\mu.$$


 * If $$ X_1,X_2 \in \Sigma $$ and $$X_1 \subseteq X_2, $$ then


 * $$\int_{X_1} f\,d\mu \leq \int_{X_2} f\,d\mu.$$

Proof. Denote by $$\operatorname{SF}(h)$$ the set of simple $$(\Sigma, \operatorname{\mathcal B}_{\R_{\geq 0}})$$-measurable functions $$s:X\to [0,\infty)$$ such that $$0\leq s\leq h$$ everywhere on $$X.$$

1. Since $$f \leq g,$$ we have $$ \operatorname{SF}(f) \subseteq \operatorname{SF}(g), $$ hence
 * $$\int_X f\,d\mu = \sup_{s\in {\rm SF}(f)}\int_X s\,d\mu \leq \sup_{s\in {\rm SF}(g)}\int_X s\,d\mu = \int_X g\,d\mu.$$

2. The functions $$f\cdot {\mathbf 1}_{X_1}, f\cdot {\mathbf 1}_{X_2},$$ where $${\mathbf 1}_{X_i}$$ is the indicator function of $$X_i$$, are easily seen to be measurable and $$f\cdot{\mathbf 1}_{X_1}\le f\cdot{\mathbf 1}_{X_2}$$. Now apply 1.

Lebesgue integral as measure
Lemma 2. Let $$(\Omega,\Sigma,\mu)$$ be a measurable space. Consider a simple $$(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})$$-measurable non-negative function $$s:\Omega\to{\mathbb R_{\geq 0}}$$. For a measurable subset $$S \in \Sigma$$, define
 * $$\nu_s(S)=\int_Ss\,d\mu.$$

Then $$\nu_s$$ is a measure on $$(\Omega, \Sigma)$$.

proof (lemma 2)
Write $$s=\sum^n_{k=1}c_k\cdot {\mathbf 1}_{A_k},$$ with $$c_k\in{\mathbb R}_{\geq 0}$$ and measurable sets $$A_k\in\Sigma$$. Then
 * $$\nu_s(S)=\sum_{k =1}^n c_k \mu(S\cap A_k).$$

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 $$\nu_A(S) = \mu(A \cap S)$$ is countably additive for all $$A \in \Sigma$$. But this follows directly from the countable additivity of $$\mu$$.

Continuity from below
Lemma 3. Let $$\mu$$ be a measure, and $$S = \bigcup^\infty_{i=1}S_i$$, where

S_1\subseteq\cdots\subseteq S_i\subseteq S_{i+1}\subseteq\cdots\subseteq S $$ is a non-decreasing chain with all its sets $$\mu$$-measurable. Then
 * $$\mu(S)=\sup_i\mu(S_i).$$

proof (lemma 3)
Set $$S_0 = \emptyset$$, then we decompose $$ S = \coprod_{1 \le i } S_i \setminus S_{i -1} $$ as a countable disjoint union of measurable sets and likewise $$S_k = \coprod_{1\le i \le k } S_i \setminus S_{i -1} $$ as a finite disjoint union. Therefore $$\mu(S_k) = \sum_{i=1}^k \mu (S_i \setminus S_{i -1})$$, and $$\mu(S) = \sum_{i = 1}^\infty \mu(S_i \setminus S_{i-1})$$ so $$\mu(S) = \sup_k \mu(S_k)$$.

Proof of theorem
Set $$ f = \sup_k f_k$$. Denote by $$\operatorname{SF}(f)$$ the set of simple $$(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})$$-measurable functions $$s:X\to [0,\infty)$$ ($$\infty$$ nor included!) such that $$0\leq s\leq f$$ on $$X$$.

Step 1. The function $$f$$ is $$ (\Sigma, \operatorname{\mathcal B}_{\bar\R_{\geq 0}}) $$–measurable, and the integral $$\textstyle \int_X f \,d\mu $$ is well-defined (albeit possibly infinite)

From $$0 \le f_k(x) \le \infty$$ we get $$0 \le f(x) \le \infty$$. Hence we have to show that $$f$$ is $$(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})$$-measurable. To see this, it suffices to prove that $$f^{-1}([0,t])$$ is $$\Sigma $$-measurable for all $$0 \le t \le \infty$$, 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, $$ f(x) = \sup_k f_k(x) \leq t$$ if and only if $$f_k(x)\leq t$$ for all $$k$$. Since we already know that $$f\ge 0$$ and $$f_k\ge 0$$ we conclude that
 * $$f^{-1}([0, t]) = \bigcap_k f_k^{-1}([0,t]).$$

Hence $$f^{-1}([0, t])$$ is a measurable set, being the countable intersection of the measurable sets $$f_k^{-1}([0,t])$$.

Since $$f \ge 0$$ the integral is well defined (but possibly infinite) as
 * $$ \int_X f \,d\mu = \sup_{s \in SF(f)}\int_X s \, d\mu$$.

Step 2. We have the inequality


 * $$\sup_k \int_X f_k \,d\mu \le \int_X f \,d\mu $$

This is equivalent to $$\int_X f_k(x) \, d\mu \le \int_X f(x)\, d\mu$$ for all $$k$$ which follows directly from $$f_k(x) \le f(x)$$ and "monotonicity of the integral" (lemma 1).

step 3 We have the reverse inequality


 * $$ \int_X f \,d\mu \le \sup_k \int_X f_k \,d\mu $$.

By the definition of integral as a supremum step 3 is equivalent to
 * $$ \int_X s\,d\mu\leq\sup_k\int_X f_k\,d\mu$$

for every $$s\in\operatorname{SF}(f)$$. It is tempting to prove $$ \int_X s\,d\mu\leq \int_X f_k \,d\mu$$ 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

(1-\varepsilon) \int_X s \, d\mu = \int_X (1 - \varepsilon) s \, d\mu \le \sup_k \int_X f_k \, d\mu $$ for every simple function $$s\in\operatorname{SF}(f)$$ and every $$0 <\varepsilon \ll 1$$ where for the equality we used that the left hand side of the inequality is a finite sum. This we will prove.

Given $$s\in\operatorname{SF}(f)$$ and $$0 <\varepsilon \ll 1$$, define
 * $$B^{s,\varepsilon}_k=\{x\in X\mid (1 - \varepsilon) s(x)\leq f_k(x)\}\subseteq X.$$

We claim the sets $$B^{s,\varepsilon}_k$$ have the following properties:
 * 1) $$B^{s,\varepsilon}_k$$ is $$\Sigma$$-measurable.
 * 2) $$B^{s,\varepsilon}_k\subseteq B^{s,\varepsilon}_{k+1}$$
 * 3) $$ X=\bigcup_k B^{s,\varepsilon}_k$$

Assuming the claim, by the definition of $$B^{s,\varepsilon}_k$$ and "monotonicity of the Lebesgue integral" (lemma 1) we have
 * $$\int_{B^{s,\varepsilon}_k}(1-\varepsilon) s\,d\mu\leq\int_{B^{s,\varepsilon}_k} f_k\,d\mu \leq\int_X f_k\,d\mu.

$$ Hence by "Lebesgue integral of a simple function as measure" (lemma 2), and "continuity from below" (lemma 3) we get:
 * $$ \sup_k \int_{B^{s,\varepsilon}_k} (1-\varepsilon)s\,d\mu = \int_X (1- \varepsilon)s \, d\mu

\le \sup_k \int_X f_k \, d\mu. $$ which we set out to prove. Thus it remains to prove the claim. Ad 1: Write $$s=\sum_{1 \le i \le m}c_i\cdot{\mathbf 1}_{A_i}$$, for  non-negative constants $$c_i \in \R_{\geq 0}$$, and measurable sets $$A_i\in\Sigma$$, which we may assume are pairwise disjoint and with union $$\textstyle X=\coprod^m_{i=1}A_i$$. Then for $$ x\in A_i$$ we have $$(1-\varepsilon)s(x)\leq f_k(x)$$ if and only if $$ f_k(x) \in [( 1- \varepsilon)c_i, \,\infty],$$ so
 * $$B^{s,\varepsilon}_k=\coprod^m_{i=1}\Bigl(f^{-1}_k\Bigl([(1-\varepsilon)c_i,\infty]\Bigr)\cap A_i\Bigr)$$

which is measurable since the $$f_k$$ are measurable.

Ad 2: For $$ x \in B^{s,\varepsilon}_k$$ we have $$(1 - \varepsilon)s(x) \le f_k(x)\le f_{k+1}(x)$$ so $$x \in B^{s,\varepsilon}_{k + 1}.$$

Ad 3: Fix $$x \in X$$. If $$s(x) = 0$$ then $$(1 - \varepsilon)s(x) = 0 \le f_1(x)$$, hence $$x \in B^{s,\varepsilon}_1$$. Otherwise, $$s(x) > 0$$ and $$(1-\varepsilon)s(x) < s(x) \le f(x) = \sup_k f(x)$$ so $$(1- \varepsilon)s(x) < f_{N_x}(x)$$ for $$N_x$$ sufficiently large, hence $$x \in B^{s,\varepsilon}_{N_x}$$.

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 $$(\Omega, \Sigma, \mu)$$ be a measure space and $$X \in \Sigma$$. Again, $$\{f_k\}_{k=1}^\infty$$ will be a sequence of $$(\Sigma, \mathcal{B}_{\R_{\geq 0}})$$-measurable non-negative functions $$f_k:X\to [0,+\infty]$$. However, we do not assume they are pointwise non-decreasing. Instead, we assume that $\{f_k(x)\}_{k=1}^\infty$ converges for almost every $$x$$, we define $$f$$ to be the pointwise limit of $$\{f_k\}_{k=1}^\infty$$, and we assume additionally that $$f_k \le f$$ pointwise almost everywhere for all $$k$$. Then $$f$$ is $$(\Sigma, \mathcal{B}_{\R_{\geq 0}})$$-measurable, and $\lim_{k\to\infty} \int_X f_k \,d\mu$ exists, and $$\lim_{k\to\infty} \int_X f_k \,d\mu = \int_X f \,d\mu.$$

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 $f = \sup_k f_k = \lim_{k \to \infty} f_k = \liminf_{k \to \infty}f_k$  almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has $$\int_X f\,d\mu = \int_X \liminf_k f_k\,d\mu \le \liminf \int_X f_k\,d\mu$$ by Fatou's lemma, and then, by standard properties of limits and $$\int f_k \,d\mu \le \int f d\mu$$ (monotonicity), $$\liminf \int_X f_k\,d\mu \le \limsup \int_X f_k\,d\mu \le \int_X f\,d\mu.$$ Therefore $$\int_X f \, d\mu = \liminf_{k \to\infty} \int_X f_k\,d\mu = \limsup_{k \to\infty} \int_X f_k\,d\mu = \lim_{k \to\infty} \int_X f_k \, d\mu = \sup_k \int_X f_k\,d\mu.$$