# Monotone convergence theorem

Short description: 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 non-decreasing or non-increasing) 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 $\displaystyle{ (a_n)_{n\in\mathbb{N}} }$ be such a sequence, and let $\displaystyle{ \{ a_n \} }$ be the set of terms of $\displaystyle{ (a_n)_{n\in\mathbb{N}} }$. By assumption, $\displaystyle{ \{ a_n \} }$ is non-empty and bounded above. By the least-upper-bound property of real numbers, $\displaystyle{ c = \sup_n \{a_n\} }$ exists and is finite. Now, for every $\displaystyle{ \varepsilon \gt 0 }$, there exists $\displaystyle{ N }$ such that $\displaystyle{ a_N \gt c - \varepsilon }$, since otherwise $\displaystyle{ c - \varepsilon }$ is an upper bound of $\displaystyle{ \{ a_n \} }$, which contradicts the definition of $\displaystyle{ c }$. Then since $\displaystyle{ (a_n)_{n\in\mathbb{N}} }$ is increasing, and $\displaystyle{ c }$ is its upper bound, for every $\displaystyle{ n \gt N }$, we have $\displaystyle{ |c - a_n| \leq |c - a_N| \lt \varepsilon }$. Hence, by definition, the limit of $\displaystyle{ (a_n)_{n\in\mathbb{N}} }$ is $\displaystyle{ \sup_n \{a_n\}. }$

### 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 $\displaystyle{ (a_n)_{n\in\mathbb{N}} }$ is a monotone sequence of real numbers (i.e., if an ≤ an+1 for every n ≥ 1 or an ≥ an+1 for every n ≥ 1), then this sequence has a limit if and only if the sequence is bounded.

### Proof

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

## Convergence of a monotone series

### Theorem

If for all natural numbers j and k, aj,k is a non-negative real number and aj,k ≤ aj+1,k, then:168

$\displaystyle{ \lim_{j\to\infty} \sum_k a_{j,k} = \sum_k \lim_{j\to\infty} a_{j,k}. }$

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 (which is also its supremum). The series has a convergent sum if and only if the (weakly increasing) sequence of row sums is bounded and therefore convergent.

As an example, consider the infinite series of rows

$\displaystyle{ \left( 1+ \frac1 n\right)^n = \sum_{k=0}^n \binom nk \frac 1 {n^k} = \sum_{k=0}^n \frac1{k!} \times \frac nn \times \frac{n-1}n\times\cdots\times\frac{n-k+1}n, }$

where n approaches infinity (the limit of this series is e). Here the matrix entry in row n and column k is

$\displaystyle{ \binom nk \frac 1 {n^k} =\frac1{k!}\times\frac nn\times\frac{n-1}n\times\cdots\times\frac{n-k+1}n; }$

the columns (fixed k) are indeed weakly increasing with n and bounded (by 1/k!), 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 $\displaystyle{ (1+1/n)^n }$ by taking the sum of the column limits, namely $\displaystyle{ \frac1{k!} }$.

## Beppo Levi's lemma

The following result is due to Beppo Levi, who proved a slight generalization in 1906 of an earlier result by Henri Lebesgue. In what follows, $\displaystyle{ \operatorname{\mathcal B}_{\R_{\geq 0}} }$ denotes the $\displaystyle{ \sigma }$-algebra of Borel sets on $\displaystyle{ [0,+\infty] }$. By definition, $\displaystyle{ \operatorname{\mathcal B}_{\R_{\geq 0}} }$ contains the set $\displaystyle{ \{+\infty\} }$ and all Borel subsets of $\displaystyle{ \R_{\geq 0}. }$

### Theorem

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

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

Set the pointwise limit of the sequence $\displaystyle{ \{f_{n}\} }$ to be $\displaystyle{ f }$. That is, for every $\displaystyle{ x\in X }$,

$\displaystyle{ f(x):= \lim_{k\to\infty} f_k(x). }$

Then $\displaystyle{ f }$ is $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable and

$\displaystyle{ \lim_{k\to\infty} \int_X f_k \,d\mu = \int_X f \,d\mu. }$

Remark 1. The integrals may be finite or infinite.

Remark 2. The theorem remains true if its assumptions hold $\displaystyle{ \mu }$-almost everywhere. In other words, it is enough that there is a null set $\displaystyle{ N }$ such that the sequence $\displaystyle{ \{f_n(x)\} }$ non-decreases for every $\displaystyle{ {x\in X\setminus N}. }$ To see why this is true, we start with an observation that allowing the sequence $\displaystyle{ \{ f_n \} }$ to pointwise non-decrease almost everywhere causes its pointwise limit $\displaystyle{ f }$ to be undefined on some null set $\displaystyle{ N }$. On that null set, $\displaystyle{ f }$ 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 $\displaystyle{ {\mu(N)=0}, }$ we have, for every $\displaystyle{ k, }$

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

provided that $\displaystyle{ f }$ is $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable.(section 21.38) (These equalities follow directly from the definition of Lebesgue integral for a non-negative function).

Remark 3. Under assumptions of the theorem,

1. $\displaystyle{ \textstyle f(x) = \liminf_k f_k(x) = \limsup_k f_k(x) = \sup_k f_k(x) }$
2. $\displaystyle{ \textstyle \liminf_k \int_X f_k \,d\mu = \textstyle \limsup_k \int_X f_k \,d\mu = \lim_k \int_X f_k \,d\mu = \sup_k \int_X f_k \,d\mu }$

(Note that the second chain of equalities follows from Remark 5).

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 (monotonicity of Lebesgue integral). In the proof below, we apply the monotonic property of Lebesgue integral to non-negative functions only. Specifically (see Remark 4), let the functions $\displaystyle{ f,g : X \to [0,+\infty] }$ be $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable.

• If $\displaystyle{ f \leq g }$ everywhere on $\displaystyle{ X, }$ then
$\displaystyle{ \int_X f\,d\mu \leq \int_X g\,d\mu. }$
• If $\displaystyle{ X_1,X_2 \in \Sigma }$ and $\displaystyle{ X_1 \subseteq X_2, }$ then
$\displaystyle{ \int_{X_1} f\,d\mu \leq \int_{X_2} f\,d\mu. }$

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

1. Since $\displaystyle{ f \leq g, }$ we have

$\displaystyle{ \operatorname{SF}(f) \subseteq \operatorname{SF}(g). }$

By definition of Lebesgue integral and the properties of supremum,

$\displaystyle{ \int_X f\,d\mu = \sup_{s\in {\rm SF}(f)}\int_X s\,d\mu \leq \sup_{s\in {\rm SF}(g)}\int_X s\,d\mu = \int_X g\,d\mu. }$

2. Let $\displaystyle{ {\mathbf 1}_{X_1} }$ be the indicator function of the set $\displaystyle{ X_1. }$ It can be deduced from the definition of Lebesgue integral that

$\displaystyle{ \int_{X_2} f\cdot {\mathbf 1}_{X_1} \,d\mu = \int_{X_1} f \,d\mu }$

if we notice that, for every $\displaystyle{ s \in {\rm SF}(f\cdot {\mathbf 1}_{X_1}), }$ $\displaystyle{ s=0 }$ outside of $\displaystyle{ X_1. }$ Combined with the previous property, the inequality $\displaystyle{ f\cdot {\mathbf 1}_{X_1} \leq f }$ implies

$\displaystyle{ \int_{X_1} f \,d\mu = \int_{X_2} f\cdot {\mathbf 1}_{X_1} \,d\mu \leq \int_{X_2} f \,d\mu. }$

### 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 $\displaystyle{ (\Omega,\Sigma,\mu) }$ be a measurable space. Consider a simple $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable non-negative function $\displaystyle{ s:\Omega\to{\mathbb R_{\geq 0}} }$. For a subset $\displaystyle{ S\subseteq\Omega }$, define

$\displaystyle{ \nu(S)=\int_Ss\,d\mu. }$

Then $\displaystyle{ \nu }$ is a measure on $\displaystyle{ \Omega }$.

###### Proof

Monotonicity follows from Remark 5. Here, we will only prove countable additivity, leaving the rest up to the reader. Let $\displaystyle{ S=\bigcup^\infty_{i=1}S_i }$, where all the sets $\displaystyle{ S_i }$ are pairwise disjoint. Due to simplicity,

$\displaystyle{ s=\sum^n_{i=1}c_i\cdot {\mathbf 1}_{A_i}, }$

for some finite non-negative constants $\displaystyle{ c_i\in{\mathbb R}_{\geq 0} }$ and pairwise disjoint sets $\displaystyle{ A_i\in\Sigma }$ such that $\displaystyle{ \bigcup^n_{i=1}A_i=\Omega }$. By definition of Lebesgue integral,

\displaystyle{ \begin{align} \nu(S) &=\sum^n_{i=1}c_i\cdot\mu(S\cap A_i)\\ &=\sum^n_{i=1}c_i\cdot\mu\left(\left(\bigcup^\infty_{j=1} S_j\right)\cap A_i\right)\\ &=\sum^n_{i=1}c_i\cdot\mu\left(\bigcup^\infty_{j=1}(S_j\cap A_i)\right) \end{align} }

Since all the sets $\displaystyle{ S_j\cap A_i }$ are pairwise disjoint, the countable additivity of $\displaystyle{ \mu }$ gives us

$\displaystyle{ \sum^n_{i=1} c_i\cdot\mu \left(\bigcup^\infty_{j=1}(S_j\cap A_i)\right)=\sum^n_{i=1}c_i\cdot\sum^\infty_{j=1} \mu(S_j\cap A_i). }$

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,

\displaystyle{ \begin{align} \sum^n_{i=1}c_i\cdot\sum^\infty_{j=1}\mu(S_j\cap A_i)&=\sum^\infty_{j=1}\sum^n_{i=1}c_i\cdot \mu(S_j\cap A_i)\\ &=\sum^\infty_{j=1}\int_{S_j} s\,d\mu\\ &=\sum^\infty_{j=1}\nu(S_j), \end{align} }

as required.

##### "Continuity from below"

The following property is a direct consequence of the definition of measure.

Lemma 2. Let $\displaystyle{ \mu }$ be a measure, and $\displaystyle{ S = \bigcup^\infty_{i=1}S_i }$, where

$\displaystyle{ S_1\subseteq\cdots\subseteq S_i\subseteq S_{i+1}\subseteq\cdots\subseteq S }$

is a non-decreasing chain with all its sets $\displaystyle{ \mu }$-measurable. Then

$\displaystyle{ \mu(S)=\lim_i\mu(S_i). }$

#### Proof of theorem

Step 1. We begin by showing that $\displaystyle{ f }$ is $\displaystyle{ (\Sigma, \operatorname{\mathcal B}_{\R_{\geq 0}}) }$–measurable.(section 21.3)

Note. If we were using Fatou's lemma, the measurability would follow easily from Remark 3(a).

To do this without using Fatou's lemma, it is sufficient to show that the inverse image of an interval $\displaystyle{ [0,t] }$ under $\displaystyle{ f }$ is an element of the sigma-algebra $\displaystyle{ \Sigma }$ on $\displaystyle{ X }$, because (closed) intervals generate the Borel sigma algebra on the reals. Since $\displaystyle{ [0,t] }$ is a closed interval, and, for every $\displaystyle{ k }$, $\displaystyle{ 0\le f_k(x) \le f(x) }$,

$\displaystyle{ 0\leq f(x)\leq t\quad \Leftrightarrow\quad \Bigl[\forall k\quad 0\leq f_k(x)\leq t\Bigr]. }$

Thus,

$\displaystyle{ \{x\in X \mid 0\leq f(x)\leq t\} = \bigcap_k \{x\in X \mid 0\leq f_k(x)\leq t\}. }$

Being the inverse image of a Borel set under a $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable function $\displaystyle{ f_k }$, each set in the countable intersection is an element of $\displaystyle{ \Sigma }$. Since $\displaystyle{ \sigma }$-algebras are, by definition, closed under countable intersections, this shows that $\displaystyle{ f }$ is $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable, and the integral $\displaystyle{ \textstyle \int_X f \,d\mu }$ is well defined (and possibly infinite).

Step 2. We will first show that $\displaystyle{ \textstyle\int_X f \,d\mu \geq \lim_k \int_X f_k \,d\mu. }$

The definition of $\displaystyle{ f }$ and monotonicity of $\displaystyle{ \{f_k\} }$ imply that $\displaystyle{ f(x)\geq f_k(x) }$, for every $\displaystyle{ k }$ and every $\displaystyle{ x\in X }$. By monotonicity (or, more precisely, its narrower version established in Remark 5; see also Remark 4) of Lebesgue integral,

$\displaystyle{ \int_X f\,d\mu\geq\int_X f_k\,d\mu, }$

and

$\displaystyle{ \int_X f\,d\mu\geq\lim_k\int_X f_k\,d\mu. }$

Note that the limit on the right exists (finite or infinite) because, due to monotonicity (see Remark 5 and Remark 4), the sequence is non-decreasing.

End of Step 2.

We now prove the reverse inequality. We seek to show that

$\displaystyle{ \int_X f \,d\mu \leq \lim_k \int_X f_k \,d\mu }$.

Proof using Fatou's lemma. Per Remark 3, the inequality we want to prove is equivalent to

$\displaystyle{ \int_X \liminf_k f_k(x) \,d\mu \leq \liminf_k \int_X f_k \,d\mu. }$

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

Step 3. Given a simple function $\displaystyle{ s\in\operatorname{SF}(f) }$ and a real number $\displaystyle{ t\in (0,1) }$, define

$\displaystyle{ B^{s,t}_k=\{x\in X\mid t\cdot s(x)\leq f_k(x)\}\subseteq X. }$

Then $\displaystyle{ B^{s,t}_k\in\Sigma }$, $\displaystyle{ B^{s,t}_k\subseteq B^{s,t}_{k+1} }$, and $\displaystyle{ \textstyle X=\bigcup_k B^{s,t}_k }$.

Step 3a. To prove the first claim, let $\displaystyle{ \textstyle s=\sum^m_{i=1}c_i\cdot{\mathbf 1}_{A_i} }$, for some finite collection of pairwise disjoint measurable sets $\displaystyle{ A_i\in\Sigma }$ such that $\displaystyle{ \textstyle X=\cup^m_{i=1}A_i }$, some (finite) non-negative constants $\displaystyle{ c_i\in {\mathbb R}_{\geq 0} }$, and $\displaystyle{ {\mathbf 1}_{A_i} }$ denoting the indicator function of the set $\displaystyle{ A_i }$.

For every $\displaystyle{ x\in A_i, }$ $\displaystyle{ t\cdot s(x)\leq f_k(x) }$ holds if and only if $\displaystyle{ f_k(x) \in [t\cdot c_i, +\infty]. }$ Given that the sets $\displaystyle{ A_i }$ are pairwise disjoint,

$\displaystyle{ B^{s,t}_k=\bigcup^m_{i=1}\Bigl(f^{-1}_k\Bigl([t\cdot c_i,+\infty]\Bigr)\cap A_i\Bigr). }$

Since the pre-image $\displaystyle{ f^{-1}_k\Bigl([t\cdot c_i,+\infty]\Bigr) }$ of the Borel set $\displaystyle{ [t\cdot c_i,+\infty] }$ under the measurable function $\displaystyle{ f_k }$ is measurable, and $\displaystyle{ \sigma }$-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 $\displaystyle{ k }$ and every $\displaystyle{ x\in X }$, $\displaystyle{ f_k(x)\leq f_{k+1}(x). }$

Step 3c. To prove the third claim, we show that $\displaystyle{ \textstyle X\subseteq\bigcup_k B^{s,t}_k }$.

Indeed, if, to the contrary, $\displaystyle{ \textstyle X\not\subseteq\bigcup_k B^{s,t}_k }$, then an element

$\displaystyle{ \textstyle x_0\in X\setminus\bigcup_k B^{s,t}_k=\bigcap_k(X\setminus B^{s,t}_k) }$

exists such that $\displaystyle{ f_k(x_0)\lt t\cdot s(x_0) }$, for every $\displaystyle{ k }$. Taking the limit as $\displaystyle{ k\to\infty }$, we get

$\displaystyle{ f(x_0)\leq t\cdot s(x_0)\lt s(x_0). }$

But by initial assumption, $\displaystyle{ s\leq f }$. This is a contradiction.

Step 4. For every simple $\displaystyle{ (\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}}) }$-measurable non-negative function $\displaystyle{ s_2 }$,

$\displaystyle{ \lim_n\int_{B^{s,t}_n}s_2\,d\mu=\int_Xs_2\,d\mu. }$

To prove this, define $\displaystyle{ \textstyle\nu(S)=\int_S s_2\,d\mu }$. By Lemma 1, $\displaystyle{ \nu(S) }$ is a measure on $\displaystyle{ \Omega }$. By "continuity from below" (Lemma 2),

$\displaystyle{ \lim_n\int_{B^{s,t}_n}s_2\,d\mu=\lim_n\nu(B^{s,t}_n)=\nu(X)=\int_Xs_2\,d\mu, }$

as required.

Step 5. We now prove that, for every $\displaystyle{ s\in\operatorname{SF}(f) }$,

$\displaystyle{ \int_X s\,d\mu\leq\lim_k\int_X f_k\,d\mu. }$

Indeed, using the definition of $\displaystyle{ B^{s,t}_k }$, the non-negativity of $\displaystyle{ f_k }$, and the monotonicity of Lebesgue integral (see Remark 5 and Remark 4), we have

$\displaystyle{ \int_{B^{s,t}_k}t\cdot s\,d\mu\leq\int_{B^{s,t}_k} f_k\,d\mu\leq\int_X f_k\,d\mu, }$

for every $\displaystyle{ k\geq 1 }$. In accordance with Step 4, as $\displaystyle{ k\to\infty }$, the inequality becomes

$\displaystyle{ t\int_X s\,d\mu\leq\lim_k\int_X f_k\,d\mu. }$

Taking the limit as $\displaystyle{ t\uparrow 1 }$ yields

$\displaystyle{ \int_X s\,d\mu\leq\lim_k\int_X f_k\,d\mu, }$

as required.

Step 6. We are now able to prove the reverse inequality, i.e.

$\displaystyle{ \int_X f \,d\mu \leq \lim_k \int_X f_k \,d\mu. }$

Indeed, by non-negativity, $\displaystyle{ f_+ = f }$ and $\displaystyle{ f_- = 0. }$ For the calculation below, the non-negativity of $\displaystyle{ f }$ is essential. Applying the definition of Lebesgue integral and the inequality established in Step 5, we have

$\displaystyle{ \int_X f \,d\mu=\sup_{s\in\operatorname{SF}(f)}\int_X s\,d\mu\leq\lim_k\int_X f_k\,d\mu. }$

The proof is complete.