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

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 $(a_{n})_{n\in \mathbb {N} }$ is a monotone sequence of real numbers (i.e., if a_n ≤ a_n+1 for every n ≥ 1 or a_n ≥ a_n+1 for every n ≥ 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 lemmas.
• "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

Theorem

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

$\lim _{j\to \infty }\sum _{k}a_{j,k}=\sum _{k}\lim _{j\to \infty }a_{j,k}.$

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

$\left(1+{\frac {1}{n}}\right)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {1}{n^{k}}}=\sum _{k=0}^{n}{\frac {1}{k!}}\times {\frac {n}{n}}\times {\frac {n-1}{n}}\times \cdots \times {\frac {n-k+1}{n}},$

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

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

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 $(1+1/n)^{n}$ by taking the sum of the column limits, namely ${\frac {1}{k!}}$.

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, $\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}}$ denotes the $\sigma$-algebra of Borel sets on $[0,+\infty ]$. By definition, $\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}}$ contains the set $\{+\infty \}$ and all Borel subsets of $\mathbb {R} _{\geq 0}.$

Theorem

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

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

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

$f(x):=\sup _{k\to \infty }f_{k}(x).$

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

$\sup _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}f\,d\mu .$

Remark 1. The integrals and the supremum may be finite or infinite.

Remark 2. Under the assumptions of the theorem,

1. $\textstyle f(x)=\sup _{k}f_{k}(x)=\limsup _{k}f_{k}(x)=\liminf _{k}f_{k}(x)=\lim _{k}f_{k}(x)$
2. $\textstyle \sup _{k}\int _{X}f_{k}\,d\mu =\liminf _{k}\int _{X}f_{k}\,d\mu =\textstyle \limsup _{k}\int _{X}f_{k}\,d\mu =\lim _{k}\int _{X}f_{k}\,d\mu$

(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 $(\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})$-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 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 $f,g:X\to [0,+\infty ]$ be $(\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})$-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 $(\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})$-measurable functions $s:X\to [0,\infty )$ such that $0\leq s\leq h$ everywhere on $X.$

1. Since $f\leq g,$ we have

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

By definition of the Lebesgue integral and the properties of supremum,

$\int _{X}f\,d\mu =\sup _{s\in {\rm {SF}}(f)}\int _{X}s\,d\mu \leq \sup _{s\in {\rm {SF}}(g)}\int _{X}s\,d\mu =\int _{X}g\,d\mu .$

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

$\int _{X_{2}}f\cdot {\mathbf {1} }_{X_{1}}\,d\mu =\int _{X_{1}}f\,d\mu$

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

$\int _{X_{1}}f\,d\mu =\int _{X_{2}}f\cdot {\mathbf {1} }_{X_{1}}\,d\mu \leq \int _{X_{2}}f\,d\mu .$

Lebesgue integral as measure

Lemma 1. Let $(\Omega ,\Sigma ,\mu )$ be a measurable space. Consider a simple $(\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})$-measurable non-negative function $s:\Omega \to {\mathbb {R} _{\geq 0}}$. For a subset $S\subseteq \Omega$, define

$\nu (S)=\int _{S}s\,d\mu .$

Then $\nu$ is a measure on $\Omega$.

Monotonicity follows from lemma 1. Here, we will only prove countable additivity, leaving the rest up to the reader. Let $S=\coprod _{i=1}^{\infty }S_{i}$, be a decomposition of $S$ as a countable pairwise disjoint union of measurable subsets $S_{i}$. Write

$s=\sum _{k=1}^{n}c_{k}\cdot {\mathbf {1} }_{A_{k}},$

for a finite number $n$ of non-negative constants $c_{k}\in {\mathbb {R} }_{\geq 0}$ and measurable sets $A_{k}\in \Sigma$. By definition of the Lebesgue integral

${\begin{aligned}\nu (S)&=\sum _{k=1}^{n}c_{k}\cdot \mu (S\cap A_{k})\\&=\sum _{k=1}^{n}c_{k}\cdot \mu \left(\left(\coprod _{i=1}^{\infty }S_{i}\right)\cap A_{k}\right)\\&=\sum _{k=1}^{n}c_{k}\cdot \mu \left(\coprod _{i=1}^{\infty }(S_{i}\cap A_{k})\right)\\&=\sum _{k=1}^{n}c_{k}\cdot \sum _{i=1}^{\infty }\mu (S_{i}\cap A_{k})\\&=\sum _{i=1}^{\infty }\sum _{k=1}^{n}c_{k}\cdot \mu (S_{i}\cap A_{k})\\&=\sum _{i=1}^{\infty }\nu (S_{i})\end{aligned}}$

as required. Here we used that for fixed $k$, all the sets $S_{i}\cap A_{k}$ are pairwise disjoint, the countable additivity of $\mu$ and the fact that since all the summands are non-negative, the sum of the series, be it finite or infinite, is independent of the order of the summation and so summations can be interchanged.

"Continuity from below"

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

Lemma 2. Let $\mu$ be a measure, and $S=\bigcup _{i=1}^{\infty }S_{i}$, 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 of theorem

The proof can be based on Fatou's lemma instead of a direct proof, because Fatou's lemma can be proved independent of the monotone convergence theorem. The measurability follows directly and the crucial "inverse Inequality" (step 3) is a direct consequence of the inequality

$\int _{X}\liminf _{k}f_{k}(x)\,d\mu \leq \liminf _{k}\int _{X}f_{k}\,d\mu .$

However a the monotone convergence theorem is some ways more primitive than Fatou's lemma and the following is a direct proof.

Denote by $\operatorname {SF} (f)$ the set of simple $(\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})$-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 $(\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})$–measurable, and the integral $\textstyle \int _{X}f\,d\mu$ is well-defined (albeit possibly infinite)

Since from $0\leq f_{k}\leq \infty$ we get $0\leq f(x)\leq \infty$, so it suffices to show that $f$ is measurable. To see this it suffices to prove that the inverse image of an interval $[0,t]$ under $f$ is an element of the sigma-algebra $\Sigma$ on $X$. This is because intervals $[0,t]$ with $0\leq t\leq \infty$ generate the Borel sigma algebra on the extended non negative reals $[0,\infty ]$ by countable intersection and union. Since the $f_{k}(x)$ is a non decreasing sequence

$f(x)=\sup _{k}f_{k}(x)\leq t\quad \Leftrightarrow \quad f_{k}(x)\leq t\ \ \forall k.$

Therefore since $f\geq 0$ and $f_{k}\geq 0$

$f^{-1}([0,t])=\bigcap _{k}f_{k}^{-1}([0,t]).$

Hence $f^{-1}([0,t])$ is a measurable set, being the countable intersection of measurable sets, each the inverse image of a Borel set under a $(\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})$-measurable function $f_{k}$. This shows that $f$ is $(\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})$-measurable.

Since $f\geq 0$ the integral is defined as a sup

$\int _{X}f\,d\mu =\sup {s\in SF(f)}s(x)\,d\mu$

over simple functions and is thus well-defined but the sup may possibly be infinite.

Step 2. We have the inequality

$\sup _{k}\int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu$

This follows directly from the monotonicity of the integral and the definition of supremum: since $f_{k}(x)\leq f(x)$ we have $\int _{X}f_{k}\,d\mu .\leq \int _{X}f\,d\mu$, hence $\sup _{k}\int _{X}f_{k}\,d\mu .\leq \int _{X}f\,d\mu$.

step 3 We have the reverse inequality

$\sup _{k}\int _{X}f_{k}\,d\mu \geq \int _{X}f\,d\mu$.

We need some extra machinery.

Given a simple function $s\in \operatorname {SF} (f)$ and a positive real number $0<t<1$, define

$B_{k}^{s,t}=\{x\in X\mid t\cdot s(x)\leq f_{k}(x)\}\subseteq X.$

Then $B_{k}^{s,t}\subseteq B_{k+1}^{s,t}$ because the sequence $f_{k}(x)$ is non decreasing. Moreover $\textstyle X=\bigcup _{k}B_{k}^{s,t}$ because either $f(x)=\sup _{k}f_{k}(x)=0$ and $s(x)=f_{k}(x)=0$ for all $k$, or $ts(x)<f(x)$ and $ts(x)\leq f_{k}(x)$ for $k\gg 0$.

step 3(a). The set $B_{k}^{s,t}$ is measurable.

To see this from first principles, write $s=\sum _{i=1}^{m}c_{i}\cdot {\mathbf {1} }_{A_{i}}$, for some finite collection of non-negative constants $c_{i}\in {\mathbb {R} }_{\geq 0}$, and measurable sets $A_{i}\in \Sigma$ which we may assume are pairwise disjoint and with union $\textstyle X=\coprod _{i=1}^{m}A_{i}$ (here ${\mathbf {1} }_{A_{i}}$ denotes the indicator function of the set $A_{i}$). Then if $x\in A_{i}$ we have $t\cdot s(x)\leq f_{k}(x)$ if and only if $f_{k}(x)\in [t\cdot c_{i},+\infty ].$. Therefore

$B_{k}^{s,t}=\coprod _{i=1}^{m}{\Bigl (}f_{k}^{-1}{\Bigl (}[t\cdot c_{i},+\infty ]{\Bigr )}\cap A_{i}{\Bigr )}.$

which since the $f_{k}$ are measurable, is a union of (disjoint) measurable sets

Step 3(b). For every simple $(\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})$-measurable non-negative function $s_{2}$,

$\sup _{k}\int _{B_{k}^{s,t}}s_{2}\,d\mu =\int _{X}s_{2}\,d\mu .$

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

$\sup _{k}\int _{B_{k}^{s,t}}s_{2}\,d\mu =\sup _{k}\nu (B_{n}^{s,t})=\nu (\cup _{k}B_{k}^{s,t})=\nu (X)=\int _{X}s_{2}\,d\mu ,$

as required.

Step 3(c). We now prove that, for every $s\in \operatorname {SF} (f)$,

$\int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .$

Indeed, using the definition of $B_{k}^{s,t}$, the monotonicity of the Lebesgue integral and the non-negativity of $f_{k}$ we have

$\int _{B_{k}^{s,t}}t\cdot s\,d\mu \leq \int _{B_{k}^{s,t}}f_{k}\,d\mu \leq \int _{X}f_{k}\,d\mu ,$

for every $k\geq 1$. Hence by step 3(b):

$\int _{X}t\cdot s\,d\mu =\sup _{k}\int _{B_{k}^{s,t}}t\cdot s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu$

Since the left hand side is just a finite sum we get

$t\int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu$

and taking the limit as $t\uparrow 1$ yields

$\int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu ,$

as required.

Step 3(d). The reverse inequality, i.e.

$\int _{X}f\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .$

Indeed by step 3(c)

$\int _{X}f\,d\mu =\sup _{s\in SF(f)}s\,d\mu \leq \sup _{k}f_{k}\,d\mu$

The proof 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 $(\Sigma ,{\mathcal {B}}_{\mathbb {R} _{\geq 0}})$-measurable non-negative functions $f_{k}:X\to [0,+\infty ]$. However, we do not assume they are pointwise non-decreasing. Instead, we assume that ${\textstyle \{f_{k}(x)\}_{k=1}^{\infty }}$ converges for almost every $x$, we define $f$ to be the pointwise limit of $\{f_{k}\}_{k=1}^{\infty }$, and we assume additionally that $f_{k}\leq f$ pointwise almost everywhere for all $k$. Then $f$ is $(\Sigma ,{\mathcal {B}}_{\mathbb {R} _{\geq 0}})$-measurable, and ${\textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu }$ exists, and

$$\lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}f\,d\mu .$$

As before, measurability follows from the fact that ${\textstyle f=\sup _{k}f_{k}}$ almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has

$$\int _{X}f\,d\mu =\int _{X}\liminf _{k}f_{k}\,d\mu \leq \liminf \int _{X}f_{k}\,d\mu$$

by Fatou's lemma, and then, by standard properties of limits and monotonicity,

$$\liminf \int _{X}f_{k}\,d\mu \leq \limsup \int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu .$$

Therefore ${\textstyle \liminf \int _{X}f_{k}\,d\mu =\limsup \int _{X}f_{k}\,d\mu }$, and both are equal to ${\textstyle \int _{X}f\,d\mu }$. It follows that ${\textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu }$ exists and equals ${\textstyle \int _{X}f\,d\mu }$.

See also

• Infinite series
• Dominated convergence theorem