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 increasing or decreasing) 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 $\{a_{n}\}$ be such a sequence. By assumption, $\{a_{n}\}$ is non-empty and bounded above. By the least-upper-bound property of real numbers, $c=\sup _{n}\{a_{n}\}$ 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 to the definition of $c$. Then since $\{a_{n}\}$ 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 $\{a_{n}\}$ is $\sup _{n}\{a_{n}\}.$

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, and is left as an exercise to the reader.

Theorem

If $\{a_{n}\}$ 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 definition of limit, every sequence $\{a_{n}\}$ 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+1,k ≤ a_j,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

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

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

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

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 ${\frac {1}{k!}}$.

Monotone convergence theorem for Lebesgue integral

The following result is due to Henri Lebesgue and Beppo Levi. In what follows, $\operatorname {\cal {B}} _{\mathbb {R} _{\geq 0}}$ denotes the $\sigma$-algebra of Borel sets on $[0,+\infty )$.

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 $(\Sigma ,\operatorname {\cal {B}} _{\mathbb {R} _{\geq 0}})$-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 limit of the sequence $\{f_{n}\}$ to be $f$. That is, for every $x\in X$,

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

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

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

Remark 1. The integrals may be finite or infinite.

Remark 2. The theorem remains true if its assumptions hold $\mu$-almost everywhere. In other words, it is enough that there is a subset ${N\in \Sigma }$ such that ${\mu (N)=0}$, and the sequence $\{f_{n}(x)\}$ non-decreases for every ${x\in X\setminus N}$. Indeed, since ${\mu (N)=0}$, for every $k$, we would then have

$\int _{X}f_{k}\,d\mu =\int _{X\setminus N}f_{k}\,d\mu$ and $\int _{X}f\,d\mu =\int _{X\setminus N}f\,d\mu ,$

provided that $f$ is $(\Sigma ,\operatorname {\cal {B}} _{\mathbb {R} _{\geq 0}})$-measurable.

Remark 3. Under assumptions of the theorem,

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

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 $(\Omega ,\Sigma ,\mu )$ be a measurable space. Consider a simple $(\Sigma ,\operatorname {\cal {B}} _{\mathbb {R} _{\geq 0}})$-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 $\Omega$.

Proof

We will only prove countable additivity, leaving the rest up to the reader. Let $S=\cup _{i=1}^{\infty }S_{i}$, where all the sets $S_{i}$ are pairwise disjoint. Due to simplicity,

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

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

${\begin{aligned}\nu (S)&=\\&=\sum _{i=1}^{n}c_{i}\cdot \mu (S\cap A_{i})\\&=\sum _{i=1}^{n}c_{i}\cdot \mu {\bigl (}(\cup _{j=1}^{\infty }S_{j})\cap A_{i}{\bigr )}\\&=\sum _{i=1}^{n}c_{i}\cdot \mu {\bigl (}\cup _{j=1}^{\infty }(S_{j}\cap A_{i}){\bigr )}\end{aligned}}$

Since all the sets $S_{j}\cap A_{i}$ are pairwise disjoint, the countable additivity of $\mu$ gives us

$\sum _{i=1}^{n}c_{i}\cdot \mu {\bigl (}\cup _{j=1}^{\infty }(S_{j}\cap A_{i}){\bigr )}=\sum _{i=1}^{n}c_{i}\cdot \sum _{j=1}^{\infty }\mu (S_{j}\cap A_{i}).$

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,

${\begin{aligned}\sum _{i=1}^{n}c_{i}\cdot \sum _{j=1}^{\infty }\mu (S_{j}\cap A_{i})&=\sum _{j=1}^{\infty }\sum _{i=1}^{n}c_{i}\cdot \mu (S_{j}\cap A_{i})\\&=\sum _{j=1}^{\infty }\int _{S_{j}}s\,d\mu \\&=\sum _{j=1}^{\infty }\nu (S_{j}),\end{aligned}}$

as required.

Lemma about monotonic chain of sets

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

$S_{1}\subseteq \ldots \subseteq S_{i}\subseteq S_{i+1}\subseteq \ldots \subseteq S$

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

$\mu (S)=\lim _{i}\mu (S_{i})$.

Proof

Let $S_{0}=\emptyset$. The sequence $\{T_{i}\}_{i=1}^{\infty }$, where $T_{i}=S_{i}\setminus S_{i-1}$, consists of pairwise disjoint sets, $S=\cup _{i=1}^{\infty }T_{i}$, and $S_{n}=\cup _{i=1}^{n}T_{i}$. By countable additivity,

${\begin{aligned}\mu (S)&=\\&=\mu (\cup _{i=1}^{\infty }T_{i})\\&=\sum _{i=1}^{\infty }\mu (T_{i})\\&=\lim _{n}\sum _{i=1}^{n}\mu (T_{i})\\&=\lim _{n}\mu (\cup _{i=1}^{n}T_{i})\\&=\lim _{n}\mu (S_{n}),\end{aligned}}$

as required.

Proof of theorem

Step 1. We begin by showing that $f$ is $\Sigma$–measurable.

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 $[0,t]$ under $f$ is an element of the sigma-algebra $\Sigma$ on $X$, because (closed) intervals generate the Borel sigma algebra on the reals. Since $[0,t]$ is a closed interval, and, for every $k$, $0\leq f_{k}(x)\leq f(x)$,

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

Thus,

$\{x\in X\mid 0\leq f(x)\leq t\}=\bigcap _{k}\{x\in X\mid 0\leq f_{k}(x)\leq t\}.$

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

Step 2. We will first show that $\textstyle \int _{X}f\,d\mu \geq \lim _{k}\int _{X}f_{k}\,d\mu .$

The definition of $f$ and monotonicity of $\{f_{k}\}$ imply that $f(x)\geq f_{k}(x)$, for every $k$ and every $x\in X$. It follows that

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

and

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

Note that the limit on the right exists (finite or infinite) because, due to monotonicity of Lebesgue integral, the sequence is non-decreasing.

End of Step 2.

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

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

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

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

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

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

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

Then $B_{k}^{s,t}\in \Sigma$, $B_{k}^{s,t}\subseteq B_{k+1}^{s,t}$, and $\textstyle X=\bigcup _{k}B_{k}^{s,t}$.

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

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

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

Step 3c. To prove the third claim, we show that $\textstyle X\subseteq \bigcup _{k}B_{k}^{s,t}$.

Indeed, if, to the contrary, $\textstyle X\not \subseteq \bigcup _{k}B_{k}^{s,t}$, then an element

$\textstyle x_{0}\in X\setminus \bigcup _{k}B_{k}^{s,t}=\bigcap _{k}(X\setminus B_{k}^{s,t})$

exists such that $f_{k}(x_{0})<t\cdot s(x_{0})$, for every $k$. Taking the limit as $k\to \infty$, get

$f(x_{0})\leq t\cdot s(x_{0})<s(x_{0})$.

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

Step 4. For every simple $(\Sigma ,\operatorname {\cal {B}} _{\mathbb {R} _{\geq 0}})$-measurable non-negative function $s_{2}$,

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

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

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

as required.

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

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

Indeed, using the definition of $B_{k}^{s,t}$, the non-negativity of $f_{k}$, and the monotonicity of Lebesgue integral, 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 $k\geq 1$. In accordance with Step 4, as $k\to \infty$, the inequality becomes

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

Taking the limit as $t\uparrow 1$ yields

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

as required.

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

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

Indeed, applying the definition of Lebesgue integral and the inequality established in Step 5, we have

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

as needed. The proof is complete.

See also

• Infinite series
• Dominated convergence theorem