Monotone convergence theorem

Template:Short description Template:Cleanup 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 <math>a_1 \le a_2 \le a_3 \le ...\le K</math> 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 <math>0 \le a_{i,1} \le a_{i,2} \le \cdots </math>, 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 <math>0 \le f_1(x) \le f_2(x) \le \cdots</math>, 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

<math>a_1 \le a_2 \le a_3 \le...\le K < \infty,</math>

the limit <math>\lim_{n \to \infty} a_n</math> exists and equals its supremum:

<math>\lim_{n \to \infty} a_n = \sup_n a_n \le K.</math>

(B) For a non-increasing and bounded-below sequence of real numbers

<math>a_1 \ge a_2 \ge a_3 \ge \cdots \ge L > -\infty,</math>

the limit <math> \lim_{n \to \infty} a_n</math> exists and equals its infimum:

<math>\lim_{n \to \infty} a_n = \inf_n a_n \ge L</math>.

Proof

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

<math>|c - a_n| = c -a_n \leq c - a_N = |c -a_N|< \varepsilon. </math>

Hence, by definition <math> \lim_{n \to \infty} a_n = c =\sup_n a_n</math>.

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

Theorem

If <math>(a_n)_{n\in\mathbb{N}}</math> is a monotone sequence of real numbers, i.e., if <math>a_n \le a_{n+1}</math> for every <math>n \ge 1</math> or <math>a_n \ge a_{n+1}</math> for every <math>n \ge 1</math>, then this sequence has a finite limit if and only if the sequence is bounded.<ref>A generalisation of this theorem was given by Template:Cite journal</ref>

Proof

• "If"-direction: The proof follows directly from the proposition.
• "Only If"-direction: By (ε, δ)-definition of limit, every sequence <math>(a_n)_{n\in\mathbb{N}}</math> with a finite limit <math>L</math> 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 <math>\infty</math> and <math> -\infty</math> added.

<math> \bar\R = \R \cup \{-\infty, \infty\}</math>

In the extended real numbers every set has a supremum (resp. infimum) which of course may be <math>\infty</math> (resp. <math>-\infty</math>) if the set is unbounded. An important use of the extended reals is that any set of non negative numbers <math> a_i \ge 0, i \in I </math> has a well defined summation order independent sum

<math> \sum_{i \in I} a_i = \sup_{J \subset I,\ |J|< \infty} \sum_{j \in J} a_j \in \bar \R_{\ge 0}</math>

where <math>\bar\R_{\ge 0} = [0, \infty] \subset \bar \R</math> are the upper extended non negative real numbers. For a series of non negative numbers

<math>\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,</math>

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 <math>a_{i,k} \ge 0 </math> be a sequence of non-negative real numbers indexed by natural numbers <math>i</math> and <math>k</math>. Suppose that <math>a_{i,k} \le a_{i,k+1}</math> for all <math>i, k</math>. Then<ref>See for instance Template:Cite book</ref>Template:Rp

<math>\sup_k \sum_i a_{i,k} = \sum_i \sup_k a_{i,k} \in \bar\R_{\ge 0}.</math>

Proof

Since <math>a_{i,k} \le \sup_k a_{i,k}</math> we have <math>\sum_i a_{i,k} \le \sum_i \sup_k a_{i,k}</math> so <math>\sup_k \sum_i a_{i,k} \le \sum_i \sup_k a_{i,k} </math>.

Conversely, we can interchange sup and sum for finite sums by reverting to the limit definition, so <math>\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}</math> hence <math>\sum_{i = 1}^\infty \sup_k a_{i,k} \le \sup_k \sum_{i =1}^\infty a_{i,k}</math>.

Examples

Matrices

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

e

Consider the expansion

<math> \left( 1+ \frac1k\right)^k = \sum_{i=0}^k \binom ki \frac1{k^i}

</math> Now set

<math> 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

</math> for <math> i \le k </math> and <math> a_{i,k} = 0</math> for <math> i > k </math>, then <math>0\le a_{i,k} \le a_{i,k+1}</math> with <math>\sup_k a_{i,k} = \frac 1{i!}<\infty </math> and

<math>\left( 1+ \frac1k\right)^k = \sum_{i =0}^\infty a_{i,k}</math>.

The right hand side is a non decreasing sequence in <math>k</math>, therefore

<math> \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</math>.

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. <ref name="BigRudin">Template:Cite book </ref> <ref>Template:Citation</ref>

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

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

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

<math> 0 \leq \ldots\le f_k(x) \leq f_{k+1}(x)\leq\ldots\leq \infty. </math>

Then the pointwise supremum

<math> \sup_k f_k : x \mapsto \sup_k f_k(x) </math>

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

<math>\sup_k \int_X f_k \,d\mu = \int_X \sup_k f_k \,d\mu.</math>

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, Template:Ordered list

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

<math> \lim_{k \to \infty} \int_X f_k(x) \, d\mu(x) = \int_X \lim_{k\to \infty} f_k(x) \, d\mu(x)

</math> with the tacit understanding that the limits are allowed to be infinite.

Remark 3. The theorem remains true if its assumptions hold <math>\mu</math>-almost everywhere. In other words, it is enough that there is a null set <math>N</math> such that the sequence <math>\{f_n(x)\}</math> non-decreases for every <math>{x\in X\setminus N}.</math> To see why this is true, we start with an observation that allowing the sequence <math>\{ f_n \}</math> to pointwise non-decrease almost everywhere causes its pointwise limit <math>f</math> to be undefined on some null set <math>N</math>. On that null set, <math>f</math> 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 <math>{\mu(N)=0},</math> we have, for every <math>k,</math>

<math> \int_X f_k \,d\mu = \int_{X \setminus N} f_k \,d\mu</math> and <math>\int_X f \,d\mu = \int_{X \setminus N} f \,d\mu, </math>

provided that <math>f</math> is <math>(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable.<ref name="SCHECHTER1997">See for instance Template:Cite book</ref>Template:Rp (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 <math>f,g : X \to [0,+\infty]</math> be <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-measurable.

• If <math>f \leq g</math> everywhere on <math>X,</math> then

<math>\int_X f\,d\mu \leq \int_X g\,d\mu.</math>

• If <math> X_1,X_2 \in \Sigma </math> and <math>X_1 \subseteq X_2, </math> then

<math>\int_{X_1} f\,d\mu \leq \int_{X_2} f\,d\mu.</math>

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

1. Since <math>f \leq g,</math> we have <math> \operatorname{SF}(f) \subseteq \operatorname{SF}(g), </math> hence

<math>\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.</math>

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

Lebesgue integral as measure

Lemma 2. Let <math>(\Omega,\Sigma,\mu)</math> be a measurable space, and let <math>s:\Omega\to{\mathbb R_{\geq 0}}</math> . be a simple <math>(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable non-negative function. For a measurable subset <math>A \in \Sigma</math>, define

<math>\nu_s(A)=\int_A s(x)\,d\mu.</math>

Then <math>\nu_s</math> is a measure on <math>(\Omega, \Sigma)</math>.

Proof (lemma 2)

Write <math>s=\sum^n_{k=1}c_k\cdot {\mathbf 1}_{A_k},</math> with <math>c_k\in{\mathbb R}_{\geq 0}</math> and measurable sets <math>A_k\in\Sigma</math>. Then

<math>\nu_s(A)=\sum_{k =1}^n c_k \mu(A\cap A_k).</math>

Since finite positive linear combinations of countably additive set functions are countably additive, to prove countable additivity of <math>\nu_s</math> it suffices to prove that, the set function defined by <math>\nu_B(A) = \mu(B \cap A)</math> is countably additive for all <math>A \in \Sigma</math>. But this follows directly from the countable additivity of <math>\mu</math>.

Continuity from below

Lemma 3. Let <math>\mu</math> be a measure, and <math>A = \bigcup^\infty_{i=1}A_i</math>, where

<math>

A_1\subseteq\cdots\subseteq A_i\subseteq A_{i+1}\subseteq\cdots\subseteq A </math> is a non-decreasing chain with all its sets <math>\mu</math>-measurable. Then

<math>\mu(A)=\sup_i\mu(A_i).</math>

proof (lemma 3)

Set <math>A_0 = \emptyset</math>, then we decompose <math> A = \coprod_{1 \le i } A_i \setminus A_{i -1} </math> as a countable disjoint union of measurable sets and likewise <math>A_k = \coprod_{1\le i \le k } A_i \setminus A_{i -1} </math> as a finite disjoint union. Therefore <math>\mu(A_k) = \sum_{i=1}^k \mu (A_i \setminus A_{i -1})</math>, and <math>\mu(A) = \sum_{i = 1}^\infty \mu(A_i \setminus A_{i-1})</math> so <math>\mu(A) = \sup_k \mu(A_k)</math>.

Proof of theorem

Set <math> f(x) = \sup_k f_k(x)</math> for the pointwise supremum at <math>x \in X</math>.

Step 1. The function <math>f</math> is <math> (\Sigma, \operatorname{\mathcal B}_{\bar\R_{\geq 0}}) </math>–measurable, and the integral <math>\textstyle \int_X f \,d\mu </math> is well-defined (albeit possibly infinite)<ref name="SCHECHTER1997"/>Template:Rp

proof: From <math>0 \le f_k(x) \le \infty</math> we get <math>0 \le f(x) \le \infty</math>. Hence we have to show that <math>f</math> is <math>(\Sigma,\operatorname{\mathcal B}_{\bar\R_{\geq 0}})</math>-measurable. For this, it suffices to prove that <math>f^{-1}([0,t])</math> is <math>\Sigma </math>-measurable for all <math>0 \le t \le \infty</math>, because the intervals <math>[0,t]</math> generate the Borel sigma algebra on <math>[0,\infty]</math> by taking countable unions, complements and countable intersections.

Now since the <math>f_k(x)</math> is a non decreasing sequence, <math> f(x) = \sup_k f_k(x) \leq t</math> if and only if <math>f_k(x)\leq t</math> for all <math>k</math>. Hence

<math>f^{-1}([0, t]) = \bigcap_{k =1}^\infty f_k^{-1}([0,t]).</math>

is a measurable set, being the countable intersection of the measurable sets <math>f_k^{-1}([0,t])</math>.

Since <math>f \ge 0</math> the integral is well defined (but possibly infinite) as

<math> \int_X f \,d\mu = \sup_{s \in SF(f)}\int_X s \, d\mu</math>.

Step 2. We have the inequality

<math>\sup_k \int_X f_k \,d\mu \le \int_X f \,d\mu </math>

proof: We have <math>f_k(x) \le f(x)</math> for all <math>x \in X</math>, so <math>\int_X f_k(x) \, d\mu \le \int_X f(x)\, d\mu</math> by "monotonicity of the integral" (lemma 1). Then step 2 follows by the definition of suprememum.

step 3 We have the reverse inequality

<math> \int_X f \,d\mu \le \sup_k \int_X f_k \,d\mu </math>.

proof: Denote by <math>\operatorname{SF}(f)</math> the set of simple <math>(\Sigma,\operatorname{\mathcal B}_{\R_{\geq 0}})</math>-measurable functions <math>s:X\to [0,\infty)</math> such that <math>0\leq s\leq f</math> on <math>X</math>. We need an "epsilon of room" to manoeuvre. For <math>s\in\operatorname{SF}(f)</math> and <math>0 <\varepsilon < 1</math>, define

<math>B^{s,\varepsilon}_k=\{x\in X\mid (1 - \varepsilon) s(x)\leq f_k(x)\}\subseteq X.</math>

We claim the sets <math>B^{s,\varepsilon}_k</math> have the following properties:

1. <math>B^{s,\varepsilon}_k</math> is <math>\Sigma</math>-measurable.
2. <math>B^{s,\varepsilon}_k\subseteq B^{s,\varepsilon}_{k+1}</math>
3. <math> X=\bigcup_{k = 1}^\infty B^{s,\varepsilon}_k</math>

Assuming the claim, by the definition of <math>B^{s,\varepsilon}_k</math> and "monotonicity of the Lebesgue integral" (lemma 1) we have

<math>\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 \le \sup_k \int_X f_k \, d\mu

</math> Hence by "Lebesgue integral of a simple function as measure" (lemma 2), and "continuity from below" (lemma 3) we get:

<math> (1- \varepsilon)\int_X s \, d\mu =

\int_X (1- \varepsilon)s \, d\mu = \sup_k \int_{B^{s,\varepsilon}_k} (1-\varepsilon)s\,d\mu \le \sup_k \int_X f_k \, d\mu. </math> so

<math>(1-\varepsilon)\int_X f d\mu \le \sup_k\int_X f_k d\mu</math>

which since <math>\varepsilon</math> is arbitrary, proves step 3.

Ad 1: Write <math>s=\sum_{1 \le i \le m}c_i\cdot{\mathbf 1}_{A_i}</math>, for non-negative constants <math>c_i \in \R_{\geq 0}</math>, and measurable sets <math>A_i\in\Sigma</math>, which we may assume are pairwise disjoint and with union <math>\textstyle X=\coprod^m_{i=1}A_i</math>. Then for <math> x\in A_i</math> we have <math>(1-\varepsilon)s(x)\leq f_k(x)</math> if and only if <math> f_k(x) \in [( 1- \varepsilon)c_i, \,\infty],</math> so

<math>B^{s,\varepsilon}_k=\coprod^m_{i=1}\Bigl(f^{-1}_k\Bigl([(1-\varepsilon)c_i,\infty]\Bigr)\cap A_i\Bigr)</math>

which is measurable since the <math>f_k</math> are measurable.

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

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

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.<ref>coudy (https://mathoverflow.net/users/6129/coudy), Do you know important theorems that remain unknown?, URL (version: 2018-06-05): https://mathoverflow.net/q/296540</ref> As before, let <math>(\Omega, \Sigma, \mu)</math> be a measure space and <math>X \in \Sigma</math>. Again, <math>\{f_k\}_{k=1}^\infty</math> will be a sequence of <math>(\Sigma, \mathcal{B}_{\R_{\geq 0}})</math>-measurable non-negative functions <math>f_k:X\to [0,+\infty]</math>. However, we do not assume they are pointwise non-decreasing. Instead, we assume that <math display="inline">\{f_k(x)\}_{k=1}^\infty</math> converges for almost every <math>x</math>, we define <math>f</math> to be the pointwise limit of <math>\{f_k\}_{k=1}^\infty</math>, and we assume additionally that <math>f_k \le f</math> pointwise almost everywhere for all <math>k</math>. Then <math>f</math> is <math>(\Sigma, \mathcal{B}_{\R_{\geq 0}})</math>-measurable, and <math display="inline">\lim_{k\to\infty} \int_X f_k \,d\mu</math> exists, and <math display="block">\lim_{k\to\infty} \int_X f_k \,d\mu = \int_X f \,d\mu.</math>

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 <math display="inline">f = \sup_k f_k = \lim_{k \to \infty} f_k = \liminf_{k \to \infty}f_k</math> almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has <math display="block">\int_X f\,d\mu = \int_X \liminf_k f_k\,d\mu \le \liminf \int_X f_k\,d\mu</math> by Fatou's lemma, and then, since <math>\int f_k \,d\mu \le \int f_{k + 1} \,d\mu \le \int f d\mu</math> (monotonicity), <math display="block">\liminf \int_X f_k\,d\mu \le \limsup_k \int_X f_k\,d\mu = \sup_k \int_X f_k\,d\mu \le \int_X f\,d\mu.</math> Therefore <math display="block">\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.</math>

See also

• Infinite series
• Dominated convergence theorem