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In the mathematical field of
real analysis In mathematics, the branch of real analysis studies the behavior of real numbers, sequences and series of real numbers, and real functions. Some particular properties of real-valued sequences and functions that real analysis studies include co ...
, the monotone convergence theorem is any of a number of related theorems proving the good
convergence Convergence may refer to: Arts and media Literature *''Convergence'' (book series), edited by Ruth Nanda Anshen *Convergence (comics), "Convergence" (comics), two separate story lines published by DC Comics: **A four-part crossover storyline that ...
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 In mathematics, the infimum (abbreviated inf; : infima) of a subset S of a partially ordered set P is the greatest element in P that is less than or equal to each element of S, if such an element exists. If the infimum of S exists, it is unique, ...
. 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_ \le a_ \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 In mathematics, the concept of a measure is a generalization and formalization of geometrical measures (length, area, volume) and other common notions, such as magnitude (mathematics), magnitude, mass, and probability of events. These seemingl ...
due to Lebesgue and Beppo Levi that says that for sequences of non-negative pointwise-increasing
measurable function In mathematics, and in particular measure theory, a measurable function is a function between the underlying sets of two measurable spaces that preserves the structure of the spaces: the preimage of any measurable set is measurable. This is in ...
s 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

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 :a_1 \le a_2 \le a_3 \le...\le K < \infty, the limit \lim_ a_n exists and equals its
supremum In mathematics, the infimum (abbreviated inf; : infima) of a subset S of a partially ordered set P is the greatest element in P that is less than or equal to each element of S, if such an element exists. If the infimum of S exists, it is unique, ...
: :\lim_ 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_ a_n exists and equals its infimum: :\lim_ a_n = \inf_n a_n \ge L.


Proof

Let \_ be the set of values of (a_n)_ . By assumption, \ is non-empty and bounded above by K. By the least-upper-bound property of real numbers, c = \sup_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 \, contradicting the definition of the supremum c. Then since (a_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_ a_n = c =\sup_n a_n. The proof of the (B) part is analogous or follows from (A) by considering \_.


Theorem

If (a_n)_ is a monotone
sequence In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. Like a set, it contains members (also called ''elements'', or ''terms''). The number of elements (possibly infinite) is cal ...
of
real number In mathematics, a real number is a number that can be used to measure a continuous one- dimensional quantity such as a duration or temperature. Here, ''continuous'' means that pairs of values can have arbitrarily small differences. Every re ...
s, i.e., if a_n \le a_ for every n \ge 1 or a_n \ge a_ for every n \ge 1, then this sequence has a finite limit
if and only if In logic and related fields such as mathematics and philosophy, "if and only if" (often shortened as "iff") is paraphrased by the biconditional, a logical connective between statements. The biconditional is true in two cases, where either bo ...
the sequence is bounded.


Proof

* "If"-direction: The proof follows directly from the proposition. * "Only If"-direction: By
(ε, δ)-definition of limit Although the function is not defined at zero, as becomes closer and closer to zero, becomes arbitrarily close to 1. In other words, the limit of as approaches zero, equals 1. In mathematics, the limit of a function is a fundame ...
, every sequence (a_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 \ In the extended real numbers every set has a
supremum In mathematics, the infimum (abbreviated inf; : infima) of a subset S of a partially ordered set P is the greatest element in P that is less than or equal to each element of S, if such an element exists. If the infimum of S exists, it is unique, ...
(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_ a_i = \sup_ \sum_ a_j \in \bar \R_ where \bar\R_ = , \infty\subset \bar \R are the upper extended non negative real numbers. For a series of non negative numbers :\sum_^\infty a_i = \lim_ \sum_^k a_i = \sup_k \sum_^k a_i = \sup_ \sum_ a_j = \sum_ 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.


Monotone convergence of non negative sums

Let a_ \ge 0 be a sequence of non-negative real numbers indexed by natural numbers i and k. Suppose that a_ \le a_ for all i, k. Then :\sup_k \sum_i a_ = \sum_i \sup_k a_ \in \bar\R_.


Proof

Since a_ \le \sup_k a_ we have \sum_i a_ \le \sum_i \sup_k a_ so \sup_k \sum_i a_ \le \sum_i \sup_k a_ . Conversely, we can interchange sup and sum for finite sums by reverting to the limit definition, so \sum_^N \sup_k a_ = \sup_k \sum_^N a_ \le \sup_k \sum_^\infty a_ hence \sum_^\infty \sup_k a_ \le \sup_k \sum_^\infty a_.


Examples


Matrices

The theorem states that if you have an infinite matrix of non-negative real numbers a_ \ge 0 such that the rows are weakly increasing and each is bounded a_ \le K_i where the bounds are summable \sum_i K_i <\infty then, for each column, the non decreasing column sums \sum_i a_ \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_ which element wise is the supremum over the row.


''e''

Consider the expansion : \left( 1+ \frac1k\right)^k = \sum_^k \binom ki \frac1 Now set : a_ = \binom ki \frac1 = \frac1 \cdot \frac kk \cdot \frack\cdot \cdots \frack for i \le k and a_ = 0 for i > k , then 0\le a_ \le a_ with \sup_k a_ = \frac 1<\infty and :\left( 1+ \frac1k\right)^k = \sum_^\infty a_. The right hand side is a non decreasing sequence in k, therefore : \lim_ \left( 1+ \frac1k\right)^k = \sup_k \sum_^\infty a_ = \sum_^\infty \sup_k a_ = \sum_^\infty \frac1 = 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_ denotes the \sigma-algebra of Borel sets on the upper extended non negative real numbers ,+\infty/math>. By definition, \operatorname_ contains the set \ and all Borel subsets of \R_.


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

Let (\Omega,\Sigma,\mu) be a
measure space A measure space is a basic object of measure theory, a branch of mathematics that studies generalized notions of volumes. It contains an underlying set, the subsets of this set that are feasible for measuring (the -algebra) and the method that ...
, and X\in\Sigma a measurable set. Let \^\infty_ be a pointwise non-decreasing sequence of (\Sigma,\operatorname_)- measurable non-negative functions, i.e. each function f_k:X\to ,+\infty/math> is (\Sigma,\operatorname_)-measurable and for every and every , : 0 \leq \ldots\le f_k(x) \leq f_(x)\leq\ldots\leq \infty. Then the pointwise supremum : \sup_k f_k : x \mapsto \sup_k f_k(x) is a (\Sigma,\operatorname_)-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, 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 : \lim_ \int_X f_k(x) \, d\mu(x) = \int_X \lim_ f_k(x) \, d\mu(x) with the tacit understanding that the limits are allowed to be infinite. 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 In mathematical analysis, a null set is a Lebesgue measurable set of real numbers that has measure zero. This can be characterized as a set that can be covered by a countable union of intervals of arbitrarily small total length. The notio ...
N such that the sequence \ non-decreases for every . To see why this is true, we start with an observation that allowing the sequence \ 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 , we have, for every k, : \int_X f_k \,d\mu = \int_ f_k \,d\mu and \int_X f \,d\mu = \int_ f \,d\mu, provided that f is (\Sigma,\operatorname_)-measurable.See for instance (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 ,+\infty/math> be (\Sigma,\operatorname_)-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_ f\,d\mu \leq \int_ f\,d\mu. Proof. Denote by \operatorname(h) the set of
simple Simple or SIMPLE may refer to: *Simplicity, the state or quality of being simple Arts and entertainment * ''Simple'' (album), by Andy Yorke, 2008, and its title track * "Simple" (Florida Georgia Line song), 2018 * "Simple", a song by John ...
(\Sigma, \operatorname_)-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(f) \subseteq \operatorname(g), hence :\int_X f\,d\mu = \sup_\int_X s\,d\mu \leq \sup_\int_X s\,d\mu = \int_X g\,d\mu. 2. The functions f\cdot _, f\cdot _, where _ is the indicator function of X_i, are easily seen to be measurable and f\cdot_\le f\cdot_. Now apply 1.


=Lebesgue integral as measure

= Lemma 2. Let (\Omega,\Sigma,\mu) be a measurable space, and let s:\Omega\to . be a simple (\Sigma,\operatorname_)-measurable non-negative function. For a measurable subset A \in \Sigma, define :\nu_s(A)=\int_A s(x)\,d\mu. Then \nu_s is a measure on (\Omega, \Sigma).


Proof (lemma 2)

Write s=\sum^n_c_k\cdot _, with c_k\in_ and measurable sets A_k\in\Sigma. Then :\nu_s(A)=\sum_^n c_k \mu(A\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_B(A) = \mu(B \cap A) 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 A = \bigcup^\infty_A_i, where : A_1\subseteq\cdots\subseteq A_i\subseteq A_\subseteq\cdots\subseteq A is a non-decreasing chain with all its sets \mu-measurable. Then :\mu(A)=\sup_i\mu(A_i).


proof (lemma 3)

Set A_0 = \emptyset, then we decompose A = \coprod_ A_i \setminus A_ as a countable disjoint union of measurable sets and likewise A_k = \coprod_ A_i \setminus A_ as a finite disjoint union. Therefore \mu(A_k) = \sum_^k \mu (A_i \setminus A_), and \mu(A) = \sum_^\infty \mu(A_i \setminus A_) so \mu(A) = \sup_k \mu(A_k).


Proof of theorem

Set f(x) = \sup_k f_k(x) for the pointwise supremum at x \in X. Step 1. The function f is (\Sigma, \operatorname_) –measurable, and the integral \textstyle \int_X f \,d\mu is well-defined (albeit possibly infinite) ''proof:'' 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_)-measurable. For this, it suffices to prove that f^( ,t is \Sigma -measurable for all 0 \le t \le \infty, because the intervals [0,t] generate the Borel sigma algebra on [0,\infty] by taking countable unions, complements and countable intersections. 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. Hence :f^( , t = \bigcap_^\infty f_k^( ,t. is a measurable set, being the countable intersection of the measurable sets f_k^( ,t. Since f \ge 0 the integral is well defined (but possibly infinite) as : \int_X f \,d\mu = \sup_\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 ''proof:'' We have f_k(x) \le f(x) for all x \in X, so \int_X f_k(x) \, d\mu \le \int_X f(x)\, d\mu by "monotonicity of the integral" (lemma 1). Then step 2 follows by the definition of suprememum. step 3 We have the reverse inequality : \int_X f \,d\mu \le \sup_k \int_X f_k \,d\mu . ''proof:'' Denote by \operatorname(f) the set of simple (\Sigma,\operatorname_)-measurable functions s:X\to [0,\infty) such that 0\leq s\leq f on X. We need an "epsilon of room" to manoeuvre. For s\in\operatorname(f) and 0 <\varepsilon < 1, define :B^_k=\\subseteq X. We claim the sets B^_k have the following properties: #B^_k is \Sigma-measurable. # B^_k\subseteq B^_ # X=\bigcup_^\infty B^_k Assuming the claim, by the definition of B^_k and "monotonicity of the Lebesgue integral" (lemma 1) we have :\int_(1-\varepsilon) s\,d\mu\leq\int_ f_k\,d\mu \leq\int_X f_k\,d\mu \le \sup_k \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: : (1- \varepsilon)\int_X s \, d\mu = \int_X (1- \varepsilon)s \, d\mu = \sup_k \int_ (1-\varepsilon)s\,d\mu \le \sup_k \int_X f_k \, d\mu. so :(1-\varepsilon)\int_X f d\mu \le \sup_k\int_X f_k d\mu which since \varepsilon is arbitrary, proves step 3. Ad 1: Write s=\sum_c_i\cdot_, for non-negative constants c_i \in \R_, and measurable sets A_i\in\Sigma, which we may assume are pairwise disjoint and with union \textstyle X=\coprod^m_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^_k=\coprod^m_\Bigl(f^_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^_k we have (1 - \varepsilon)s(x) \le f_k(x)\le f_(x) so x \in B^_. Ad 3: Fix x \in X. If s(x) = 0 then (1 - \varepsilon)s(x) = 0 \le f_1(x), hence x \in B^_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_(x) for N_x sufficiently large, hence x \in B^_. 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.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 As before, let (\Omega, \Sigma, \mu) be a
measure space A measure space is a basic object of measure theory, a branch of mathematics that studies generalized notions of volumes. It contains an underlying set, the subsets of this set that are feasible for measuring (the -algebra) and the method that ...
and X \in \Sigma. Again, \_^\infty will be a sequence of (\Sigma, \mathcal_)- measurable non-negative functions f_k:X\to ,+\infty/math>. However, we do not assume they are pointwise non-decreasing. Instead, we assume that \_^\infty converges for almost every x, we define f to be the pointwise limit of \_^\infty, and we assume additionally that f_k \le f pointwise almost everywhere for all k. Then f is (\Sigma, \mathcal_)-measurable, and \lim_ \int_X f_k \,d\mu exists, and \lim_ \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_ f_k = \liminf_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, since \int f_k \,d\mu \le \int f_ \,d\mu \le \int f d\mu (monotonicity), \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. Therefore \int_X f \, d\mu = \liminf_ \int_X f_k\,d\mu = \limsup_ \int_X f_k\,d\mu = \lim_ \int_X f_k \, d\mu = \sup_k \int_X f_k\,d\mu.


See also

* Infinite series * Dominated convergence theorem


Notes

{{Measure theory Articles containing proofs Theorems in calculus Sequences and series Theorems in real analysis Theorems in measure theory it:Passaggio al limite sotto segno di integrale#Integrale di Lebesgue