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
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
, 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
, taking the integral and the supremum can be interchanged with the result being finite if either one is finite.
Bounded above monotone non decreasing sequences are convergent in the real numbers because suprema exist there. Note that the proposition is wrong when considered in the rational numbers.
Proposition
(A) For a non-decreasing and bounded-above sequence of real numbers
![{\displaystyle a_{1}\leq a_{2}\leq a_{3}\leq ...\leq K<\infty ,}](//wikimedia.org/api/rest_v1/media/math/render/svg/b6542bc383804c1b16c03cd1b1264877bc2c480f)
the limit
exists and equals its supremum:
![{\displaystyle \lim _{n\to \infty }a_{n}=\sup _{n}a_{n}\leq K.}](//wikimedia.org/api/rest_v1/media/math/render/svg/148a531bbbae076182b1b7096cc51f181b6283d5)
(B) For a non-increasing and bounded-below sequence of real numbers
![{\displaystyle a_{1}\geq a_{2}\geq a_{3}\geq \cdots \geq L>-\infty ,}](//wikimedia.org/api/rest_v1/media/math/render/svg/8407838fa39b04d7fcf8c630098d8b26e6e790e9)
the limit
exists and equals its infimum:
.
Proof
Let
be the set of values of
. By assumption,
is non-empty and bounded above by
. By the least-upper-bound property of real numbers,
exists and
. Now, for every
, there exists
such that
, since otherwise
is a strictly smaller upper bound of
, contradicting the definition of the supremum
. Then since
is non decreasing, and
is an upper bound, for every
, we have
![{\displaystyle |c-a_{n}|=c-a_{n}\leq c-a_{N}=|c-a_{N}|<\varepsilon .}](//wikimedia.org/api/rest_v1/media/math/render/svg/1da92d245f81acddb76098c3f83313ea37eba223)
Hence, by definition
.
The proof of the (B) part is analogous or follows from (A) by considering
.
Proof
- "If"-direction: The proof follows directly from the proposition.
- "Only If"-direction: By (ε, δ)-definition of limit, every sequence
with a finite limit
is necessarily bounded.
There is a variant of the proposition above where we allow unbounded sequences in the extended real numbers, the real numbers with
and
added.
![{\displaystyle {\bar {\mathbb {R} }}=\mathbb {R} \cup \{-\infty ,\infty \}}](//wikimedia.org/api/rest_v1/media/math/render/svg/1753670d83b6830812f1718a483b7f3b85aa254e)
In the extended real numbers every set has a supremum (resp. infimum) which of course may be
(resp.
) if the set is unbounded. An important use of the extended reals is that any set of non negative numbers
has a well defined summation order independent sum
![{\displaystyle \sum _{i\in I}a_{i}=\sup _{J\subset I,\ |J|<\infty }\sum _{j\in J}a_{j}\in {\bar {\mathbb {R} }}_{\geq 0}}](//wikimedia.org/api/rest_v1/media/math/render/svg/0a67aad878df028e4e20e05d6ad9321f6cf482a2)
where
are the upper extended non negative real numbers. For a series of non negative numbers
![{\displaystyle \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 \mathbb {N} ,|J|<\infty }\sum _{j\in J}a_{j}=\sum _{i\in \mathbb {N} }a_{i},}](//wikimedia.org/api/rest_v1/media/math/render/svg/b2e512c0fac2207cc044fa3043dae75b9d686be8)
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.
Theorem (monotone convergence of non negative sums)
Let
be a sequence of non-negative real numbers indexed by natural numbers
and
. Suppose that
for all
. Then[2]: 168
![{\displaystyle \sup _{k}\sum _{i}a_{i,k}=\sum _{i}\sup _{k}a_{i,k}\in {\bar {\mathbb {R} }}_{\geq 0}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/ea7de8cfc73fd0d49357bbd1b15f3f2f709a9526)
Remark
The suprema and the sums may be finite or infinite but the left hand side is finite if and only if the right hand side is.
proof: Since
we have
so
. Conversely, we can interchange sup and sum for finite sums so
hence
.
The theorem states that if you have an infinite matrix of non-negative real numbers
such that
- the rows are weakly increasing and each is bounded
where the bounds are summable ![{\displaystyle \sum _{i}K_{i}<\infty }](//wikimedia.org/api/rest_v1/media/math/render/svg/b399205ed4b88e14c742fd8eba3d9e4549b5dbfe)
then
- for each column, the non decreasing column sums
are bounded hence convergent, and the limit of the column sums is equal to the sum of the "limit column"
which element wise is the supremum over the row.
As an example, consider the expansion
![{\displaystyle \left(1+{\frac {1}{k}}\right)^{k}=\sum _{i=0}^{k}{\binom {k}{i}}{\frac {1}{k^{i}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/3ccc4d7038d85bfcff17df19f797e243308fb228)
Now set
![{\displaystyle a_{i,k}={\binom {k}{i}}{\frac {1}{k^{i}}}={\frac {1}{i!}}\cdot {\frac {k}{k}}\cdot {\frac {k-1}{k}}\cdot \cdots {\frac {k-i+1}{k}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/c604706df888dcafefab06cd783dc5c4d212d62b)
for
and
for
, then
with
and
.
The right hand side is a non decreasing sequence in
, therefore
.
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.
[3]
[4]
Let
denotes the
-algebra of Borel sets on the upper extended non negative real numbers
. By definition,
contains the set
and all Borel subsets of ![{\displaystyle \mathbb {R} _{\geq 0}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/2479184346ebb17b862a8e8c5136a6d9826bfe17)
Theorem (monotone convergence theorem for non-negative measurable functions)
Let
be a measure space, and
a measurable set. Let
be a pointwise non-decreasing sequence of
-measurable non-negative functions, i.e. each function
is
-measurable and for every
and every
,
![{\displaystyle 0\leq \ldots \leq f_{k}(x)\leq f_{k+1}(x)\leq \ldots \leq \infty .}](//wikimedia.org/api/rest_v1/media/math/render/svg/6623b3619059a1533be8f090d34d01ceb00f4499)
Then the pointwise supremum
![{\displaystyle \sup _{k}f_{k}:x\mapsto \sup _{k}f_{k}(x)}](//wikimedia.org/api/rest_v1/media/math/render/svg/104bc7735e058375f179735517dd3a5d4191b2f6)
is a
-measurable function and
![{\displaystyle \sup _{k}\int _{X}f_{k}\,d\mu =\int _{X}\sup _{k}f_{k}\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/2341012abde1111f7136974dbcbdd74c67e42553)
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,
![{\displaystyle \textstyle \lim _{k\to \infty }f_{k}(x)=\sup _{k}f_{k}(x)=\limsup _{k\to \infty }f_{k}(x)=\liminf _{k\to \infty }f_{k}(x)}](//wikimedia.org/api/rest_v1/media/math/render/svg/e08575732571064a0d1664a66625134fc7dbacc3)
![{\displaystyle \textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\sup _{k}\int _{X}f_{k}\,d\mu =\liminf _{k\to \infty }\int _{X}f_{k}\,d\mu =\limsup _{k\to \infty }\int _{X}f_{k}\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/41b45833b49efdef6121e1a7df186aa83aa02cbe)
(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
![{\displaystyle \lim _{k\to \infty }\int _{X}f_{k}(x)\,d\mu (x)=\int _{X}\lim _{k\to \infty }f_{k}(x)\,d\mu (x)}](//wikimedia.org/api/rest_v1/media/math/render/svg/96a5acab438c5500654770a8a9f842f58374290b)
with the tacit understanding that the limits are allowed to be infinite.
Remark 3. The theorem remains true if its assumptions hold
-almost everywhere. In other words, it is enough that there is a null set
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
to be undefined on some null set
. On that null set,
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 ![{\displaystyle k,}](//wikimedia.org/api/rest_v1/media/math/render/svg/44e185ab9c990830d5055fa3ae698a4225ce67e0)
and ![{\displaystyle \int _{X}f\,d\mu =\int _{X\setminus N}f\,d\mu ,}](//wikimedia.org/api/rest_v1/media/math/render/svg/8287691fb6b704cdc8a8f2d47bab93854d6788fc)
provided that
is
-measurable.[5]: section 21.38 (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.
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
be
-measurable.
- If
everywhere on
then
![{\displaystyle \int _{X}f\,d\mu \leq \int _{X}g\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/d65426a52e9938f23f702c92be4da32a3692ea32)
- If
and
then
![{\displaystyle \int _{X_{1}}f\,d\mu \leq \int _{X_{2}}f\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/895a94011808abb570d4ad1cce53ec206d1b74f3)
Proof. Denote by
the set of simple
-measurable functions
such that
everywhere on ![{\displaystyle X.}](//wikimedia.org/api/rest_v1/media/math/render/svg/5ba76c5a460c4a0bb1639a193bc1830f0a773e03)
1. Since
we have
hence
![{\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 .}](//wikimedia.org/api/rest_v1/media/math/render/svg/a3f92e3b4ccc161d79f5000888b29eae162828d3)
2. The functions
where
is the indicator function of
, are easily seen to be measurable and
. Now apply 1.
Lebesgue integral as measure
Lemma 2. Let
be a measurable space. Consider a simple
-measurable non-negative function
. For a measurable subset
, define
![{\displaystyle \nu _{s}(S)=\int _{S}s\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/79e02a8904903c462a66b465be893f8ccf20b5ce)
Then
is a measure on
.
Set
.
Denote by
the set of simple
-measurable functions
(
nor included!) such that
on
.
Step 1. The function
is
–measurable, and the integral
is well-defined (albeit possibly infinite)[5]: section 21.3
From
we get
. Hence we have to show that
is
-measurable. To see this, it suffices to prove that
is
-measurable for all
, because the intervals
generate the Borel sigma algebra on the extended non negative reals
by complementing and taking countable intersections, complements and countable unions.
Now since the
is a non decreasing sequence,
if and only if
for all
. Since we already know that
and
we conclude that
![{\displaystyle f^{-1}([0,t])=\bigcap _{k}f_{k}^{-1}([0,t]).}](//wikimedia.org/api/rest_v1/media/math/render/svg/7e5b2332a9699620bc200902103e9b12f9e485db)
Hence
is a measurable set,
being the countable intersection of the measurable sets
.
Since
the integral is well defined (but possibly infinite) as
.
Step 2. We have the inequality
![{\displaystyle \sup _{k}\int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu }](//wikimedia.org/api/rest_v1/media/math/render/svg/ea36e5489363e4c55c20c070eee049c71eace02e)
This is equivalent to
for all
which follows directly from
and "monotonicity of the integral" (lemma 1).
step 3 We have the reverse inequality
.
By the definition of integral as a supremum step 3 is equivalent to
![{\displaystyle \int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }](//wikimedia.org/api/rest_v1/media/math/render/svg/1e4fa4353d811639c6d7fd06115b78b989ba20c1)
for every
.
It is tempting to prove
for
sufficiently large, but this does not work e.g. if
is itself simple and the
. However, we can get ourself an "epsilon of room" to manoeuvre and avoid this problem.
Step 3 is also equivalent to
![{\displaystyle (1-\varepsilon )\int _{X}s\,d\mu =\int _{X}(1-\varepsilon )s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }](//wikimedia.org/api/rest_v1/media/math/render/svg/f06e95c686082b64f4e814d3532c523cef82753f)
for every simple function
and every
where for the equality we used that the left hand side of the inequality is a finite sum. This we will prove.
Given
and
, define
![{\displaystyle B_{k}^{s,\varepsilon }=\{x\in X\mid (1-\varepsilon )s(x)\leq f_{k}(x)\}\subseteq X.}](//wikimedia.org/api/rest_v1/media/math/render/svg/519eb5d30660d3f46a237a715a473bd366240383)
We claim the sets
have the following properties:
is
-measurable.
![{\displaystyle B_{k}^{s,\varepsilon }\subseteq B_{k+1}^{s,\varepsilon }}](//wikimedia.org/api/rest_v1/media/math/render/svg/98fad1f5b02cd0cff0f8d98f54b0382897434e67)
![{\displaystyle X=\bigcup _{k}B_{k}^{s,\varepsilon }}](//wikimedia.org/api/rest_v1/media/math/render/svg/724b6aa7b42b924e9658dcb5101a01360d11f36d)
Assuming the claim, by the definition of
and "monotonicity of the Lebesgue integral" (lemma 1) we have
![{\displaystyle \int _{B_{k}^{s,\varepsilon }}(1-\varepsilon )s\,d\mu \leq \int _{B_{k}^{s,\varepsilon }}f_{k}\,d\mu \leq \int _{X}f_{k}\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/d8dc848628c33f27f9dcbe5e039dbe49129b5982)
Hence by "Lebesgue integral of a simple function as measure" (lemma 2), and "continuity from below" (lemma 3) we get:
![{\displaystyle \sup _{k}\int _{B_{k}^{s,\varepsilon }}(1-\varepsilon )s\,d\mu =\int _{X}(1-\varepsilon )s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/e9e511706891b11c04de2598c95fba4f104e806a)
which we set out to prove. Thus it remains to prove the claim.
Ad 1: Write
, for non-negative constants
, and measurable sets
, which we may assume are pairwise disjoint and with union
. Then for
we have
if and only if
so
![{\displaystyle B_{k}^{s,\varepsilon }=\coprod _{i=1}^{m}{\Bigl (}f_{k}^{-1}{\Bigl (}[(1-\varepsilon )c_{i},\infty ]{\Bigr )}\cap A_{i}{\Bigr )}}](//wikimedia.org/api/rest_v1/media/math/render/svg/03f2aec9cc3ba0107083aa8f356460c162d57746)
which is measurable since the
are measurable.
Ad 2: For
we have
so ![{\displaystyle x\in B_{k+1}^{s,\varepsilon }.}](//wikimedia.org/api/rest_v1/media/math/render/svg/9b497bae5dbfa01c42b0fa85bed7e5a429d9597e)
Ad 3: Fix
. If
then
, hence
. Otherwise,
and
so
for
sufficiently large, hence
.
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.[6] As before, let
be a measure space and
. Again,
will be a sequence of
-measurable non-negative functions
. However, we do not assume they are pointwise non-decreasing. Instead, we assume that
converges for almost every
, we define
to be the pointwise limit of
, and we assume additionally that
pointwise almost everywhere for all
. Then
is
-measurable, and
exists, and ![{\displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}f\,d\mu .}](//wikimedia.org/api/rest_v1/media/math/render/svg/ea889a2a5b30f2e55bc3436e221549b2dbdf6547)
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
almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has
by Fatou's lemma, and then, since
(monotonicity),
Therefore
![{\displaystyle \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 .}](//wikimedia.org/api/rest_v1/media/math/render/svg/a7f1d5e29366506d9d87a4f8618a9c2c1af881ee)