Theorems on the convergence of bounded monotonic sequences
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. Likewise for sequences of non-negative pointwise-increasing (measurable) functions
, taking the integral and the supremum can be interchanged.
Convergence of a monotone sequence of real numbers
Lemma 1
For a non-decreasing and bounded-above sequence of real numbers
![{\displaystyle a_{1}\leq a_{2}\leq a_{3}\leq ...\leq K<\infty ,}](https://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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/148a531bbbae076182b1b7096cc51f181b6283d5)
Lemma 2
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 ,}](https://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 .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1da92d245f81acddb76098c3f83313ea37eba223)
Hence, by definition
.
The proof of lemma 2 is analogous or follows from lemma 1 by considering
.
Theorem
If
is a monotone sequence of real numbers, i.e., if
for every
or
for every
, then this sequence has a finite limit if and only if the sequence is bounded.[1]
Proof
- "If"-direction: The proof follows directly from the lemmas.
- "Only If"-direction: By (ε, δ)-definition of limit, every sequence
with a finite limit
is necessarily bounded.
Convergence of a monotone series
There is a variant of Lemma 1 and 2 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 \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecd1b48c9cc3576e13e36ae4853df3743d99559c)
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}}](https://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},}](https://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}.}](https://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 }](https://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}}}}](https://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}}.}](https://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
.
Beppo Levi's lemma
The following result is a generalisation of the monotone convergence of non negative sums theorem above. It is due to Beppo Levi, who proved a slight generalization in 1906 of an earlier result by Henri Lebesgue.[3] In what follows,
denotes the
-algebra of Borel sets on the upper extended non negative real numbers
. By definition,
contains the set
and all Borel subsets of
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 f_{k}(x)\leq f_{k+1}(x)\leq \infty .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/644a9faa878d9cf88cfd2e75a8556af339aa96a1)
Then the pointwise supremum
![{\displaystyle \sup _{k}f_{k}:x\mapsto \sup _{k}f_{k}(x)}](https://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 .}](https://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)}](https://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 .}](https://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).
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
and ![{\displaystyle \int _{X}f\,d\mu =\int _{X\setminus N}f\,d\mu ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8287691fb6b704cdc8a8f2d47bab93854d6788fc)
provided that
is
-measurable.[4]: 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.
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
be
-measurable.
- If
everywhere on
then
![{\displaystyle \int _{X}f\,d\mu \leq \int _{X}g\,d\mu .}](https://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 .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/895a94011808abb570d4ad1cce53ec206d1b74f3)
Proof. Denote by
the set of simple
-measurable functions
such that
everywhere on
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 .}](https://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 1. Let
be a measurable space. Consider a simple
-measurable non-negative function
. For a subset
, define
![{\displaystyle \nu (S)=\int _{S}s\,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1dbfcf785adbdbd2b64bcadb6a0cec386bc97c05)
Then
is a measure on
.
Monotonicity follows from lemma 1. To prove countable additivity, let
be a decomposition of
as a countable pairwise-disjoint union of measurable subsets
. Write
with
and measurable sets
. By the countable additivity of
and the order independence of summation for non negative numbers,
![{\displaystyle {\begin{aligned}\nu (\coprod _{i=1}^{\infty }S)&=\sum _{k=1}^{n}c_{k}\cdot \mu (\coprod _{i=1}^{\infty }(S_{i}\cap A_{k}))\\&=\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd684110a797ba54e463164ba12659ba716bfde1)
as required.
"Continuity from below"
Lemma 2. Let
be a measure, and
, where
![{\displaystyle S_{1}\subseteq \cdots \subseteq S_{i}\subseteq S_{i+1}\subseteq \cdots \subseteq S}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23635477c8c56f05aa5a5b262584106944203292)
is a non-decreasing chain with all its sets
-measurable. Then
![{\displaystyle \mu (S)=\sup _{i}\mu (S_{i}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70f0d544730dce177d29dadef27fb9742640cf1e)
proof (lemma 2)
Set
, then we decompose
and likewise
as a union of disjoint measurable sets. Therefore
, and
so
.
Proof of theorem
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)[4]: 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, 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]).}](https://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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea36e5489363e4c55c20c070eee049c71eace02e)
This is equivalent to
for all
which follows directly from
and the monotonicity of the integral.
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 }](https://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 simple and the
. We need an "epsilon of room" to manoeuvre.
Given a simple function
and an
, define
![{\displaystyle B_{k}^{s,\varepsilon }=\{x\in X\mid (1-\varepsilon )s(x)\leq f_{k}(x)\}\subseteq X.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/519eb5d30660d3f46a237a715a473bd366240383)
step 3(a). We have
is
-measurable. ![{\displaystyle B_{k}^{s,\varepsilon }\subseteq B_{k+1}^{s,\varepsilon }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98fad1f5b02cd0cff0f8d98f54b0382897434e67)
![{\displaystyle X=\bigcup _{k}B_{k}^{s,\varepsilon }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/724b6aa7b42b924e9658dcb5101a01360d11f36d)
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 )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03f2aec9cc3ba0107083aa8f356460c162d57746)
which is measurable because the
are measurable.
Ad 2: the sequence
is non decreasing.
Ad 3: Fix
. Either
so
hence
, or
so
for
sufficiently large hence
.
Now by the definition of
and the monotonicity of the Lebesgue integral 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 .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8dc848628c33f27f9dcbe5e039dbe49129b5982)
Hence by lemma 2, "continuity from below" and (3(a).3):
![{\displaystyle \int _{X}(1-\varepsilon )s\,d\mu =\sup _{k}\int _{B_{k}^{s,\varepsilon }}(1-\varepsilon )s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5e4f215e2a75a5c64d70a2e9f238877890acb29)
The left hand side is a finite sum and the inequality can be rewritten as
![{\displaystyle (1-\varepsilon )\int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9161181df4af0033b304ce13e7ba7caffaecb85)
which gives step 3 by first taking the supremum over
and then the supremum over
.
The proof of Beppo Levi's theorem is complete.
Relaxing the monotonicity assumption
Under similar hypotheses to Beppo Levi's theorem, it is possible to relax the hypothesis of monotonicity.[5] 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 .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea889a2a5b30f2e55bc3436e221549b2dbdf6547)
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 and the following is a direct proof.
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
![{\displaystyle \int _{X}f\,d\mu =\int _{X}\liminf _{k}f_{k}\,d\mu \leq \liminf \int _{X}f_{k}\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a691d204d484064a5ee6839f7d5450f027cc15f0)
by Fatou's lemma, and then, by standard properties of limits and monotonicity,
![{\displaystyle \liminf \int _{X}f_{k}\,d\mu \leq \limsup \int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79cf52cff1c89a32b2c304b4e536c68047c75846)
Therefore
![{\textstyle \liminf \int _{X}f_{k}\,d\mu =\limsup \int _{X}f_{k}\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e57c8c93d43fd391c26713baaa87d9a52888242b)
, and both are equal to
![{\textstyle \int _{X}f\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/17d570dc3705c29db9d2a99375824079f9bc7b58)
. It follows that
![{\textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/11c271467c77d9e3428124486859810654bc67a8)
exists and equals
![{\textstyle \int _{X}f\,d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/17d570dc3705c29db9d2a99375824079f9bc7b58)
.
See also
Notes
- ^ A generalisation of this theorem was given by Bibby, John (1974). "Axiomatisations of the average and a further generalisation of monotonic sequences". Glasgow Mathematical Journal. 15 (1): 63–65. doi:10.1017/S0017089500002135.
- ^ See for instance Yeh, J. (2006). Real Analysis: Theory of Measure and Integration. Hackensack, NJ: World Scientific. ISBN 981-256-653-8.
- ^ Schappacher, Norbert; Schoof, René (1996), "Beppo Levi and the arithmetic of elliptic curves" (PDF), The Mathematical Intelligencer, 18 (1): 60, doi:10.1007/bf03024818, MR 1381581, S2CID 125072148, Zbl 0849.01036
- ^ a b See for instance Schechter, Erik (1997). Handbook of Analysis and Its Foundations. San Diego: Academic Press. ISBN 0-12-622760-8.
- ^ 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
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