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.

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Convergence of a monotone sequence of real numbersEdit

Lemma 1Edit

If a sequence of real numbers is increasing and bounded above, then its supremum is the limit.

ProofEdit

Let   be such a sequence. By assumption,   is non-empty and bounded above. By the least-upper-bound property of real numbers,   exists and is finite. Now, for every  , there exists   such that  , since otherwise   is an upper bound of  , which contradicts to the definition of  . Then since   is increasing, and   is its upper bound, for every  , we have  . Hence, by definition, the limit of   is  

Lemma 2Edit

If a sequence of real numbers is decreasing and bounded below, then its infimum is the limit.

ProofEdit

The proof is similar to the proof for the case when the sequence is increasing and bounded above,

TheoremEdit

If   is a monotone sequence of real numbers (i.e., if an ≤ an+1 for every n ≥ 1 or an ≥ an+1 for every n ≥ 1), then this sequence has a finite limit if and only if the sequence is bounded.[1]

ProofEdit

  • "If"-direction: The proof follows directly from the lemmas.
  • "Only If"-direction: By definition of limit, every sequence   with a finite limit   is necessarily bounded.

Convergence of a monotone seriesEdit

TheoremEdit

If for all natural numbers j and k, aj,k is a non-negative real number and aj,k ≤ aj+1,k, then[2]:168

 

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

 

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

 

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   by taking the sum of the column limits, namely  .

Beppo Levi's monotone convergence theorem for Lebesgue integralEdit

The following result is due to Beppo Levi and Henri Lebesgue. In what follows,   denotes the  -algebra of Borel sets on  . By definition,   contains the set   and all Borel subsets of  

TheoremEdit

Let   be a measure space, and  . Consider a pointwise non-decreasing sequence   of  -measurable non-negative functions  , i.e., for every   and every  ,

 

Set the pointwise limit of the sequence   to be  . That is, for every  ,

 

Then   is  -measurable and

 

Remark 1. The integrals may be finite or infinite.

Remark 2. 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  

provided that   is  -measurable.[3](section 21.38) (These equalities follow directly from the definition of Lebesgue integral for a non-negative function).

Remark 3. Under assumptions of the theorem,

  1.  
  2.  

(Note that the second chain of equalities follows from Remark 5).

Remark 4. The proof below does not use any properties of Lebesgue integral except those established here. The theorem, thus, can be used to prove other basic properties, such as linearity, pertaining to Lebesgue integration.

Remark 5 (monotonicity of Lebesgue integral). In the proof below, we apply the monotonic property of Lebesgue integral to non-negative functions only. Specifically (see Remark 4), let the functions   be  -measurable.

  • If   everywhere on   then
 
  • If   and   then
 

Proof. Denote   the set of simple  -measurable functions   such that   everywhere on  

1. Since   we have

 

By definition of Lebesgue integral and the properties of supremum,

 

2. Let   be the indicator function of the set   It can be deduced from the definition of Lebesgue integral that

 

if we notice that, for every     outside of   Combined with the previous property, the inequality   implies

 

ProofEdit

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 resultsEdit

Lebesgue integral as measureEdit

Lemma 1. Let   be a measurable space. Consider a simple  -measurable non-negative function  . For a subset  , define

 .

Then   is a measure on  .

ProofEdit

Monotonicity follows from Remark 5. Here, we will only prove countable additivity, leaving the rest up to the reader. Let  , where all the sets   are pairwise disjoint. Due to simplicity,

 ,

for some finite non-negative constants   and pairwise disjoint sets   such that  . By definition of Lebesgue integral,

 

Since all the sets   are pairwise disjoint, the countable additivity of   gives us

 

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,

 

as required.

"Continuity from below"Edit

The following property is a direct consequence of the definition of measure.

Lemma 2. Let   be a measure, and  , where

 

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

 .

Proof of theoremEdit

Step 1. We begin by showing that   is  –measurable.[3](section 21.3)

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   under   is an element of the sigma-algebra   on  , because (closed) intervals generate the Borel sigma algebra on the reals. Since   is a closed interval, and, for every  ,  ,

 

Thus,

 

Being the inverse image of a Borel set under a  -measurable function  , each set in the countable intersection is an element of  . Since  -algebras are, by definition, closed under countable intersections, this shows that   is  -measurable, and the integral   is well defined (and possibly infinite).

Step 2. We will first show that  

The definition of   and monotonicity of   imply that  , for every   and every  . By monotonicity (or, more precisely, its narrower version established in Remark 5; see also Remark 4) of Lebesgue integral,

 

and

 

Note that the limit on the right exists (finite or infinite) because, due to monotonicity (see Remark 5 and Remark 4), the sequence is non-decreasing.

End of Step 2.

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

 .

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

 .

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   the set of simple  -measurable functions   such that   on  .

Step 3. Given a simple function   and a real number  , define

 

Then  ,  , and  .

Step 3a. To prove the first claim, let  , for some finite collection of pairwise disjoint measurable sets   such that  , some (finite) non-negative constants  , and   denoting the indicator function of the set  .

For every     holds if and only if   Given that the sets   are pairwise disjoint,

 .

Since the pre-image   of the Borel set   under the measurable function   is measurable, and  -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   and every  ,  

Step 3c. To prove the third claim, we show that  .

Indeed, if, to the contrary,  , then an element

 

exists such that  , for every  . Taking the limit as  , we get

 .

But by initial assumption,  . This is a contradiction.

Step 4. For every simple  -measurable non-negative function  ,

 .

To prove this, define  . By Lemma 1,   is a measure on  . By "continuity from below" (Lemma 2),

 ,

as required.

Step 5. We now prove that, for every  ,

 .

Indeed, using the definition of  , the non-negativity of  , and the monotonicity of Lebesgue integral (see Remark 5 and Remark 4), we have

 ,

for every  . In accordance with Step 4, as  , the inequality becomes

 

Taking the limit as   yields

 ,

as required.

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

 .

Indeed, by non-negativity,   and   For the calculation below, the non-negativity of   is essential. Applying the definition of Lebesgue integral and the inequality established in Step 5, we have

 

The proof is complete.

See alsoEdit

NotesEdit

  1. ^ 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.
  2. ^ See for instance Yeh, J. (2006). Real Analysis: Theory of Measure and Integration. Hackensack, NJ: World Scientific. ISBN 981-256-653-8.
  3. ^ a b See for instance Schechter, Erik (1997). Handbook of Analysis and Its Foundations. San Diego: Academic Press. ISBN 0-12-622760-8.