Monotone convergence theorem – Wikipedia

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 convergence of monotonic sequences (sequences that are non-decreasing or non-increasing) 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.

Convergence of a monotone sequence of real numbers[edit]

Lemma 1[edit]

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

Proof[edit]

Let

(an)n∈N{displaystyle (a_{n})_{nin mathbb {N} }}

be such a sequence, and let

{an}{displaystyle {a_{n}}}

be the set of terms of

(an)n∈N{displaystyle (a_{n})_{nin mathbb {N} }}

. By assumption,

{an}{displaystyle {a_{n}}}

is non-empty and bounded above. By the least-upper-bound property of real numbers,

c=supn{an}{textstyle c=sup _{n}{a_{n}}}

exists and is finite. Now, for every

ε>0{displaystyle varepsilon >0}

N{displaystyle N}

such that

aN>c−ε{displaystyle a_{N}>c-varepsilon }

c−ε{displaystyle c-varepsilon }

is an upper bound of

{an}{displaystyle {a_{n}}}

, which contradicts the definition of

c{displaystyle c}

. Then since

(an)n∈N{displaystyle (a_{n})_{nin mathbb {N} }}

is increasing, and

c{displaystyle c}

is its upper bound, for every

n>N{displaystyle n>N}

|c−an|≤|c−aN|<ε{displaystyle |c-a_{n}|leq |c-a_{N}|

. Hence, by definition, the limit of

(an)n∈N{displaystyle (a_{n})_{nin mathbb {N} }}

is

supn{an}.{textstyle sup _{n}{a_{n}}.}

Lemma 2[edit]

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

Proof[edit]

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

Theorem[edit]

If

(an)n∈N{displaystyle (a_{n})_{nin mathbb {N} }}

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

Proof[edit]

• “If”-direction: The proof follows directly from the lemmas.
• “Only If”-direction: By (ε, δ)-definition of limit, every sequence
  (an)n∈N{displaystyle (a_{n})_{nin mathbb {N} }}
  with a finite limit L{displaystyle L}
  is necessarily bounded.

Convergence of a monotone series[edit]

Theorem[edit]

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

limj→∞∑kaj,k=∑klimj→∞aj,k.{displaystyle lim _{jto infty }sum _{k}a_{j,k}=sum _{k}lim _{jto infty }a_{j,k}.}

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

(1+1n)n=∑k=0n(nk)1nk=∑k=0n1k!×nn×n−1n×⋯×n−k+1n,{displaystyle left(1+{frac {1}{n}}right)^{n}=sum _{k=0}^{n}{binom {n}{k}}{frac {1}{n^{k}}}=sum _{k=0}^{n}{frac {1}{k!}}times {frac {n}{n}}times {frac {n-1}{n}}times cdots times {frac {n-k+1}{n}},}

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

(nk)1nk=1k!×nn×n−1n×⋯×n−k+1n;{displaystyle {binom {n}{k}}{frac {1}{n^{k}}}={frac {1}{k!}}times {frac {n}{n}}times {frac {n-1}{n}}times cdots times {frac {n-k+1}{n}};}

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

(1+1/n)n{displaystyle (1+1/n)^{n}}

by taking the sum of the column limits, namely 

1k!{displaystyle {frac {1}{k!}}}

.

Beppo Levi’s lemma[edit]

The following result is due to Beppo Levi, who proved a slight generalization in 1906 of an earlier result by Henri Lebesgue.^[3] In what follows,

BR≥0{displaystyle operatorname {mathcal {B}} _{mathbb {R} _{geq 0}}}

denotes the

σ{displaystyle sigma }

-algebra of Borel sets on

[0,+∞]{displaystyle [0,+infty ]}

. By definition,

BR≥0{displaystyle operatorname {mathcal {B}} _{mathbb {R} _{geq 0}}}

contains the set

{+∞}{displaystyle {+infty }}

and all Borel subsets of

R≥0.{displaystyle mathbb {R} _{geq 0}.}

Theorem[edit]

Let

(Ω,Σ,μ){displaystyle (Omega ,Sigma ,mu )}

be a measure space, and

X∈Σ{displaystyle Xin Sigma }

. Consider a pointwise non-decreasing sequence

{fk}k=1∞{displaystyle {f_{k}}_{k=1}^{infty }}

of

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable non-negative functions

fk:X→[0,+∞]{displaystyle f_{k}:Xto [0,+infty ]}

, i.e., for every

k≥1{displaystyle {kgeq 1}}

and every

x∈X{displaystyle {xin X}}

,

0≤fk(x)≤fk+1(x)≤∞.{displaystyle 0leq f_{k}(x)leq f_{k+1}(x)leq infty .}

Set the pointwise limit of the sequence

{fn}{displaystyle {f_{n}}}

to be

f{displaystyle f}

. That is, for every

x∈X{displaystyle xin X}

,

f(x):=limk→∞fk(x).{displaystyle f(x):=lim _{kto infty }f_{k}(x).}

Then

f{displaystyle f}

is

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable and

limk→∞∫Xfkdμ=∫Xfdμ.{displaystyle lim _{kto infty }int _{X}f_{k},dmu =int _{X}f,dmu .}

Remark 1. The integrals may be finite or infinite.

Remark 2. The theorem remains true if its assumptions hold

μ{displaystyle mu }

-almost everywhere. In other words, it is enough that there is a null set

N{displaystyle N}

such that the sequence

{fn(x)}{displaystyle {f_{n}(x)}}

non-decreases for every

x∈X∖N.{displaystyle {xin Xsetminus N}.}

To see why this is true, we start with an observation that allowing the sequence

{fn}{displaystyle {f_{n}}}

to pointwise non-decrease almost everywhere causes its pointwise limit

f{displaystyle f}

to be undefined on some null set

N{displaystyle N}

. On that null set,

f{displaystyle 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

μ(N)=0,{displaystyle {mu (N)=0},}

we have, for every

k,{displaystyle k,}

∫Xfkdμ=∫X∖Nfkdμ{displaystyle int _{X}f_{k},dmu =int _{Xsetminus N}f_{k},dmu }

and ∫Xfdμ=∫X∖Nfdμ,{displaystyle int _{X}f,dmu =int _{Xsetminus N}f,dmu ,}

provided that

f{displaystyle f}

is

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable.^{[4]: 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,

(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

f,g:X→[0,+∞]{displaystyle f,g:Xto [0,+infty ]}

be

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable.

• If
  f≤g{displaystyle fleq g}
  everywhere on X,{displaystyle X,}
  then

∫Xfdμ≤∫Xgdμ.{displaystyle int _{X}f,dmu leq int _{X}g,dmu .}

• If
  X1,X2∈Σ{displaystyle X_{1},X_{2}in Sigma }
  and X1⊆X2,{displaystyle X_{1}subseteq X_{2},}
  then

∫X1fdμ≤∫X2fdμ.{displaystyle int _{X_{1}}f,dmu leq int _{X_{2}}f,dmu .}

Proof. Denote

SF⁡(h){displaystyle operatorname {SF} (h)}

the set of simple

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable functions

s:X→[0,∞){displaystyle s:Xto [0,infty )}

such that

0≤s≤h{displaystyle 0leq sleq h}

everywhere on

X.{displaystyle X.}

1. Since

f≤g,{displaystyle fleq g,}

we have

SF⁡(f)⊆SF⁡(g).{displaystyle operatorname {SF} (f)subseteq operatorname {SF} (g).}

By definition of Lebesgue integral and the properties of supremum,

∫Xfdμ=sups∈SF(f)∫Xsdμ≤sups∈SF(g)∫Xsdμ=∫Xgdμ.{displaystyle int _{X}f,dmu =sup _{sin {rm {SF}}(f)}int _{X}s,dmu leq sup _{sin {rm {SF}}(g)}int _{X}s,dmu =int _{X}g,dmu .}

2. Let

1X1{displaystyle {mathbf {1} }_{X_{1}}}

be the indicator function of the set

X1.{displaystyle X_{1}.}

It can be deduced from the definition of Lebesgue integral that

∫X2f⋅1X1dμ=∫X1fdμ{displaystyle int _{X_{2}}fcdot {mathbf {1} }_{X_{1}},dmu =int _{X_{1}}f,dmu }

if we notice that, for every

s∈SF(f⋅1X1),{displaystyle sin {rm {SF}}(fcdot {mathbf {1} }_{X_{1}}),}

s=0{displaystyle s=0}

outside of

X1.{displaystyle X_{1}.}

Combined with the previous property, the inequality

f⋅1X1≤f{displaystyle fcdot {mathbf {1} }_{X_{1}}leq f}

implies

∫X1fdμ=∫X2f⋅1X1dμ≤∫X2fdμ.{displaystyle int _{X_{1}}f,dmu =int _{X_{2}}fcdot {mathbf {1} }_{X_{1}},dmu leq int _{X_{2}}f,dmu .}

Proof[edit]

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[edit]

Lebesgue integral as measure[edit]

Lemma 1. Let

(Ω,Σ,μ){displaystyle (Omega ,Sigma ,mu )}

be a measurable space. Consider a simple

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable non-negative function

s:Ω→R≥0{displaystyle s:Omega to {mathbb {R} _{geq 0}}}

. For a subset

S⊆Ω{displaystyle Ssubseteq Omega }

, define

ν(S)=∫Ssdμ.{displaystyle nu (S)=int _{S}s,dmu .}

Then

ν{displaystyle nu }

is a measure on

Ω{displaystyle Omega }

.

Proof[edit]

Monotonicity follows from Remark 5. Here, we will only prove countable additivity, leaving the rest up to the reader. Let

S=⋃i=1∞Si{displaystyle S=bigcup _{i=1}^{infty }S_{i}}

, where all the sets

Si{displaystyle S_{i}}

are pairwise disjoint. Due to simplicity,

s=∑i=1nci⋅1Ai,{displaystyle s=sum _{i=1}^{n}c_{i}cdot {mathbf {1} }_{A_{i}},}

for some finite non-negative constants

ci∈R≥0{displaystyle c_{i}in {mathbb {R} }_{geq 0}}

and pairwise disjoint sets

Ai∈Σ{displaystyle A_{i}in Sigma }

such that

⋃i=1nAi=Ω{displaystyle bigcup _{i=1}^{n}A_{i}=Omega }

. By definition of Lebesgue integral,

ν(S)=∑i=1nci⋅μ(S∩Ai)=∑i=1nci⋅μ((⋃j=1∞Sj)∩Ai)=∑i=1nci⋅μ(⋃j=1∞(Sj∩Ai)){displaystyle {begin{aligned}nu (S)&=sum _{i=1}^{n}c_{i}cdot mu (Scap A_{i})\&=sum _{i=1}^{n}c_{i}cdot mu left(left(bigcup _{j=1}^{infty }S_{j}right)cap A_{i}right)\&=sum _{i=1}^{n}c_{i}cdot mu left(bigcup _{j=1}^{infty }(S_{j}cap A_{i})right)end{aligned}}}

Since all the sets

Sj∩Ai{displaystyle S_{j}cap A_{i}}

are pairwise disjoint, the countable additivity of

μ{displaystyle mu }

gives us

∑i=1nci⋅μ(⋃j=1∞(Sj∩Ai))=∑i=1nci⋅∑j=1∞μ(Sj∩Ai).{displaystyle sum _{i=1}^{n}c_{i}cdot mu left(bigcup _{j=1}^{infty }(S_{j}cap A_{i})right)=sum _{i=1}^{n}c_{i}cdot sum _{j=1}^{infty }mu (S_{j}cap A_{i}).}

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,

∑i=1nci⋅∑j=1∞μ(Sj∩Ai)=∑j=1∞∑i=1nci⋅μ(Sj∩Ai)=∑j=1∞∫Sjsdμ=∑j=1∞ν(Sj),{displaystyle {begin{aligned}sum _{i=1}^{n}c_{i}cdot sum _{j=1}^{infty }mu (S_{j}cap A_{i})&=sum _{j=1}^{infty }sum _{i=1}^{n}c_{i}cdot mu (S_{j}cap A_{i})\&=sum _{j=1}^{infty }int _{S_{j}}s,dmu \&=sum _{j=1}^{infty }nu (S_{j}),end{aligned}}}

as required.

“Continuity from below”[edit]

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

Lemma 2. Let

μ{displaystyle mu }

be a measure, and

S=⋃i=1∞Si{displaystyle S=bigcup _{i=1}^{infty }S_{i}}

, where

S1⊆⋯⊆Si⊆Si+1⊆⋯⊆S{displaystyle S_{1}subseteq cdots subseteq S_{i}subseteq S_{i+1}subseteq cdots subseteq S}

is a non-decreasing chain with all its sets

μ{displaystyle mu }

-measurable. Then

μ(S)=limiμ(Si).{displaystyle mu (S)=lim _{i}mu (S_{i}).}

Proof of theorem[edit]

Step 1. We begin by showing that

f{displaystyle f}

is

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

–measurable.^{[4]: 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

[0,t]{displaystyle [0,t]}

under

f{displaystyle f}

is an element of the sigma-algebra

Σ{displaystyle Sigma }

on

X{displaystyle X}

, because (closed) intervals generate the Borel sigma algebra on the reals. Since

[0,t]{displaystyle [0,t]}

is a closed interval, and, for every

k{displaystyle k}

,

0≤fk(x)≤f(x){displaystyle 0leq f_{k}(x)leq f(x)}

,

0≤f(x)≤t⇔[∀k0≤fk(x)≤t].{displaystyle 0leq f(x)leq tquad Leftrightarrow quad {Bigl [}forall kquad 0leq f_{k}(x)leq t{Bigr ]}.}

Thus,

{x∈X∣0≤f(x)≤t}=⋂k{x∈X∣0≤fk(x)≤t}.{displaystyle {xin Xmid 0leq f(x)leq t}=bigcap _{k}{xin Xmid 0leq f_{k}(x)leq t}.}

Being the inverse image of a Borel set under a

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable function

fk{displaystyle f_{k}}

, each set in the countable intersection is an element of

Σ{displaystyle Sigma }

. Since

σ{displaystyle sigma }

-algebras are, by definition, closed under countable intersections, this shows that

f{displaystyle f}

is

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable, and the integral

∫Xfdμ{displaystyle textstyle int _{X}f,dmu }

is well defined (and possibly infinite).

Step 2. We will first show that

∫Xfdμ≥limk∫Xfkdμ.{displaystyle textstyle int _{X}f,dmu geq lim _{k}int _{X}f_{k},dmu .}

The definition of

f{displaystyle f}

and monotonicity of

{fk}{displaystyle {f_{k}}}

imply that

f(x)≥fk(x){displaystyle f(x)geq f_{k}(x)}

, for every

k{displaystyle k}

and every

x∈X{displaystyle xin X}

. By monotonicity (or, more precisely, its narrower version established in Remark 5; see also Remark 4) of Lebesgue integral,

∫Xfdμ≥∫Xfkdμ,{displaystyle int _{X}f,dmu geq int _{X}f_{k},dmu ,}

and

∫Xfdμ≥limk∫Xfkdμ.{displaystyle int _{X}f,dmu geq lim _{k}int _{X}f_{k},dmu .}

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

∫Xfdμ≤limk∫Xfkdμ{displaystyle int _{X}f,dmu leq lim _{k}int _{X}f_{k},dmu }

.

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

∫Xlim infkfk(x)dμ≤lim infk∫Xfkdμ.{displaystyle int _{X}liminf _{k}f_{k}(x),dmu leq liminf _{k}int _{X}f_{k},dmu .}

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

SF⁡(f){displaystyle operatorname {SF} (f)}

the set of simple

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable functions

s:X→[0,∞){displaystyle s:Xto [0,infty )}

such that

0≤s≤f{displaystyle 0leq sleq f}

on

X{displaystyle X}

.

Step 3. Given a simple function

s∈SF⁡(f){displaystyle sin operatorname {SF} (f)}

and a real number

t∈(0,1){displaystyle tin (0,1)}

, define

Bks,t={x∈X∣t⋅s(x)≤fk(x)}⊆X.{displaystyle B_{k}^{s,t}={xin Xmid tcdot s(x)leq f_{k}(x)}subseteq X.}

Then

Bks,t∈Σ{displaystyle B_{k}^{s,t}in Sigma }

,

Bks,t⊆Bk+1s,t{displaystyle B_{k}^{s,t}subseteq B_{k+1}^{s,t}}

, and

X=⋃kBks,t{displaystyle textstyle X=bigcup _{k}B_{k}^{s,t}}

.

Step 3a. To prove the first claim, let

s=∑i=1mci⋅1Ai{displaystyle textstyle s=sum _{i=1}^{m}c_{i}cdot {mathbf {1} }_{A_{i}}}

, for some finite collection of pairwise disjoint measurable sets

Ai∈Σ{displaystyle A_{i}in Sigma }

such that

X=∪i=1mAi{displaystyle textstyle X=cup _{i=1}^{m}A_{i}}

, some (finite) non-negative constants

ci∈R≥0{displaystyle c_{i}in {mathbb {R} }_{geq 0}}

, and

1Ai{displaystyle {mathbf {1} }_{A_{i}}}

denoting the indicator function of the set

Ai{displaystyle A_{i}}

.

For every

x∈Ai,{displaystyle xin A_{i},}

t⋅s(x)≤fk(x){displaystyle tcdot s(x)leq f_{k}(x)}

holds if and only if

fk(x)∈[t⋅ci,+∞].{displaystyle f_{k}(x)in [tcdot c_{i},+infty ].}

Given that the sets

Ai{displaystyle A_{i}}

are pairwise disjoint,

Bks,t=⋃i=1m(fk−1([t⋅ci,+∞])∩Ai).{displaystyle B_{k}^{s,t}=bigcup _{i=1}^{m}{Bigl (}f_{k}^{-1}{Bigl (}[tcdot c_{i},+infty ]{Bigr )}cap A_{i}{Bigr )}.}

Since the pre-image

fk−1([t⋅ci,+∞]){displaystyle f_{k}^{-1}{Bigl (}[tcdot c_{i},+infty ]{Bigr )}}

of the Borel set

[t⋅ci,+∞]{displaystyle [tcdot c_{i},+infty ]}

under the measurable function

fk{displaystyle f_{k}}

is measurable, and

σ{displaystyle sigma }

-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

k{displaystyle k}

and every

x∈X{displaystyle xin X}

,

fk(x)≤fk+1(x).{displaystyle f_{k}(x)leq f_{k+1}(x).}

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

X⊆⋃kBks,t{displaystyle textstyle Xsubseteq bigcup _{k}B_{k}^{s,t}}

.

Indeed, if, to the contrary,

X⊈⋃kBks,t{displaystyle textstyle Xnot subseteq bigcup _{k}B_{k}^{s,t}}

, then an element

x0∈X∖⋃kBks,t=⋂k(X∖Bks,t){displaystyle textstyle x_{0}in Xsetminus bigcup _{k}B_{k}^{s,t}=bigcap _{k}(Xsetminus B_{k}^{s,t})}

exists such that

fk(x0)<t⋅s(x0){displaystyle f_{k}(x_{0})

, for every

k{displaystyle k}

. Taking the limit as

k→∞{displaystyle kto infty }

, we get

f(x0)≤t⋅s(x0)<s(x0).{displaystyle f(x_{0})leq tcdot s(x_{0})

But by initial assumption,

s≤f{displaystyle sleq f}

. This is a contradiction.

Step 4. For every simple

(Σ,BR≥0){displaystyle (Sigma ,operatorname {mathcal {B}} _{mathbb {R} _{geq 0}})}

-measurable non-negative function

s2{displaystyle s_{2}}

,

limn∫Bns,ts2dμ=∫Xs2dμ.{displaystyle lim _{n}int _{B_{n}^{s,t}}s_{2},dmu =int _{X}s_{2},dmu .}

To prove this, define

ν(S)=∫Ss2dμ{displaystyle textstyle nu (S)=int _{S}s_{2},dmu }

. By Lemma 1,

ν(S){displaystyle nu (S)}

is a measure on

Ω{displaystyle Omega }

. By “continuity from below” (Lemma 2),

limn∫Bns,ts2dμ=limnν(Bns,t)=ν(X)=∫Xs2dμ,{displaystyle lim _{n}int _{B_{n}^{s,t}}s_{2},dmu =lim _{n}nu (B_{n}^{s,t})=nu (X)=int _{X}s_{2},dmu ,}

as required.

Step 5. We now prove that, for every

s∈SF⁡(f){displaystyle sin operatorname {SF} (f)}

,

∫Xsdμ≤limk∫Xfkdμ.{displaystyle int _{X}s,dmu leq lim _{k}int _{X}f_{k},dmu .}

Indeed, using the definition of

Bks,t{displaystyle B_{k}^{s,t}}

, the non-negativity of

fk{displaystyle f_{k}}

, and the monotonicity of Lebesgue integral (see Remark 5 and Remark 4), we have

∫Bks,tt⋅sdμ≤∫Bks,tfkdμ≤∫Xfkdμ,{displaystyle int _{B_{k}^{s,t}}tcdot s,dmu leq int _{B_{k}^{s,t}}f_{k},dmu leq int _{X}f_{k},dmu ,}

for every

k≥1{displaystyle kgeq 1}

. In accordance with Step 4, as

k→∞{displaystyle kto infty }

, the inequality becomes

t∫Xsdμ≤limk∫Xfkdμ.{displaystyle tint _{X}s,dmu leq lim _{k}int _{X}f_{k},dmu .}

Taking the limit as

t↑1{displaystyle tuparrow 1}

yields

∫Xsdμ≤limk∫Xfkdμ,{displaystyle int _{X}s,dmu leq lim _{k}int _{X}f_{k},dmu ,}

as required.

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

∫Xfdμ≤limk∫Xfkdμ.{displaystyle int _{X}f,dmu leq lim _{k}int _{X}f_{k},dmu .}

Indeed, by non-negativity,

f+=f{displaystyle f_{+}=f}

and

f−=0.{displaystyle f_{-}=0.}

For the calculation below, the non-negativity of

f{displaystyle f}

is essential. Applying the definition of Lebesgue integral and the inequality established in Step 5, we have

∫Xfdμ=sups∈SF⁡(f)∫Xsdμ≤limk∫Xfkdμ.{displaystyle int _{X}f,dmu =sup _{sin operatorname {SF} (f)}int _{X}s,dmu leq lim _{k}int _{X}f_{k},dmu .}

The proof is complete.

See also[edit]