Monotone convergence theorem

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, ie. sequences that are non-increasing , or non-decreasing. In its simplest form it says that a non decreasing sequence ${\displaystyle a_{1}\leq a_{2}\leq a_{3}\leq ...\leq K}$ of numbers that are bounded above will converge to its smallest upper bound, its supremum (in the same way, if a sequence is decreasing and is bounded below it will convergence to its largest lower bound, its infimum).

For sums of non negative increasing sequences ${\displaystyle 0\leq a_{i,1}\leq a_{i,2}\leq \cdots }$ it says that taking sum and supremum can be interchanged. Likewise for sequences of non negative point wise increasing (measurable) functions ${\displaystyle 0\leq f_{1}(x)\leq f_{2}(x)\leq ...}$ taking integral and supremum can be interchanged.

Convergence of a monotone sequence of real numbers[edit]

Lemma 1[edit]

For a non decreasing and bounded above sequence of real numbers

${\displaystyle a_{1}\leq a_{2}\leq a_{3}\leq ...\leq K<\infty }$

the ${\displaystyle \lim _{n}a_{n}}$ exists and equals the supremum:

${\displaystyle \lim _{n}a_{n}=\sup _{n}a_{n}\leq K}$

Lemma 2[edit]

If a non increasing and bounded below sequence of real numbers

${\displaystyle a_{1}\geq a_{2}\geq a_{3}\geq \cdots \geq L>-\infty }$

the limit ${\displaystyle \lim _{n}a_{n}}$ exists and equals its infimum:

${\displaystyle \lim _{n}a_{n}=\inf _{n}a_{n}\geq L}$.

Proof[edit]

Let ${\displaystyle \{a_{n}\}_{n\in \mathbb {N} }}$ be the set of values of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$. By assumption, ${\displaystyle \{a_{n}\}}$ is non-empty and bounded above by ${\displaystyle K}$. By the least-upper-bound property of real numbers, ${\textstyle c=\sup _{n}\{a_{n}\}}$ exists and ${\displaystyle c\leq K}$. Now, for every ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle N}$ such that ${\displaystyle c\geq a_{N}>c-\varepsilon }$, since otherwise ${\displaystyle c-\varepsilon }$ is a strictly smaller upper bound of ${\displaystyle \{a_{n}\}}$, contradicting the definition of the supremum ${\displaystyle c}$. Then since ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is non decreasing, and ${\displaystyle c}$ is an upper bound, for every ${\displaystyle n>N}$, we have

${\displaystyle |c-a_{n}|=c-a_{n}\leq c-a_{N}=|c-a_{N}|<\varepsilon .}$

Hence, by definition ${\displaystyle \lim _{n}a_{n}=c=\sup _{n}a_{n}}$.

The proof of lemma 2 is analogous or follows from lemma 1 by considering ${\displaystyle \{-a_{n}\}_{n\in \mathbb {N} }}$.

Theorem[edit]

If ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is a monotone sequence of real numbers (i.e., if ${\displaystyle a_{n}\leq a_{n+1}}$ for every ${\displaystyle n\geq 1}$ or ${\displaystyle a_{n}\geq a_{n+1}}$ for every ${\displaystyle n\geq 1}$ 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 ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ with a finite limit ${\displaystyle L}$ is necessarily bounded.

Convergence of a monotone series[edit]

There is a variant of Lemma 1 and 2 where we allow unbounded sequences in the extended real numbers, the real numbers with ${\displaystyle \infty }$ and ${\displaystyle -\infty }$ added.

${\displaystyle {\bar {\mathbb {R} }}=\mathbb {R} \cup \{\infty ,-\infty \}}$

In the extended real numbers every set has a supremum (resp. infimum) which of course may be ${\displaystyle \infty }$ (resp. ${\displaystyle -\infty }$) if the set is unbounded. An important use of the extended reals is that any set of non negative numbers ${\displaystyle a_{i}\geq 0,i\in I}$ has a well defined sum

${\displaystyle \sum _{i}a_{i}=\sup _{J\subset I,\ |J|<\infty }\sum _{j\in J}a_{j}\in {\bar {\mathbb {R} }}_{\geq 0}}$

where ${\displaystyle {\bar {\mathbb {R} }}_{\geq 0}=[0,\infty ]\subset {\bar {\mathbb {R} }}}$ are the upper extended non negative real numbers.

Theorem (monotone convergence of non negative sums)[edit]

Let ${\displaystyle a_{i,k}\geq 0}$ be a sequence of non-negative real numbers indexed by natural numbers ${\displaystyle i}$ and ${\displaystyle k}$. Suppose that ${\displaystyle a_{i,k}\leq a_{i,k+1}}$ for all ${\displaystyle i,k}$. Then^[2]: 168

${\displaystyle \sup _{k\to \infty }\sum _{i}a_{i,k}=\sum _{i}\sup _{k\to \infty }a_{i,k}\in {\bar {\mathbb {R} }}_{\geq 0}.}$

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.

The theorem states that if you have an infinite matrix of non-negative real numbers ${\displaystyle a_{i,k}\geq 0}$ such that

• the rows are weakly increasing and each is bounded ${\displaystyle a_{i,k}\leq K_{i}}$ where the bounds are summable ${\displaystyle \sum _{i}K_{i}<\infty }$

then

• for each column, the non decreasing column sums ${\displaystyle \sum _{i}a_{i,k}\leq \sum K_{i}}$ are bounded hence convergent, and the limit of the column sums is equal to the sum of the "limit column" ${\displaystyle \sup _{k}a_{i,k}}$ which element wise is the supremum over the row.

As an example, consider the expansion

${\displaystyle \left(1+{\frac {1}{k}}\right)^{k}=\sum _{i=0}^{k}{\binom {k}{i}}{\frac {1}{k^{i}}}}$

Now set

${\displaystyle a_{i,k}={\binom {k}{i}}{\frac {1}{k^{i}}}={\frac {1}{i!}}\cdot {\frac {k}{k}}\cdot {\frac {k-1}{k}}\cdot \cdots {\frac {k-i+1}{k}}.}$

for ${\displaystyle i\leq k}$ and ${\displaystyle a_{i,k}=0}$ for ${\displaystyle i>k}$, then ${\displaystyle 0\leq a_{i,k}\leq a_{i,k+1}}$ with ${\displaystyle \sup _{k}a_{i,k}={\frac {1}{i!}}<\infty }$ and

${\displaystyle \left(1+{\frac {1}{k}}\right)^{k}=\sum _{i=0}^{\infty }a_{i,k}}$.

The right hand side is a non decreasing sequence in ${\displaystyle k}$, therefore

${\displaystyle \lim _{k\to \infty }\left(1+{\frac {1}{k}}\right)^{k}=\sup _{k\to \infty }\sum _{i=0}^{\infty }a_{i,k}=\sum _{i=0}^{\infty }\sup _{k\to \infty }a_{i,k}=\sum _{i=0}^{\infty }{\frac {1}{i!}}=e}$.

Beppo Levi's lemma[edit]

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, ${\displaystyle \operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}}}$ denotes the ${\displaystyle \sigma }$-algebra of Borel sets on the upper extended non negative real numbers ${\displaystyle [0,+\infty ]}$. By definition, ${\displaystyle \operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}}}$ contains the set ${\displaystyle \{+\infty \}}$ and all Borel subsets of ${\displaystyle \mathbb {R} _{\geq 0}.}$

Theorem (monotone convergence theorem for non negative measurable functions)[edit]

Let ${\displaystyle (\Omega ,\Sigma ,\mu )}$ be a measure space, and ${\displaystyle X\in \Sigma }$ a measurable set. Let ${\displaystyle \{f_{k}\}_{k=1}^{\infty }}$ be pointwise non-decreasing sequence of ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable non-negative functions i.e. each function ${\displaystyle f_{k}:X\to [0,+\infty ]}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable and for every ${\displaystyle {k\geq 1}}$ and every ${\displaystyle {x\in X}}$,

${\displaystyle 0\leq f_{k}(x)\leq f_{k+1}(x)\leq \infty .}$

Then the pointwise supremum

${\displaystyle \sup _{k\to \infty }f_{k}:x\mapsto \sup _{k\to \infty }f_{k}(x)}$

is a ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable function and

${\displaystyle \sup _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}\sup _{k\to \infty }f_{k}\,d\mu .}$

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).

Remark 3. The theorem remains true if its assumptions hold ${\displaystyle \mu }$-almost everywhere. In other words, it is enough that there is a null set ${\displaystyle N}$ such that the sequence ${\displaystyle \{f_{n}(x)\}}$ non-decreases for every ${\displaystyle {x\in X\setminus N}.}$ To see why this is true, we start with an observation that allowing the sequence ${\displaystyle \{f_{n}\}}$ to pointwise non-decrease almost everywhere causes its pointwise limit ${\displaystyle f}$ to be undefined on some null set ${\displaystyle N}$. On that null set, ${\displaystyle f}$ 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 ${\displaystyle {\mu (N)=0},}$ we have, for every ${\displaystyle k,}$

${\displaystyle \int _{X}f_{k}\,d\mu =\int _{X\setminus N}f_{k}\,d\mu }$ and ${\displaystyle \int _{X}f\,d\mu =\int _{X\setminus N}f\,d\mu ,}$

provided that ${\displaystyle f}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-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[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]

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 ${\displaystyle f,g:X\to [0,+\infty ]}$ be ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable.

• If ${\displaystyle f\leq g}$ everywhere on ${\displaystyle X,}$ then

${\displaystyle \int _{X}f\,d\mu \leq \int _{X}g\,d\mu .}$

• If ${\displaystyle X_{1},X_{2}\in \Sigma }$ and ${\displaystyle X_{1}\subseteq X_{2},}$ then

${\displaystyle \int _{X_{1}}f\,d\mu \leq \int _{X_{2}}f\,d\mu .}$

Proof. Denote by ${\displaystyle \operatorname {SF} (h)}$ the set of simple ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable functions ${\displaystyle s:X\to [0,\infty )}$ such that ${\displaystyle 0\leq s\leq h}$ everywhere on ${\displaystyle X.}$

1. Since ${\displaystyle f\leq g,}$ we have

${\displaystyle \operatorname {SF} (f)\subseteq \operatorname {SF} (g).}$

By definition of the Lebesgue integral and the properties of supremum,

${\displaystyle \int _{X}f\,d\mu =\sup _{s\in {\rm {SF}}(f)}\int _{X}s\,d\mu \leq \sup _{s\in {\rm {SF}}(g)}\int _{X}s\,d\mu =\int _{X}g\,d\mu .}$

2. Let ${\displaystyle {\mathbf {1} }_{X_{1}}}$ be the indicator function of the set ${\displaystyle X_{1}.}$ It can be deduced from the definition of the Lebesgue integral that

${\displaystyle \int _{X_{2}}f\cdot {\mathbf {1} }_{X_{1}}\,d\mu =\int _{X_{1}}f\,d\mu }$

if we notice that, for every ${\displaystyle s\in {\rm {SF}}(f\cdot {\mathbf {1} }_{X_{1}}),}$ ${\displaystyle s=0}$ outside of ${\displaystyle X_{1}.}$ Combined with the previous property, the inequality ${\displaystyle f\cdot {\mathbf {1} }_{X_{1}}\leq f}$ implies

${\displaystyle \int _{X_{1}}f\,d\mu =\int _{X_{2}}f\cdot {\mathbf {1} }_{X_{1}}\,d\mu \leq \int _{X_{2}}f\,d\mu .}$

Lebesgue integral as measure[edit]

Lemma 1. Let ${\displaystyle (\Omega ,\Sigma ,\mu )}$ be a measurable space. Consider a simple ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable non-negative function ${\displaystyle s:\Omega \to {\mathbb {R} _{\geq 0}}}$. For a subset ${\displaystyle S\subseteq \Omega }$, define

${\displaystyle \nu (S)=\int _{S}s\,d\mu .}$

Then ${\displaystyle \nu }$ is a measure on ${\displaystyle \Omega }$.

Monotonicity follows from lemma 1. Here, we will only prove countable additivity, leaving the rest up to the reader. Let ${\displaystyle S=\coprod _{i=1}^{\infty }S_{i}}$, be a decomposition of ${\displaystyle S}$ as a countable pairwise disjoint union of measurable subsets ${\displaystyle S_{i}}$. Write

${\displaystyle s=\sum _{k=1}^{n}c_{k}\cdot {\mathbf {1} }_{A_{k}},}$

for a finite number ${\displaystyle n}$ of non-negative constants ${\displaystyle c_{k}\in {\mathbb {R} }_{\geq 0}}$ and measurable sets ${\displaystyle A_{k}\in \Sigma }$. By definition of the Lebesgue integral

${\displaystyle {\begin{aligned}\nu (S)&=\sum _{k=1}^{n}c_{k}\cdot \mu (S\cap A_{k})\\&=\sum _{k=1}^{n}c_{k}\cdot \mu \left(\left(\coprod _{i=1}^{\infty }S_{i}\right)\cap A_{k}\right)\\&=\sum _{k=1}^{n}c_{k}\cdot \mu \left(\coprod _{i=1}^{\infty }(S_{i}\cap A_{k})\right)\\&=\sum _{k=1}^{n}c_{k}\cdot \sum _{i=1}^{\infty }\mu (S_{i}\cap A_{k})\\&=\sum _{i=1}^{\infty }\sum _{k=1}^{n}c_{k}\cdot \mu (S_{i}\cap A_{k})\\&=\sum _{i=1}^{\infty }\nu (S_{i})\end{aligned}}}$

as required. Here we used that for fixed ${\displaystyle k}$, all the sets ${\displaystyle S_{i}\cap A_{k}}$ are pairwise disjoint, the countable additivity of ${\displaystyle \mu }$ and the fact that since all the summands are non-negative, the sum of the series, be it finite or infinite, is independent of the order of the summation and so summations can be interchanged.

"Continuity from below"[edit]

Lemma 2. Let ${\displaystyle \mu }$ be a measure, and ${\displaystyle S=\bigcup _{i=1}^{\infty }S_{i}}$, where

${\displaystyle S_{1}\subseteq \cdots \subseteq S_{i}\subseteq S_{i+1}\subseteq \cdots \subseteq S}$

is a non-decreasing chain with all its sets ${\displaystyle \mu }$-measurable. Then

${\displaystyle \mu (S)=\sup _{i}\mu (S_{i}).}$

proof (lemma 2)[edit]

Set ${\displaystyle S_{0}=\emptyset }$, then we decompose ${\displaystyle S=\coprod _{1\leq i}S_{i}\setminus S_{i-1}}$ and likewise ${\displaystyle S_{k}=\coprod _{1\leq i\leq k}S_{i}\setminus S_{i-1}}$ as a union of disjoint measurable sets. Therefore ${\displaystyle \mu (S_{k})=\sum _{i=1}^{k}\mu (S_{i}\setminus S_{i-1})}$, and ${\displaystyle \mu (S)=\sum _{i=1}^{\infty }\mu (S_{i}\setminus S_{i-1})}$ so ${\displaystyle \mu (S)=\sup _{k}\mu (S_{k})}$.

Proof of theorem[edit]

The proof can be based on Fatou's lemma instead of a direct proof, because Fatou's lemma can be proved independent of the monotone convergence theorem. The measurability follows directly and the crucial "inverse Inequality" (step 3) is a direct consequence of the inequality

${\displaystyle \int _{X}\liminf _{k}f_{k}(x)\,d\mu \leq \liminf _{k}\int _{X}f_{k}\,d\mu .}$

However the monotone convergence theorem is in some ways more primitive than Fatou's lemma and the following is a direct proof.

Set ${\displaystyle f=\sup _{k}f_{k}}$. Denote by ${\displaystyle \operatorname {SF} (f)}$ the set of simple ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable functions ${\displaystyle s:X\to [0,\infty )}$ (${\displaystyle \infty }$ nor included!) such that ${\displaystyle 0\leq s\leq f}$ on ${\displaystyle X}$.

Step 1. The function ${\displaystyle f}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$–measurable, and the integral ${\displaystyle \textstyle \int _{X}f\,d\mu }$ is well-defined (albeit possibly infinite)^{[4]: section 21.3}

Since from ${\displaystyle 0\leq f_{k}(x)\leq \infty }$ we get ${\displaystyle 0\leq f(x)\leq \infty }$, so it suffices to show that ${\displaystyle f}$ is measurable. To see this, it suffices to prove that the inverse image of an interval ${\displaystyle [0,t]}$ under ${\displaystyle f}$ is an element of the sigma-algebra ${\displaystyle \Sigma }$ on ${\displaystyle X}$. This is because intervals ${\displaystyle [0,t]}$ with ${\displaystyle 0\leq t\leq \infty }$ generate the Borel sigma algebra on the extended non negative reals ${\displaystyle [0,\infty ]}$ by countable intersection and union. Since the ${\displaystyle f_{k}(x)}$ is a non decreasing sequence

${\displaystyle f(x)=\sup _{k}f_{k}(x)\leq t\quad \Leftrightarrow \quad f_{k}(x)\leq t\ \ \forall k.}$

Therefore since ${\displaystyle f\geq 0}$ and ${\displaystyle f_{k}\geq 0}$

${\displaystyle f^{-1}([0,t])=\bigcap _{k}f_{k}^{-1}([0,t]).}$

Hence ${\displaystyle f^{-1}([0,t])}$ is a measurable set, being the countable intersection of measurable sets, each the inverse image of a Borel set under a ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable function ${\displaystyle f_{k}}$. This shows that ${\displaystyle f}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{{\bar {\mathbb {R} }}_{\geq 0}})}$-measurable.

Since ${\displaystyle f\geq 0}$ the integral is defined as a supremum over simple functions

${\displaystyle \int _{X}f\,d\mu =\sup _{s\in SF(f)}\int _{X}s\,d\mu }$

The supremum is always well defined in ${\displaystyle [0,\infty ]}$. However, the sup may be infinite.

Step 2. We have the inequality

${\displaystyle \sup _{k}\int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu }$

This follows directly from the monotonicity of the integral and the definition of supremum: since ${\displaystyle f_{k}(x)\leq f(x)}$ we have ${\displaystyle \int _{X}f_{k}\,d\mu .\leq \int _{X}f\,d\mu }$ for all ${\displaystyle k}$, hence ${\displaystyle \sup _{k}\int _{X}f_{k}\,d\mu .\leq \int _{X}f\,d\mu }$.

step 3 We have the reverse inequality

${\displaystyle \int _{X}f\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }$.

To prove step 3, by the definition of integral as a supremum it suffices to prove

${\displaystyle \int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }$

for every ${\displaystyle s\in \operatorname {SF} (f)}$. To do that we need an "epsilon of room" to manoeuvre.

Given a simple function ${\displaystyle s\in \operatorname {SF} (f)}$ and a positive real number ${\displaystyle 0<t<1}$, define

${\displaystyle B_{k}^{s,t}=\{x\in X\mid t\cdot s(x)\leq f_{k}(x)\}\subseteq X.}$

step 3(a). We have

1. ${\displaystyle B_{k}^{s,t}\subseteq B_{k+1}^{s,t}}$
2. ${\displaystyle X=\bigcup _{k}B_{k}^{s,t}}$
3. ${\displaystyle B_{k}^{s,t}\in \Sigma }$.

Ad 1: the sequence ${\displaystyle f_{k}(x)}$ is non decreasing.

Ad 2: Fix ${\displaystyle x\in X}$. Either ${\displaystyle s(x)=0}$ so ${\displaystyle t\cdot s(x)=0\leq f_{1}(x)}$ and ${\displaystyle x\in B_{1}^{s,t}}$, or ${\displaystyle ts(x)<s(x)\leq f(x)=\sup _{k}f(x)}$ so ${\displaystyle ts(x)<f_{N_{x}}(x)}$ for ${\displaystyle N_{x}}$ sufficiently large and ${\displaystyle x\in B_{N_{x}}^{s,t}}$.

Ad 3: That ${\displaystyle B_{k}^{s,t}}$ is measurable is clear if we know that ${\displaystyle f_{k}-ts}$ is measurable. However that sums and differences of measurable functions are measurable may be circular. So write ${\displaystyle s=\sum _{i=1}^{m}c_{i}\cdot {\mathbf {1} }_{A_{i}}}$, for some finite collection of non-negative constants ${\displaystyle c_{i}\in {\mathbb {R} }_{\geq 0}}$, and measurable sets ${\displaystyle A_{i}\in \Sigma }$ which we may assume are pairwise disjoint and with union ${\displaystyle \textstyle X=\coprod _{i=1}^{m}A_{i}}$ (here ${\displaystyle {\mathbf {1} }_{A_{i}}}$ denotes the indicator function of the set ${\displaystyle A_{i}}$). Then if ${\displaystyle x\in A_{i}}$ we have ${\displaystyle t\cdot s(x)\leq f_{k}(x)}$ if and only if ${\displaystyle f_{k}(x)\in [t\cdot c_{i},\,\infty ].}$ Therefore

${\displaystyle B_{k}^{s,t}=\coprod _{i=1}^{m}{\Bigl (}f_{k}^{-1}{\Bigl (}[t\cdot c_{i},\infty ]{\Bigr )}\cap A_{i}{\Bigr )}.}$

which since the ${\displaystyle f_{k}}$ are measurable, is a union of (disjoint) measurable sets, hence measurable.

Step 3(b). For every simple ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable non-negative function ${\displaystyle s_{0}}$,

${\displaystyle \sup _{k}\int _{B_{k}^{s,t}}s_{0}\,d\mu =\int _{X}s_{0}\,d\mu .}$

By Lemma 1, ${\displaystyle \nu (S)=\int _{S}s_{0}\,d\mu }$ defines a measure on ${\displaystyle (\Omega ,\Sigma )}$. Since ${\displaystyle X=\bigcup _{k}B_{k}^{s,t}}$ ( (step 3(a).2)) step 3(b) follows from "continuity from below" (Lemma 2).

Step 3(c). For every simple ${\displaystyle s\in \operatorname {SF} (f)}$,

${\displaystyle \int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .}$

Indeed, by the definition of ${\displaystyle B_{k}^{s,t}}$ and the monotonicity of the Lebesgue integral we have

${\displaystyle \int _{B_{k}^{s,t}}t\cdot s\,d\mu \leq \int _{B_{k}^{s,t}}f_{k}\,d\mu ,}$

and since ${\displaystyle f_{k}\geq 0}$ and ${\displaystyle B_{k}^{s,t}\subseteq X}$,

${\displaystyle \int _{B_{k}^{s,t}}f_{k}\,d\mu \leq \int _{X}f_{k}\,d\mu .}$

Hence by step 3(b):

${\displaystyle \int _{X}t\cdot s\,d\mu =\sup _{k}\int _{B_{k}^{s,t}}t\cdot s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .}$

The left hand side is just a finite sum and the inequality can be rewritten as

${\displaystyle t\int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }$

which gives step 3(c) by taking the supremum over ${\displaystyle t<1}$.

The proof of Beppo Levi's theorem is complete.

Relaxing the monotonicity assumption[edit]

Under similar hypotheses to Beppo Levi's theorem, it is possible to relax the hypothesis of monotonicity.^[5] As before, let ${\displaystyle (\Omega ,\Sigma ,\mu )}$ be a measure space and ${\displaystyle X\in \Sigma }$. Again, ${\displaystyle \{f_{k}\}_{k=1}^{\infty }}$ will be a sequence of ${\displaystyle (\Sigma ,{\mathcal {B}}_{\mathbb {R} _{\geq 0}})}$-measurable non-negative functions ${\displaystyle f_{k}:X\to [0,+\infty ]}$. However, we do not assume they are pointwise non-decreasing. Instead, we assume that ${\textstyle \{f_{k}(x)\}_{k=1}^{\infty }}$ converges for almost every ${\displaystyle x}$, we define ${\displaystyle f}$ to be the pointwise limit of ${\displaystyle \{f_{k}\}_{k=1}^{\infty }}$, and we assume additionally that ${\displaystyle f_{k}\leq f}$ pointwise almost everywhere for all ${\displaystyle k}$. Then ${\displaystyle f}$ is ${\displaystyle (\Sigma ,{\mathcal {B}}_{\mathbb {R} _{\geq 0}})}$-measurable, and ${\textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu }$ exists, and

As before, measurability follows from the fact that ${\textstyle f=\sup _{k}f_{k}}$ 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, by standard properties of limits and monotonicity, Therefore ${\textstyle \liminf \int _{X}f_{k}\,d\mu =\limsup \int _{X}f_{k}\,d\mu }$, and both are equal to ${\textstyle \int _{X}f\,d\mu }$. It follows that ${\textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu }$ exists and equals ${\textstyle \int _{X}f\,d\mu }$.

See also[edit]

• Infinite series
• Dominated convergence theorem

Notes[edit]