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 ${\displaystyle a_{1}\leq a_{2}\leq a_{3}\leq ...\leq K}$ 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 ${\displaystyle 0\leq a_{i,1}\leq a_{i,2}\leq \cdots }$, 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 ${\displaystyle 0\leq f_{1}(x)\leq f_{2}(x)\leq \cdots }$, taking the integral and the supremum can be interchanged with the result being finite if either one is finite.

## Convergence of a monotone sequence of real numbers

### Lemma 1

For a non-decreasing and bounded-above sequence of real numbers

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

the limit ${\displaystyle \lim _{n\to \infty }a_{n))$ exists and equals its supremum:

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

### Lemma 2

For a non-increasing and bounded-below sequence of real numbers

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

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

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

### Proof

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 ${\displaystyle \lim _{n\to \infty }a_{n}=c=\sup _{n}a_{n))$.

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

### Theorem

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

• "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

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 summation order independent sum

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

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

${\displaystyle \sum _{i=1}^{\infty }a_{i}=\lim _{k\to \infty }\sum _{i=1}^{k}a_{i}=\sup _{k}\sum _{i=1}^{k}a_{i}=\sup _{J\subset \mathbb {N} ,|J|<\infty }\sum _{j\in J}a_{j}=\sum _{i\in \mathbb {N} }a_{i},}$

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 ${\displaystyle a_{i,k}\geq 0}$ 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

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

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.

proof: Since ${\displaystyle a_{i,k}\leq \sup _{k}a_{i,k))$ we have ${\displaystyle \sum _{i}a_{i,k}\leq \sum _{i}\sup _{k}a_{i,k))$ so ${\displaystyle \sup _{k}\sum _{i}a_{i,k}\leq \sum _{i}\sup _{k}a_{i,k))$. Conversely, we can interchange sup and sum for finite sums so ${\displaystyle \sum _{i=1}^{N}\sup _{k}a_{i,k}=\sup _{k}\sum _{i=1}^{N}a_{i,k}\leq \sup _{k}\sum _{i=1}^{\infty }a_{i,k))$ hence ${\displaystyle \sum _{i=1}^{\infty }\sup _{k}a_{i,k}\leq \sup _{k}\sum _{i=1}^{\infty }a_{i,k))$.

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

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

## 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, ${\displaystyle \operatorname {\mathcal {B)) _((\bar {\mathbb {R} ))_{\geq 0))}$ 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)

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 a 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

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

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

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

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,

1. ${\displaystyle \textstyle \lim _{k\to \infty }f_{k}(x)=\sup _{k}f_{k}(x)=\limsup _{k\to \infty }f_{k}(x)=\liminf _{k\to \infty }f_{k}(x)}$
2. ${\displaystyle \textstyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\sup _{k}\int _{X}f_{k}\,d\mu =\liminf _{k\to \infty }\int _{X}f_{k}\,d\mu =\limsup _{k\to \infty }\int _{X}f_{k}\,d\mu .}$

(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

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 ${\displaystyle f,g:X\to [0,+\infty ]}$ 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 ${\displaystyle \operatorname {SF} (f)\subseteq \operatorname {SF} (g),}$ hence

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

2. The functions ${\displaystyle f\cdot {\mathbf {1} }_{X_{1)),f\cdot {\mathbf {1} }_{X_{2)),}$ where ${\displaystyle {\mathbf {1} }_{X_{i))}$ is the indicator function of ${\displaystyle X_{i))$, are easily seen to be measurable and ${\displaystyle f\cdot {\mathbf {1} }_{X_{1))\leq f\cdot {\mathbf {1} }_{X_{2))}$. Now apply 1.

##### Lebesgue integral as measure

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 measurable subset ${\displaystyle S\in \Sigma }$, define

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

Then ${\displaystyle \nu _{s))$ is a measure on ${\displaystyle (\Omega ,\Sigma )}$.

#### proof (lemma1)

Write ${\displaystyle s=\sum _{k=1}^{n}c_{k}\cdot {\mathbf {1} }_{A_{k)),}$ with ${\displaystyle c_{k}\in {\mathbb {R} }_{\geq 0))$ and measurable sets ${\displaystyle A_{k}\in \Sigma }$. Then

${\displaystyle \nu _{s}(S)=\sum _{k=1}^{n}c_{k}\mu (S\cap A_{k}).}$

Since finite positive linear combinations of measures are countably additive, to prove countable additivity of ${\displaystyle \nu _{s))$ it suffices to prove that, the set function defined by ${\displaystyle \nu _{A}(S)=\mu (A\cap S)}$ is countably additive for all ${\displaystyle A\in \Sigma }$. But this follows directly from the countable additivity of ${\displaystyle \mu }$.

##### "Continuity from below"

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)

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

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

From ${\displaystyle 0\leq f_{k}(x)\leq \infty }$ we get ${\displaystyle 0\leq f(x)\leq \infty }$. Hence we have to show that ${\displaystyle f}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B)) _((\bar {\mathbb {R} ))_{\geq 0)))}$-measurable. To see this, it suffices to prove that ${\displaystyle f^{-1}([0,t])}$ is ${\displaystyle \Sigma }$-measurable for all ${\displaystyle 0\leq t\leq \infty }$, because the intervals ${\displaystyle [0,t]}$ generate the Borel sigma algebra on the extended non negative reals ${\displaystyle [0,\infty ]}$ by complementing and taking countable intersections, unions.

Now since the ${\displaystyle f_{k}(x)}$ is a non decreasing sequence, ${\displaystyle f(x)=\sup _{k}f_{k}(x)\leq t}$ if and only if ${\displaystyle f_{k}(x)\leq t}$ for all ${\displaystyle k}$. Since we already know that ${\displaystyle f\geq 0}$ and ${\displaystyle f_{k}\geq 0}$ we conclude that

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

Hence ${\displaystyle f^{-1}([0,t])}$ is a measurable set, being the countable intersection of the measurable sets ${\displaystyle f_{k}^{-1}([0,t])}$.

Since ${\displaystyle f\geq 0}$ the integral is well defined (but possibly infinite) as

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

Step 2. We have the inequality

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

This is equivalent to ${\displaystyle \int _{X}f_{k}(x)\,d\mu \leq \int _{X}f(x)\,d\mu }$ for all ${\displaystyle k}$ which follows directly from ${\displaystyle f_{k}(x)\leq f(x)}$ and the monotonicity of the integral.

step 3 We have the reverse inequality

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

By the definition of integral as a supremum step 3 is equivalent to

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

for every ${\displaystyle s\in \operatorname {SF} (f)}$. It is tempting to prove ${\displaystyle \int _{X}s\,d\mu \leq \int _{X}f_{k}\,d\mu }$ for ${\displaystyle k>K_{s))$ sufficiently large, but this does not work e.g. if ${\displaystyle f}$ is simple and the ${\displaystyle f_{k}. We need an "epsilon of room" to manoeuvre.

Given a simple function ${\displaystyle s\in \operatorname {SF} (f)}$ and an ${\displaystyle 0<\varepsilon \ll 1}$, define

${\displaystyle B_{k}^{s,\varepsilon }=\{x\in X\mid (1-\varepsilon )s(x)\leq f_{k}(x)\}\subseteq X.}$

step 3(a). We have

1. ${\displaystyle B_{k}^{s,\varepsilon ))$ is ${\displaystyle \Sigma }$-measurable.
2. ${\displaystyle B_{k}^{s,\varepsilon }\subseteq B_{k+1}^{s,\varepsilon ))$
3. ${\displaystyle X=\bigcup _{k}B_{k}^{s,\varepsilon ))$

Ad 1: Write ${\displaystyle s=\sum _{1\leq i\leq m}c_{i}\cdot {\mathbf {1} }_{A_{i))}$, for 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 ${\displaystyle \textstyle X=\coprod _{i=1}^{m}A_{i))$. Then for ${\displaystyle x\in A_{i))$ we have ${\displaystyle (1-\varepsilon )s(x)\leq f_{k}(x)}$ if and only if ${\displaystyle f_{k}(x)\in [(1-\varepsilon )c_{i},\,\infty ],}$ so

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

which is measurable because the ${\displaystyle f_{k))$ are measurable.

Ad 2: the sequence ${\displaystyle f_{k}(x)}$ is non decreasing.

Ad 3: Fix ${\displaystyle x\in X}$. Either ${\displaystyle s(x)=0}$ so ${\displaystyle (1-\varepsilon )s(x)=0\leq f_{1}(x)}$ hence ${\displaystyle x\in B_{1}^{s,\varepsilon ))$, or ${\displaystyle (1-\varepsilon )s(x) so ${\displaystyle (1-\varepsilon )s(x) for ${\displaystyle N_{x))$ sufficiently large hence ${\displaystyle x\in B_{N_{x))^{s,\varepsilon ))$.

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

${\displaystyle \int _{B_{k}^{s,\varepsilon ))(1-\varepsilon )s\,d\mu \leq \int _{B_{k}^{s,\varepsilon ))f_{k}\,d\mu \leq \int _{X}f_{k}\,d\mu .}$

Hence by lemma 2, "continuity from below" and (3(a).3):

${\displaystyle \int _{X}(1-\varepsilon )s\,d\mu =\sup _{k}\int _{B_{k}^{s,\varepsilon ))(1-\varepsilon )s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu .}$

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

${\displaystyle (1-\varepsilon )\int _{X}s\,d\mu \leq \sup _{k}\int _{X}f_{k}\,d\mu }$

which gives step 3 by first taking the supremum over ${\displaystyle \varepsilon >0}$ and then the supremum over ${\displaystyle s\in \operatorname {SF} (f)}$.

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 ${\displaystyle (\Omega ,\Sigma ,\mu )}$ 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 ${\displaystyle \lim _{k\to \infty }\int _{X}f_{k}\,d\mu =\int _{X}f\,d\mu .}$

## 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 ${\textstyle f=\sup _{k}f_{k))$ almost everywhere. The interchange of limits and integrals is then an easy consequence of Fatou's lemma. One has ${\displaystyle \int _{X}f\,d\mu =\int _{X}\liminf _{k}f_{k}\,d\mu \leq \liminf \int _{X}f_{k}\,d\mu }$ by Fatou's lemma, and then, by standard properties of limits and monotonicity, ${\displaystyle \liminf \int _{X}f_{k}\,d\mu \leq \limsup \int _{X}f_{k}\,d\mu \leq \int _{X}f\,d\mu .}$ 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 ${\textstyle \int _{X}f\,d\mu }$.