Monotone convergence theorem

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

Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be such a sequence, and let ${\displaystyle \{a_{n}\}}$ be the set of terms of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$. By assumption, ${\displaystyle \{a_{n}\}}$ is non-empty and bounded above. By the least-upper-bound property of real numbers, ${\textstyle c=\sup _{n}\{a_{n}\}}$ exists and is finite. Now, for every ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle N}$ such that ${\displaystyle a_{N}>c-\varepsilon }$, since otherwise ${\displaystyle c-\varepsilon }$ is an upper bound of ${\displaystyle \{a_{n}\}}$, which contradicts the definition of ${\displaystyle c}$. Then since ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is increasing, and ${\displaystyle c}$ is its upper bound, for every ${\displaystyle n>N}$, we have ${\displaystyle |c-a_{n}|\leq |c-a_{N}|<\varepsilon }$. Hence, by definition, the limit of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is ${\textstyle \sup _{n}\{a_{n}\}.}$

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

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

If ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ 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]

• "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.

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

${\displaystyle \lim _{j\to \infty }\sum _{k}a_{j,k}=\sum _{k}\lim _{j\to \infty }a_{j,k}.}$

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

${\displaystyle \left(1+{\frac {1}{n}}\right)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {1}{n^{k}}}=\sum _{k=0}^{n}{\frac {1}{k!}}\times {\frac {n}{n}}\times {\frac {n-1}{n}}\times \cdots \times {\frac {n-k+1}{n}},}$

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

${\displaystyle {\binom {n}{k}}{\frac {1}{n^{k}}}={\frac {1}{k!}}\times {\frac {n}{n}}\times {\frac {n-1}{n}}\times \cdots \times {\frac {n-k+1}{n}};}$

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 ${\displaystyle (1+1/n)^{n}}$ by taking the sum of the column limits, namely ${\displaystyle {\frac {1}{k!}}}$.

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

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

Set the pointwise limit of the sequence ${\displaystyle \{f_{n}\}}$ to be ${\displaystyle f}$. That is, for every ${\displaystyle x\in X}$,

${\displaystyle f(x):=\lim _{k\to \infty }f_{k}(x).}$

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

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

Remark 1. The integrals may be finite or infinite.

Remark 2. 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 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 ${\displaystyle f,g:X\to [0,+\infty ]}$ be ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-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 ${\displaystyle \operatorname {SF} (h)}$ 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 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 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

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

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.

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 Remark 5. Here, we will only prove countable additivity, leaving the rest up to the reader. Let ${\displaystyle S=\bigcup _{i=1}^{\infty }S_{i}}$, where all the sets ${\displaystyle S_{i}}$ are pairwise disjoint. Due to simplicity,

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

for some finite non-negative constants ${\displaystyle c_{i}\in {\mathbb {R} }_{\geq 0}}$ and pairwise disjoint sets ${\displaystyle A_{i}\in \Sigma }$ such that ${\displaystyle \bigcup _{i=1}^{n}A_{i}=\Omega }$. By definition of Lebesgue integral,

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

Since all the sets ${\displaystyle S_{j}\cap A_{i}}$ are pairwise disjoint, the countable additivity of ${\displaystyle \mu }$ gives us

${\displaystyle \sum _{i=1}^{n}c_{i}\cdot \mu \left(\bigcup _{j=1}^{\infty }(S_{j}\cap A_{i})\right)=\sum _{i=1}^{n}c_{i}\cdot \sum _{j=1}^{\infty }\mu (S_{j}\cap A_{i}).}$

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,

${\displaystyle {\begin{aligned}\sum _{i=1}^{n}c_{i}\cdot \sum _{j=1}^{\infty }\mu (S_{j}\cap A_{i})&=\sum _{j=1}^{\infty }\sum _{i=1}^{n}c_{i}\cdot \mu (S_{j}\cap A_{i})\\&=\sum _{j=1}^{\infty }\int _{S_{j}}s\,d\mu \\&=\sum _{j=1}^{\infty }\nu (S_{j}),\end{aligned}}}$

as required.

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

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

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

Step 1. We begin by showing that ${\displaystyle f}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$–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 ${\displaystyle [0,t]}$ under ${\displaystyle f}$ is an element of the sigma-algebra ${\displaystyle \Sigma }$ on ${\displaystyle X}$, because (closed) intervals generate the Borel sigma algebra on the reals. Since ${\displaystyle [0,t]}$ is a closed interval, and, for every ${\displaystyle k}$, ${\displaystyle 0\leq f_{k}(x)\leq f(x)}$,

${\displaystyle 0\leq f(x)\leq t\quad \Leftrightarrow \quad {\Bigl [}\forall k\quad 0\leq f_{k}(x)\leq t{\Bigr ]}.}$

Thus,

${\displaystyle \{x\in X\mid 0\leq f(x)\leq t\}=\bigcap _{k}\{x\in X\mid 0\leq f_{k}(x)\leq t\}.}$

Being the inverse image of a Borel set under a ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable function ${\displaystyle f_{k}}$, each set in the countable intersection is an element of ${\displaystyle \Sigma }$. Since ${\displaystyle \sigma }$-algebras are, by definition, closed under countable intersections, this shows that ${\displaystyle f}$ is ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable, and the integral ${\displaystyle \textstyle \int _{X}f\,d\mu }$ is well defined (and possibly infinite).

Step 2. We will first show that ${\displaystyle \textstyle \int _{X}f\,d\mu \geq \lim _{k}\int _{X}f_{k}\,d\mu .}$

The definition of ${\displaystyle f}$ and monotonicity of ${\displaystyle \{f_{k}\}}$ imply that ${\displaystyle f(x)\geq f_{k}(x)}$, for every ${\displaystyle k}$ and every ${\displaystyle x\in X}$. By monotonicity (or, more precisely, its narrower version established in Remark 5; see also Remark 4) of Lebesgue integral,

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

and

${\displaystyle \int _{X}f\,d\mu \geq \lim _{k}\int _{X}f_{k}\,d\mu .}$

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

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

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

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

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

Step 3. Given a simple function ${\displaystyle s\in \operatorname {SF} (f)}$ and a real number ${\displaystyle t\in (0,1)}$, define

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

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

Step 3a. To prove the first claim, let ${\displaystyle \textstyle s=\sum _{i=1}^{m}c_{i}\cdot {\mathbf {1} }_{A_{i}}}$, for some finite collection of pairwise disjoint measurable sets ${\displaystyle A_{i}\in \Sigma }$ such that ${\displaystyle \textstyle X=\cup _{i=1}^{m}A_{i}}$, some (finite) non-negative constants ${\displaystyle c_{i}\in {\mathbb {R} }_{\geq 0}}$, and ${\displaystyle {\mathbf {1} }_{A_{i}}}$ denoting the indicator function of the set ${\displaystyle A_{i}}$.

For every ${\displaystyle x\in A_{i},}$ ${\displaystyle t\cdot s(x)\leq f_{k}(x)}$ holds if and only if ${\displaystyle f_{k}(x)\in [t\cdot c_{i},+\infty ].}$ Given that the sets ${\displaystyle A_{i}}$ are pairwise disjoint,

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

Since the pre-image ${\displaystyle f_{k}^{-1}{\Bigl (}[t\cdot c_{i},+\infty ]{\Bigr )}}$ of the Borel set ${\displaystyle [t\cdot c_{i},+\infty ]}$ under the measurable function ${\displaystyle f_{k}}$ is measurable, and ${\displaystyle \sigma }$-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 ${\displaystyle k}$ and every ${\displaystyle x\in X}$, ${\displaystyle f_{k}(x)\leq f_{k+1}(x).}$

Step 3c. To prove the third claim, we show that ${\displaystyle \textstyle X\subseteq \bigcup _{k}B_{k}^{s,t}}$.

Indeed, if, to the contrary, ${\displaystyle \textstyle X\not \subseteq \bigcup _{k}B_{k}^{s,t}}$, then an element

${\displaystyle \textstyle x_{0}\in X\setminus \bigcup _{k}B_{k}^{s,t}=\bigcap _{k}(X\setminus B_{k}^{s,t})}$

exists such that ${\displaystyle f_{k}(x_{0})<t\cdot s(x_{0})}$, for every ${\displaystyle k}$. Taking the limit as ${\displaystyle k\to \infty }$, we get

${\displaystyle f(x_{0})\leq t\cdot s(x_{0})<s(x_{0}).}$

But by initial assumption, ${\displaystyle s\leq f}$. This is a contradiction.

Step 4. For every simple ${\displaystyle (\Sigma ,\operatorname {\mathcal {B}} _{\mathbb {R} _{\geq 0}})}$-measurable non-negative function ${\displaystyle s_{2}}$,

${\displaystyle \lim _{n}\int _{B_{n}^{s,t}}s_{2}\,d\mu =\int _{X}s_{2}\,d\mu .}$

To prove this, define ${\displaystyle \textstyle \nu (S)=\int _{S}s_{2}\,d\mu }$. By Lemma 1, ${\displaystyle \nu (S)}$ is a measure on ${\displaystyle \Omega }$. By "continuity from below" (Lemma 2),

${\displaystyle \lim _{n}\int _{B_{n}^{s,t}}s_{2}\,d\mu =\lim _{n}\nu (B_{n}^{s,t})=\nu (X)=\int _{X}s_{2}\,d\mu ,}$

as required.

Step 5. We now prove that, for every ${\displaystyle s\in \operatorname {SF} (f)}$,

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

Indeed, using the definition of ${\displaystyle B_{k}^{s,t}}$, the non-negativity of ${\displaystyle f_{k}}$, and the monotonicity of Lebesgue integral (see Remark 5 and Remark 4), we have

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

for every ${\displaystyle k\geq 1}$. In accordance with Step 4, as ${\displaystyle k\to \infty }$, the inequality becomes

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

Taking the limit as ${\displaystyle t\uparrow 1}$ yields

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

as required.

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

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

Indeed, by non-negativity, ${\displaystyle f_{+}=f}$ and ${\displaystyle f_{-}=0.}$ For the calculation below, the non-negativity of ${\displaystyle f}$ is essential. Applying the definition of Lebesgue integral and the inequality established in Step 5, we have

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

The proof is complete.