# Monotone convergence theorem

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.

## Convergence of a monotone sequence of real numbers

### Lemma 1

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

### Proof

Let ${\displaystyle \{a_{n}\}}$ be such a sequence. By assumption, ${\displaystyle \{a_{n}\}}$ is non-empty and bounded above. By the least-upper-bound property of real numbers, ${\displaystyle c=\sup _{n}\{a_{n}\}}$ 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 to the definition of ${\displaystyle c}$. Then since ${\displaystyle \{a_{n}\}}$ 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 ${\displaystyle \{a_{n}\}}$ is ${\displaystyle \sup _{n}\{a_{n}\}.}$

### Lemma 2

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

### Proof

The proof is similar to the proof for the case when the sequence is increasing and bounded above, and is left as an exercise to the reader.

### Theorem

If ${\displaystyle \{a_{n}\}}$ 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]

### Proof

• "If"-direction: The proof follows directly from the lemmas.
• "Only If"-direction: By definition of limit, every sequence ${\displaystyle \{a_{n}\}}$ with a finite limit ${\displaystyle L}$ is necessarily bounded.

## Convergence of a monotone series

### Theorem

If for all natural numbers j and k, aj,k is a non-negative real number and aj+1,k ≤ aj,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

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

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

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

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 ${\displaystyle {\frac {1}{k!}}}$.

## Monotone convergence theorem for Lebesgue integral

The following result is due to Henri Lebesgue and Beppo Levi.

### Theorem

Let ${\displaystyle (\Omega ,\Sigma ,\mu )}$ be a measure space containing a measurable subset ${\displaystyle X\subseteq \Omega }$. Let ${\displaystyle f_{1},f_{2},\ldots }$  be a pointwise non-decreasing sequence of ${\displaystyle [0,\infty ]}$-valued ${\displaystyle \Sigma }$-measurable functions, 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 ${\displaystyle \Sigma }$-measurable and

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

Remark 1. For this result to remain true, it is enough to only require that the assumptions hold ${\displaystyle \mu }$-almost everywhere. In other words, it is enough that there be a subset ${\displaystyle {N\in \Sigma }}$ such that ${\displaystyle {\mu (N)=0}}$ and the sequence ${\displaystyle \{f_{n}(x)\}}$ non-decreases for every ${\displaystyle {x\in X\setminus N}}$. Indeed, since ${\displaystyle {\mu (N)=0}}$, for every ${\displaystyle k}$, we would then have

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

provided that ${\displaystyle f}$ is ${\displaystyle \Sigma }$-measurable.[3](section 21.38)

Remark 2. Under assumptions of the theorem,

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

### Proof

This proof does not rely on Fatou's lemma. For those interested, we do explain how that lemma might be used.

Step 1. We begin by showing that ${\displaystyle f}$ is ${\displaystyle \Sigma }$–measurable.[3](section 21.3)

Note. If we were using Fatou's lemma, the measurability would follow easily from Remark 2(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 ${\displaystyle \Sigma }$-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 ${\displaystyle \Sigma }$-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 ${\displaystyle x\in X}$. It follows that

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

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 of Lebesgue integral, the sequence is non-decreasing.

End of Step 2.

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

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

Note. If we were using Fatou's lemma, this inequality would follow easily from Remark 2, and the remaining steps would not be needed.

To prove the inequality without using Fatou's lemma, we need some extra machinery. Denote ${\displaystyle \operatorname {SF} (f)}$ the set of simple ${\displaystyle \Sigma }$-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

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

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

Indeed, for each ${\displaystyle k}$ and every ${\displaystyle x\in X}$, ${\displaystyle f_{k+1}(x)\geq f_{k}(x)}$, so the first claim follows.

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

Indeed, if, to the contrary, an element

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

exists, then, for every ${\displaystyle k}$, ${\displaystyle f_{k}(x_{0}); or after taking the limit as ${\displaystyle k\to \infty }$, ${\displaystyle f(x_{0})\leq t\cdot s(x_{0}). But by initial assumption, ${\displaystyle s\leq f}$. This is a contradiction.

Step 4. As a corollary from Step 3,

${\displaystyle \lim _{n}\mu (X\setminus B_{n}^{s,t})=0}$.

Step 5. For every simple ${\displaystyle \Sigma }$-measurable non-negative function ${\displaystyle s_{2}}$,

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

Indeed, the function ${\displaystyle s_{2}}$ is bounded from the above by a constant ${\displaystyle c}$. Therefore

${\displaystyle \left|\int _{B_{k}^{s,t}}s_{2}\,d\mu -\int _{X}s_{2}\,d\mu \right|\leq c\cdot \mu (X\setminus B_{k}^{s,t})\to 0}$,

as ${\displaystyle k\to \infty }$.

Step 6. 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, 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 5, 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 7. 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, applying the definition of Lebesgue integral and the inequality established in Step 6, we have

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

as needed. The proof is complete.

Remark 3. The inequality in Step 5 can also be written as

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

as per Remark 2.