# 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

We prove that if an increasing sequence ${\displaystyle \{a_{n}\}}$ is bounded above, then it is convergent and the limit is ${\displaystyle \sup \limits _{n}\{a_{n}\}}$.

Since ${\displaystyle \{a_{n}\}}$ is non-empty and by assumption, it is bounded above, then, 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 ${\displaystyle c}$ being ${\displaystyle \sup _{n}\{a_{n}\}}$. Then since ${\displaystyle \{a_{n}\}}$ is increasing, if ${\displaystyle n>N,|c-a_{n}|\leq |c-a_{N}|<\varepsilon }$, 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.

### 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

The proof follows directly from the lemmas.

## Convergence of a monotone series

### Theorem

If for all natural numbers j and k, aj,k is a non-negative real number and aj,k ≤ aj+1,k, then (see for instance [2] page 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

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

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 ${\displaystyle {\frac {1}{k!}}}$.

## Lebesgue's monotone convergence theorem

This theorem generalizes the previous one, and is probably the most important monotone convergence theorem. It is also known as Beppo Levi's theorem.

### Theorem

Let (X, Σ, μ) be a measure space. Let ${\displaystyle f_{1},f_{2},\ldots }$  be a pointwise non-decreasing sequence of [0, ∞]-valued Σ–measurable functions, i.e. for every k ≥ 1 and every x in X,

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

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

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

Then f is Σ–measurable and

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

Remark. If the sequence ${\displaystyle (f_{k})}$ satisfies the assumptions μ–almost everywhere, one can find a set N ∈ Σ with μ(N) = 0 such that the sequence ${\displaystyle (f_{k}(x))}$ is non-decreasing for every ${\displaystyle x\notin N}$. The result remains true because for every k,

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

provided that f is Σ–measurable (see for instance [3] section 21.38).

### Proof

We will first show that f is Σ–measurable (see for instance [3] section 21.3). To do this, it is sufficient to show that the inverse image of an interval [0, t] under f is an element of the sigma algebra Σ on X, because (closed) intervals generate the Borel sigma algebra on the reals. Let I = [0, t] be such a subinterval of [0, ∞]. Let

${\displaystyle f^{-1}(I)=\{x\in X\mid f(x)\in I\}.}$

Since I is a closed interval and ${\displaystyle \forall k,f_{k}(x)\leq f(x)}$,

${\displaystyle f(x)\in I\Leftrightarrow f_{k}(x)\in I,~\forall k\in \mathbb {N} .}$

Thus,

${\displaystyle \{x\in X\mid f(x)\in I\}=\bigcap _{k\in \mathbb {N} }\{x\in X\mid f_{k}(x)\in I\}.}$

Note that each set in the countable intersection is an element of Σ because it is the inverse image of a Borel subset under a Σ-measurable function ${\displaystyle f_{k}}$. Since sigma algebras are, by definition, closed under countable intersections, this shows that f is Σ-measurable. In general, the supremum of any countable family of measurable functions is also measurable.

Now we will prove the rest of the monotone convergence theorem. The fact that f is Σ-measurable implies that the expression ${\displaystyle \int f\,\mathrm {d} \mu }$ is well defined.

By the definition of the Lebesgue integral,

${\displaystyle \int f\,\mathrm {d} \mu =\sup \left\{\int g\,\mathrm {d} \mu \mid g\in SF,\ g\leq f\right\},}$

where SF is the set of Σ-measurable simple functions on X. Since ${\displaystyle f_{k}(x)\leq f(x)}$ at every x ∈ X, we have that

${\displaystyle \left\{\int g\,\mathrm {d} \mu \mid g\in SF,\ g\leq f_{k}\right\}\subseteq \left\{\int g\,\mathrm {d} \mu \mid g\in SF,\ g\leq f\right\}.}$

Hence, since the supremum of a subset cannot be larger than that of the whole set, we have that:

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

and the limit on the right exists, since the sequence is monotonic.

We now prove the inequality in the other direction (which also follows from Fatou's lemma), that is we seek to show that

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

It follows from the definition of integral, that there is a non-decreasing sequence (gk) of non-negative simple functions such that gk ≤ f and such that

${\displaystyle \lim _{k}\int g_{k}\,\mathrm {d} \mu =\int f\,\mathrm {d} \mu .}$

It suffices to prove that for each ${\displaystyle k\in \mathbb {N} }$,

${\displaystyle \int g_{k}\,\mathrm {d} \mu \leq \lim _{j}\int f_{j}\,\mathrm {d} \mu }$

because if this is true for each k, then the limit of the left-hand side will also be less than or equal to the right-hand side.

We will show that if gk is a simple function and

${\displaystyle \lim _{j}f_{j}(x)\geq g_{k}(x)\,}$

for every x, then

${\displaystyle \lim _{j}\int f_{j}\,\mathrm {d} \mu \geq \int g_{k}\,\mathrm {d} \mu .}$

Since the integral is linear, we may break up the function ${\displaystyle g_{k}}$ into its constant value parts, reducing to the case in which ${\displaystyle g_{k}}$ is the indicator function of an element B of the sigma algebra Σ. In this case, we assume that ${\displaystyle f_{j}}$ is a sequence of measurable functions whose supremum at every point of B is greater than or equal to one.

To prove this result, fix ε > 0 and define the sequence of measurable sets

${\displaystyle B_{n}=\{x\in B:f_{n}(x)\geq 1-\epsilon \}.\,}$

By monotonicity of the integral, it follows that for any ${\displaystyle n\in \mathbb {N} }$,

${\displaystyle \mu (B_{n})(1-\epsilon )=\int (1-\epsilon )1_{B_{n}}\,\mathrm {d} \mu \leq \int f_{n}\,\mathrm {d} \mu }$

By the assumption that ${\displaystyle \lim _{j}f_{j}(x)\geq g_{k}(x)}$, any x in B will be in ${\displaystyle B_{n}}$ for sufficiently high values of n, and therefore

${\displaystyle \bigcup _{n}B_{n}=B.}$

Thus, we have that

${\displaystyle \int g_{k}\,\mathrm {d} \mu =\int 1_{B}\,\mathrm {d} \mu =\mu (B)=\mu \left(\bigcup _{n}B_{n}\right).}$

Using the monotonicity property of measures, we can continue the above equalities as follows:

${\displaystyle \mu \left(\bigcup _{n}B_{n}\right)=\lim _{n}\mu (B_{n})\leq \lim _{n}(1-\epsilon )^{-1}\int f_{n}\,\mathrm {d} \mu .}$

Taking k → ∞, and using the fact that this is true for any positive ε, the result follows.