# Monotone convergence theorem

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Monotone convergence theorem

In mathematics, there are several theorems dubbed monotone convergence; here we present some major examples.

## Convergence of a monotone sequence of real numbers

### Theorem

If ak is a monotone sequence of real numbers (e.g., if ak ≤ ak+1), then this sequence has a finite limit if and only if the sequence is bounded.

### Proof

We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is sup n{an}.

Since {an} is non-empty and by assumption, it is bounded above, therefore, by the Least upper bound property of real numbers, c = sup n{an} exists and is finite. Now for every ε > 0, there exists aN such that aN > c − ε, since otherwise c − ε is an upper bound of {an}, which contradicts to c being sup n{an}. Then since {an} is increasing, $\forall n > N , |c - a_n| = c - a_n \leq c - a_N < \varepsilon$, hence by definition, the limit of {an} is sup n{an}.

### Remark

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

## 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  page 168) $\lim_{j\to\infty} \sum_k a_{j,k} = \sum_k \lim_{j\to\infty} a_{j,k}.$

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 $(1+1/n)^n=\sum_{k=0}^{n}\binom nk/n^k=\sum_{k=0}^{n}\frac1{k!}\times\frac nn\times\frac{n-1}n\times\cdots\times\frac{n-k+1}n$,

where n approaches infinity (the limit of this series is e). Here the matrix entry in row n and column k is $\binom nk/n^k=\frac1{k!}\times\frac nn\times\frac{n-1}n\times\cdots\times\frac{n-k+1}n;$

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 (1 + 1 / n)n by taking the sum of the column limits, namely $\frac1{k!}$.

## Lebesgue 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 (XΣμ) be a measure space. Let $f_1, f_2, \ldots$  be a pointwise non-decreasing sequence of [0, ∞]-valued Σmeasurable functions, i.e. for every k ≥ 1 and every x in X, $0 \leq f_k(x) \leq f_{k+1}(x).$

Next, set the pointwise limit of the sequence (fn) to be f. That is, for every x in X, $f(x):= \lim_{k\to\infty} f_k(x).$

Then f is Σmeasurable (see for instance  section 21.38) and $\lim_{k\to\infty} \int f_k \, \mathrm{d}\mu = \int f \, \mathrm{d}\mu.$.

Remark. If the sequence (fk) satisfies the assumptions μ–almost everywhere, one can find a set N ∈ Σ with μ(N) = 0 such that the sequence (fk(x)) is non-decreasing for every $x \notin N$. The result remains true because for every k, $\int f_k \, \mathrm{d}\mu = \int_{X \backslash N} f_k \, \mathrm{d}\mu, \ \text{and} \ \int f \, \mathrm{d}\mu = \int_{X \backslash N} f \, \mathrm{d}\mu.$

### Proof

We will first show that f is Σmeasurable. 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, ∞]. Then $f^{-1}(I) = \{x\in X \,|\, f(x)\in I \}.$

On the other hand, since I is a closed interval, $f(x)\in I \Leftrightarrow f_k(x)\in I, ~ \forall k\in \mathbb{N}.$

Thus, $\{x\in X \,|\, f(x)\in I\} = \bigcap_{k\in \mathbb{N}} \{x\in X \,|\, f_k(x)\in I\}.$

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 fk. 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 $\int f \, \mathrm{d}\mu$ is well defined.

We will start by showing that $\int f \, \mathrm{d} \mu \geq \lim_k \int f_k \, \mathrm{d} \mu.$

By the definition of the Lebesgue integral, $\int f \, \mathrm{d} \mu = \sup \{\int g \, \mathrm{d} \mu \,|\, g \in SF, \ g\leq f \},$

where SF is the set of Σ-measurable simple functions on X. Since $f_k(x)\leq f(x)$ at every x ∈ X, we have that $\left\{\int g \, \mathrm{d} \mu \,|\, g \in SF, \ g\leq f_k \right\}\subseteq \left\{\int g \, \mathrm{d} \mu \,|\, g \in SF, \ g\leq f \right\}.$

Hence, since the supremum of a subset cannot be larger than that of the whole set, we have that: $\int f \, \mathrm{d} \mu \geq \lim_k \int f_k \, \mathrm{d} \mu,$

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 $\int f \, \mathrm{d} \mu \leq \lim_k \int f_k \, \mathrm{d} \mu.$

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 $\lim_k \int g_k \, \mathrm{d} \mu = \int f \, \mathrm{d} \mu.$

It suffices to prove that for each $k\in \mathbb{N}$, $\int g_k \, \mathrm{d}\mu \leq \lim_j \int f_j \, \mathrm{d}\mu$

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 $\lim_j f_j(x) \geq g_k(x)$

for every x, then $\lim_j \int f_j \, \mathrm{d} \mu \geq \int g_k \, \mathrm{d} \mu.$

Since the integral is linear, we may break up the function gk into its constant value parts, reducing to the case in which gk is the indicator function of an element B of the sigma algebra Σ. In this case, we assume that fj 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 $B_n = \{x \in B: f_n(x) \geq 1 - \epsilon \}.$

By monotonicity of the integral, it follows that for any $n\in \mathbb{N}$, $\mu(B_n) (1 - \epsilon) = \int (1 - \epsilon) 1_{B_n} \, \mathrm{d} \mu \leq \int f_n \, \mathrm{d} \mu$

By the assumption that $\lim_j f_j(x) \geq g_k(x)$, any x in B will be in Bn for sufficiently high values of n, and therefore $\bigcup_n B_n = B.$

Thus, we have that $\int g_k \, \mathrm{d} \mu =\int 1_B \, \mathrm{d}\mu = \mu(B) = \mu(\bigcup_n B_n) .$

Using the monotonicity property of measures, we can continue the above equalities as follows: $\mu(\bigcup_n B_n)=\lim_n \mu(B_n) \leq \lim_n (1 - \epsilon)^{-1} \int f_n \, \mathrm{d}\mu.$

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

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