Monotone convergence theorem

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

1 Convergence of a monotone sequence of real numbers
- 1.1 Theorem
- 1.2 Proof
2 Convergence of a monotone series
- 2.1 Theorem
3 Lebesgue monotone convergence theorem
- 3.1 Theorem
- 3.2 Proof
4 See also
5 References

[edit] Convergence of a monotone sequence of real numbers

[edit] Theorem

If a_k is a monotone sequence of real numbers (e.g., if a_k ≤ a_k+1,) then this sequence has a limit (if we admit plus and minus infinity as possible limits.) The limit is finite if and only if the sequence is bounded. (A generalisation of this theorem was given by John Bibby (1974) “Axiomatisations of the average and a further generalisation of monotonic sequences,” Glasgow Mathematical Journal, vol. 15, pp. 63–65.)

[edit] Proof

We prove that if an increasing sequence $\langle a_n \rangle$ is bounded above, then it is convergent and the limit is $\sup_n \{a_n\}$ .

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

Similarly, if a sequence of real numbers is decreasing and bounded below, then its Infimum is the limit.

[edit] Convergence of a monotone series

[edit] Theorem

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 (see for instance ^[1] page 168)

$\lim_{j\to\infty} \sum_k a_{j,k} = \sum_k \lim_{j\to\infty} a_{j,k}$

[edit] Lebesgue monotone convergence theorem

This theorem generalizes the previous one, and is probably the most important monotone convergence theorem.

[edit] Theorem

Let ( X, A, $μ$ ) be a measure space. Let $f_1, f_2, \ldots$ be $μ$ -measurable $[0,\infty]$ -valued monotonically increasing functions. Equivalently,

$\forall x\in X, k\in \mathbb{N}, f_k(x) \leq f_{k+1}(x)$ .

Next, set the pointwise limit of the sequence $f n$ to be f. That is,

$\forall x\in X, f(x):= \lim_{k\to\infty} f_k(x)$ .

Then f is $μ$ -measurable and (see for instance ^[2] section 21.38)

$\lim_{k\to\infty} \int f_k d\mu = \int f d\mu$ .

[edit] Proof

We will first show that f is $μ$ -measurable. 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 A on X. Let I be such a subinterval of $[0,\infty]$ . Then

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

On the other hand, since [0,t] 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 A because it is the inverse image of a Borel subset under a $μ$ -measurable function $f k$ . Since sigma algebras are, by definition, closed under countable intersections, this shows the f is $μ$ -measurable. It should be noted that, in general, the supremum of any measurable functions is also measurable.

Now we will prove the rest of the monotone convergence theorem. The fact that f is $μ$ -measurable means that the expression $\int f d\mu$ is well defined.

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

By the definition of the Lebesgue integral,

$\int f d \mu = sup \{\int g d \mu | g \in SF, g\leq f \}$ ,

where SF is the set of $μ$ -measurable simple functions on X. Since $f_k(x)\leq f(x)$ at every $x\in X$ , we have that

$\left\{\int g d \mu | g \in SF, g\leq f_k \right\}\subseteq \left\{\int g d \mu | g \in SF, g\leq f \right\}.$

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

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

$\int f d \mu \leq \lim_k \int f_k d \mu.$

It follows from the definition of integral, that there is a non-decreasing sequence g_n of non-negative simple functions which converges to f pointwise almost everywhere and such that

$\lim_k \int g_k d \mu = \int f d \mu.$

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

$\int g_k d \mu \leq \lim_j \int f_j d \mu$

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 g_k is a simple function and

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

almost everywhere, then

$\lim_j \int f_j d \mu \geq \int g_k d \mu.$

Since the integral is linear, we may break up the function $g k$ into its constant value parts, reducing to the case in which $g k$ is the indicator function of an element B of the sigma algebra A. In this case, we assume that $f j$ 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

$B_n = \{x \in B: f_n(x) \geq 1 - \epsilon \}.$