Monotone convergence theorem

From Wikipedia, the free encyclopedia

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 sketch
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 μ be a measure. If f, f₁, f₂, ... are μ-measurable $[0,\infty]$ -valued functions such that for each k and x, f_k(x) ≤ f_k+1(x), and such that

$\lim_{k\to\infty} f_k=f$ (μ-almost everywhere),

then (see for instance ^[2] section 21.38)

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

[edit] Proof sketch

Let {f_k}_{k ∈ N} be a non-decreasing sequence of non-negative measurable functions and put

$f = \sup_{k \in \mathbb{N}} f_k$

By the monotonicity property of the integral, it is immediate 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

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

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

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

Therefore, it suffices to prove that for each k ∈ N,

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

We will show that if g is a simple function and

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

almost everywhere, then

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

By breaking up the function g into its constant value parts, this reduces to the case in which g is the indicator function of a set. The result we have to prove is then

Suppose A is a measurable set and {f_k}_{k ∈ N} is a nondecreasing sequence of measurable functions on E such that

$\lim_n f_n (x) \geq 1$