 Monotone convergence theorem

In mathematics, there are several theorems dubbed monotone convergence; here we present some major examples.
Contents
Convergence of a monotone sequence of real numbers
Theorem
If a_{k} is a monotone sequence of real numbers (e.g., if a_{k} ≤ a_{k+1}), then this sequence has a finite limit if and only if the sequence is bounded.^{[1]}
Proof
We prove that if an increasing sequence {a_{n}} is bounded above, then it is convergent and the limit is sup _{n}{a_{n}}.
Since {a_{n}} is nonempty 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 ε > 0, there exists a_{N} such that a_{N} > c − ε, since otherwise c − ε is an upper bound of {a_{n}}, which contradicts to c being sup _{n}{a_{n}}. Then since {a_{n}} is increasing, , hence by definition, the limit of {a_{n}} is sup _{n}{a_{n}}.
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, a_{j,k} is a nonnegative real number and a_{j,k} ≤ a_{j+1,k}, then (see for instance ^{[2]} page 168)
The theorem states that if you have an infinite matrix of nonnegative real numbers such that
 the columns are weakly increasing and bounded, and
 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

 ,
where n approaches infinity (the limit of this series is e). Here the matrix entry in row n and column k is
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 .
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 (X, Σ, μ) be a measure space. Let be a pointwise nondecreasing sequence of [0, ∞]valued Σ–measurable functions, i.e. for every k ≥ 1 and every x in X,
Next, set the pointwise limit of the sequence (f_{n}) to be f. That is, for every x in X,
Then f is Σ–measurable (see for instance ^{[3]} section 21.38) and
 .
Remark. If the sequence (f_{k}) satisfies the assumptions μ–almost everywhere, one can find a set N ∈ Σ with μ(N) = 0 such that the sequence (f_{k}(x)) is nondecreasing for every . The result remains true because for every k,
Proof
We will first show that f is Σ–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 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
On the other hand, since I is a closed interval,
Thus,
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 Σmeasurable function f_{k}. Since sigma algebras are, by definition, closed under countable intersections, this shows that f is Σmeasurable. 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 Σmeasurable implies that the expression is well defined.
We will start by showing that
By the definition of the Lebesgue integral,
where SF is the set of Σmeasurable simple functions on X. Since at every x ∈ X, we have that
Hence, since the supremum of a subset cannot be larger than that of the whole set, we have that:
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
It follows from the definition of integral, that there is a nondecreasing sequence (g_{k}) of nonnegative simple functions such that g_{k} ≤ f and such that
It suffices to prove that for each ,
because if this is true for each k, then the limit of the lefthand side will also be less than or equal to the righthand side.
We will show that if g_{k} is a simple function and
for every x, then
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 Σ. 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
By monotonicity of the integral, it follows that for any ,
By the assumption that , any x in B will be in B_{n} for sufficiently high values of n, and therefore
Thus, we have that
Using the monotonicity property of measures, we can continue the above equalities as follows:
Taking k → ∞, and using the fact that this is true for any positive ε, the result follows.
See also
 Infinite series
 Dominated convergence theorem
Notes
 ^ 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.
 ^ J Yeh (2006). Real analysis. Theory of measure and integration.
 ^ Erik Schechter (1997). Analysis and Its Foundations.
Categories: Theorems in calculus
 Sequences and series
 Theorems in real analysis
 Theorems in measure theory
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