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
If a sequence of real numbers is increasing and bounded above, then its supremum is the limit.
We prove that if an increasing sequence is bounded above, then it is convergent and the limit is .
Since is non-empty and by assumption, it is bounded above, then, by the least-upper-bound property of real numbers, exists and is finite. Now for every , there exists such that , since otherwise is an upper bound of , which contradicts to being . Then since is increasing, if . Hence, by definition, the limit of is
If a sequence of real numbers is decreasing and bounded below, then its infimum is the limit.
The proof is similar to the proof for the case when the sequence is increasing and bounded above.
- "If"-direction: The proof follows directly from the lemmas.
- "Only If"-direction: By definition of limit, every sequence with a finite limit is necessarily bounded.
Convergence of a monotone series
If for all natural numbers j and k, aj,k is a non-negative real number and aj,k ≤ aj+1,k, then:168
The theorem states that if you have an infinite matrix of non-negative 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 by taking the sum of the column limits, namely .
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.
Next, set the pointwise limit of the sequence to be f. That is, for every x in X,
Then f is Σ–measurable and
Remark. If the sequence satisfies the assumptions μ-almost everywhere, one can find a set N ∈ Σ with μ(N) = 0 such that the sequence is non-decreasing for every . The result remains true because for every k,
provided that f is Σ-measurable.(section 21.38)
We will first show that f is Σ–measurable.(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
Since I is a closed interval and ,
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 . 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 is well defined.
We will start by showing that
By the definition of the Lebesgue integral,
where SF is the set of Σ-measurable 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 non-decreasing sequence (gk) of non-negative simple functions such that gk ≤ f and such that
It suffices to prove that for each ,
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
for every x, then
Since the integral is linear, we may break up the function into its constant value parts, reducing to the case in which is the indicator function of an element B of the sigma algebra Σ. In this case, we assume that 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 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.
- A generalisation of this theorem was given by Bibby, John (1974). "Axiomatisations of the average and a further generalisation of monotonic sequences". Glasgow Mathematical Journal 15 (1): 63–65. doi:10.1017/S0017089500002135.
- See for instance Yeh, J. (2006). Real Analysis: Theory of Measure and Integration. Hackensack, NJ: World Scientific. ISBN 981-256-653-8.
- See for instance Schechter, Erik (1997). Handbook of Analysis and Its Foundations. San Diego: Academic Press. ISBN 0-12-622760-8.