 Dominated convergence theorem

In measure theory, Lebesgue's dominated convergence theorem provides sufficient conditions under which two limit processes commute, namely Lebesgue integration and almost everywhere convergence of a sequence of functions. The dominated convergence theorem does not hold for the Riemann integral because the limit of a sequence of Riemannintegrable functions is in many cases not Riemannintegrable. Its power and utility are two of the primary theoretical advantages of Lebesgue integration over Riemann integration.
It is widely used in probability theory, since it gives a sufficient condition for the convergence of expected values of random variables.
Contents
Statement of the theorem
Let {ƒ_{n}} denote a sequence of realvalued measurable functions on a measure space (S, Σ, μ). Assume that the sequence converges pointwise to a function ƒ and is dominated by some integrable function g in the sense that
for all numbers n in the index set of the sequence and all points x in S. Then the limiting function ƒ is integrable and
(By integrable we mean
 )
Remarks:
 The convergence of the sequence and domination by g can be relaxed to hold only μalmost everywhere provided the measure space (S, Σ, μ) is complete or ƒ is chosen as a measurable function which agrees μalmost everywhere with the μalmost everywhere existing pointwise limit. (These precautions are necessary, because otherwise there might exist a nonmeasurable subset of a μnull set N ∈ Σ, hence ƒ might not be measurable.)
 The condition that there is a dominating integrable function g can be relaxed to uniform integrability of the sequence {ƒ_{n}}, see Vitali convergence theorem.
Proof of the theorem
Lebesgue's dominated convergence theorem is a special case of the Fatou–Lebesgue theorem. Below is a direct proof, using Fatou’s lemma as the essential tool.
As the pointwise limit of the sequence, ƒ is also measurable and dominated by g, hence integrable. Furthermore,
for all n and
By the reverse Fatou lemma,
Using linearity and monotonicity of the Lebesgue integral,
and the theorem follows.
If the assumptions hold only μalmost everywhere, then there exists a μnull set N ∈ Σ such that the functions ƒ_{n}1_{N} satisfy the assumptions everywhere on S. Then ƒ(x) is the pointwise limit of ƒ_{n}(x) for x ∈ S \ N and ƒ(x) = 0 for x ∈ N, hence ƒ is measurable. The values of the integrals are not influenced by this μnull set N.
Discussion of the assumptions
The assumption that the sequence is dominated by some integrable g can not be dispensed with. This may be seen as follows: define ƒ_{n}(x) = n for x in the interval (0, 1/n] and ƒ_{n}(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = sup_{n} ƒ_{n}. Observe that
by the divergence of the harmonic series. Hence, the monotonicity of the Lebesgue integral tells us that there exists no integrable function which dominates the sequence on [0,1]. A direct calculation shows that integration and pointwise limit do not commute for this sequence:
because the pointwise limit of the sequence is the zero function. Note that the sequence {ƒ_{n}} is not even uniformly integrable, hence also the Vitali convergence theorem is not applicable.
Bounded convergence theorem
One corollary to the dominated convergence theorem is the bounded convergence theorem, which states that if ƒ_{1}, ƒ_{2}, ƒ_{3}, … is a sequence of uniformly bounded realvalued measurable functions which converges pointwise on a bounded measure space (S, Σ, μ) (i.e. one in which μ(S) is finite) to a function ƒ, then the limit ƒ is an integrable function and
Remark: The pointwise convergence and uniform boundedness of the sequence can be relaxed to hold only μalmost everywhere, provided the measure space (S, Σ, μ) is complete or ƒ is chosen as a measurable function which agrees μalmost everywhere with the μalmost everywhere existing pointwise limit.
Proof
Since the sequence is uniformly bounded, there is a real number M such that ƒ_{n}(x) ≤ M for all x ∈ S and for all n. Define g(x) = M for all x ∈ S. Then the sequence is dominated by g. Furthermore, g is integrable since it is a constant function on a set of finite measure. Therefore the result follows from the dominated convergence theorem.
If the assumptions hold only μalmost everywhere, then there exists a μnull set N ∈ Σ such that the functions ƒ_{n}1_{N} satisfy the assumptions everywhere on S.
Extensions
The dominated convergence theorem applies also to measurable functions with values in a Banach space, with the dominating function still being nonnegative and integrable as above.
See also
 Convergence of random variables, Convergence in mean
 Monotone convergence theorem (does not require domination by an integrable function but assumes monotonicity of the sequence instead)
 Scheffé’s lemma
 Uniform integrability
 Vitali convergence theorem (a generalization of Lebesgue's dominated convergence theorem)
References
 Bartle, R.G. (1995). The elements of integration and Lebesgue measure. Wiley Interscience.
 Royden, H.L. (1988). Real analysis. Prentice Hall.
 Williams, D. (1991). Probability with martingales. Cambridge University Press. ISBN 0521406056.
Categories: Real analysis
 Theorems in analysis
 Measure theory
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