Dominated convergence theorem

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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 Riemann-integrable functions is in many cases not Riemann-integrable. 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.

Statement of the theorem

Let {ƒn} be a sequence of real-valued measurable functions on a measure space (S, Σ, μ). Suppose 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 ƒ is integrable and

which also implies

Remarks:

  1. The statement 'g is integrable' is meant in the sense of Lebesgue; that is
  2. The convergence of the sequence and domination by g can be relaxed to hold only μ-almost everywhere provided the measure space (S, Σ, μ) is complete or ƒ is chosen as a measurable function which agrees μ-almost everywhere with the μ-almost everywhere existing pointwise limit. (These precautions are necessary, because otherwise there might exist a non-measurable subset of a μ-null set N ∈ Σ, hence ƒ might not be measurable.)
  3. The condition that there is a dominating integrable function g can be relaxed to uniform integrability of the sequence {ƒn}, see Vitali convergence theorem.

Proof of the theorem

Lebesgue's dominated convergence theorem is a special case of the Fatou–Lebesgue theorem. Below, however, is a direct proof that uses Fatou’s lemma as the essential tool.

Since ƒ is the pointwise limit of the sequence (fn) of measurable functions that is dominated by g, it is also measurable and dominated by g, hence it is integrable. Furthermore (these will be needed later),

for all n and

The second of these is trivially true f (by the very definition of f). Using linearity and monotonicity of the Lebesgue integral,

By the reverse Fatou lemma (it is here that we use the fact that |f-fn| is bounded above by an integrable function)

which implies that the limit exists and vanishes i.e.

The theorem now follows.

If the assumptions hold only μ-almost everywhere, then there exists a μ-null set N ∈ Σ such that the functions ƒn1N satisfy the assumptions everywhere on S. Then ƒ(x) is the pointwise limit of ƒn(x) for xS \ N and ƒ(x) = 0 for xN, hence ƒ is measurable. The values of the integrals are not influenced by this μ-null 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 ƒn(x) = n for x in the interval (0, 1/n] and ƒn(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = supn ƒn. 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 {ƒn} 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 ƒ1, ƒ2, ƒ3, … is a sequence of uniformly bounded real-valued 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 μ-almost everywhere, provided the measure space (S, Σ, μ) is complete or ƒ is chosen as a measurable function which agrees μ-almost everywhere with the μ-almost everywhere existing pointwise limit.

Proof

Since the sequence is uniformly bounded, there is a real number M such that |ƒn(x)| ≤ M for all xS and for all n. Define g(x) = M for all xS. 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 μ-almost everywhere, then there exists a μ-null set N ∈ Σ such that the functions ƒn1N satisfy the assumptions everywhere on S.

Dominated convergence in -spaces (corollary)

Let be a measure space, a real number and a sequence of-measurable functions .

Assume the sequence converges -almost everywhere to an -measurable function , and is dominated by a , i.e., for every holds -almost everywhere.

Then all as well as are in and the sequence converges to in the sense of , i.e.: .

Idea of the proof: Apply the original theorem to the function sequence with the dominating function .

Extensions

The dominated convergence theorem applies also to measurable functions with values in a Banach space, with the dominating function still being non-negative and integrable as above.

See also

References

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