# Dominated convergence theorem

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 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} denote a sequence of real-valued 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 $|f_n(x)| \le g(x)$

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 $\lim_{n\to\infty} \int_S f_n\,d\mu = \int_S f\,d\mu.$

(By integrable we mean $\int_S|g|\,d\mu < \infty.$)

Remarks:

1. 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.)
2. 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 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, $|f-f_n| \le 2g$

for all n and $\limsup_{n\to\infty} |f-f_n| = 0.$

By the reverse Fatou lemma, $\limsup_{n\to\infty} \int_S |f-f_n|\,d\mu \le \int_S \limsup_{n\to\infty} |f-f_n|\,d\mu = 0.$

Using linearity and monotonicity of the Lebesgue integral, $\biggl| \int_S{f\,d\mu} - \int_S{f_n\,d\mu} \biggr| = \biggl| \int_S{(f-f_n)\,d\mu} \biggr| \le \int_S{|f-f_n|\,d\mu} ,$

and the theorem 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 $\int_0^1 h(x)\,dx \ge \int_{1/m}^1{h(x)\,dx} = \sum_{n=1}^{m-1} \int_{\left(\frac1{n+1},\frac1n\right]}{n\,dx} = \sum_{n=1}^{m-1} \frac{1}{n+1} \to \infty \quad \text{as }m\to\infty$

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: $\int_0^1 \lim_{n\to\infty} f_n(x)\,dx = 0 \neq 1 = \lim_{n\to\infty}\int_0^1 f_n(x)\,dx,$

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 $\lim_{n\to\infty} \int_S{f_n\,d\mu} = \int_S{f\,d\mu}.$

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.

## 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.

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