# 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 {*ƒ _{n}*} 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

**Remarks:**

- The statement 'g is integrable' is meant in the sense of Lebesgue; that is
- 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.) - The condition that there is a dominating integrable function
*g*can be relaxed to uniform integrability of the sequence {*ƒ*}, see Vitali convergence theorem._{n}

## 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 *(f _{n})* 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-f _{n}|* 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 *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

*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 *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*. Observe that

_{n}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 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 |*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 *μ*-almost everywhere, then there exists a *μ*-null set *N* ∈ Σ such that the functions *ƒ _{n}*

**1**

_{N}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

- 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

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