In measure theory, Lebesgue's **dominated convergence theorem** provides sufficient conditions under which almost everywhere convergence of a sequence of functions implies convergence in the *L*^{1} norm. Its power and utility are two of the primary theoretical advantages of Lebesgue integration over Riemann integration.

In addition to its frequent appearance in mathematical analysis and partial differential equations, it is widely used in probability theory, since it gives a sufficient condition for the convergence of expected values of random variables.

**Lebesgue's dominated convergence theorem.**^{[1]} Let be a sequence of complex-valued measurable functions on a measure space . Suppose that the sequence converges pointwise to a function and is dominated by some integrable function in the sense that

for all numbers *n* in the index set of the sequence and all points .
Then *f* is integrable (in the Lebesgue sense) and

which also implies

**Remark 1.** The statement "*g* is integrable" means that measurable function is Lebesgue integrable; i.e.

**Remark 2.** The convergence of the sequence and domination by can be relaxed to hold only μ-almost everywhere provided the measure space (*S*, Σ, μ) is complete or 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 might not be measurable.)

**Remark 3.** If , the condition that there is a dominating integrable function can be relaxed to uniform integrability of the sequence (*f _{n}*), see Vitali convergence theorem.

**Remark 4.** While is Lebesgue integrable, it is not in general Riemann integrable. For example, order the rationals in , and let be defined on to take the value 1 on the first n rationals and 0 otherwise. Then is the Dirichlet function on , which is not Riemann integrable but is Lebesgue integrable.

Without loss of generality, one can assume that *f* is real, because one can split *f* into its real and imaginary parts (remember that a sequence of complex numbers converges if and only if both its real and imaginary counterparts converge) and apply the triangle inequality at the end.

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 *f* is the pointwise limit of the sequence (*f*_{n}) of measurable functions that are 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 (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.

Finally, since

we have that

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 *f _{n}*

DCT holds even if *f*_{n} converges to *f* in measure (finite measure) and the dominating function is non-negative almost everywhere.

The assumption that the sequence is dominated by some integrable *g* cannot be dispensed with. This may be seen as follows: define *f _{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 (*f _{n}*) is not even uniformly integrable, hence also the Vitali convergence theorem is not applicable.

One corollary to the dominated convergence theorem is the **bounded convergence theorem**, which states that if (*f*_{n}) is a sequence of uniformly bounded complex-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 *f*, then the limit *f* 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 *f* is chosen as a measurable function which agrees μ-almost everywhere with the μ-almost everywhere existing pointwise limit.

Since the sequence is uniformly bounded, there is a real number *M* such that |*f _{n}*(

If the assumptions hold only μ-almost everywhere, then there exists a μ-null set *N* ∈ Σ such that the functions *f _{n}*

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 (cf. Lp space), i.e., for every natural number we have: , μ-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 .

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. The assumption of convergence almost everywhere can be weakened to require only convergence in measure.

The dominated convergence theorem applies also to conditional expectations.^{[2]}