Dominated convergence theorem


In measure theory, Lebesgue's dominated convergence theorem gives a mild sufficient condition under which limits and integrals of a sequence of functions can be interchanged. More technically it says that if a sequence of functions is bounded in absolute value by an integrable function and is almost everywhere point wise convergent to a function then the sequence convergences in to its point wise limit, and in particular the integral of the limit is the limit of the integrals. 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   i.e.


exists for every  . Assume moreover that the sequence   is dominated by some integrable function   in the sense that


for all points   and all   in the index set. Then   are integrable (in the Lebesgue sense) and


In fact, we have the stronger,


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


Remark 2. The convergence of the sequence and domination by   can be relaxed to hold only  -almost everywhere i.e. except possibly on a measurable set   of  -measure  . In fact we can modify the functions   (hence its point wise limit  ) to be 0 on   without changing the value of the integrals. (If we insist on e.g. defining   as the limit whenever it exists, we may end up with a non-measurable subset within   where convergence is violated if the measure space is non complete, and so   might not be measurable. However, there is no harm in ignoring the limit inside the null set  ). We can thus consider the   and   as being defined except for a set of  -measure 0.

Remark 3. If  , the condition that there is a dominating integrable function   can be relaxed to uniform integrability of the sequence (fn), 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.

Remark 5 The stronger version of the dominated convergence theorem can be reformulated as: if a sequence of measurable complex functions   is almost everywhere pointwise convergent to a function   and almost everywhere bounded in absolute value by an integrable function then   in the Banach space  



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 (fn) 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 |ffn| 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 fn 1S \ N satisfy the assumptions everywhere on S. Then the function f(x) defined as the pointwise limit of fn(x) for xS \ N and by f(x) = 0 for xN, is measurable and is the pointwise limit of this modified function sequence. The values of these integrals are not influenced by these changes to the integrands on this μ-null set N, so the theorem continues to hold.

DCT holds even if fn converges to f in measure (finite measure) and the dominating function is non-negative almost everywhere.

Discussion of the assumptions


The assumption that the sequence is dominated by some integrable g cannot be dispensed with. This may be seen as follows: define fn(x) = n for x in the interval (0, 1/n] and fn(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = supn fn. 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 (fn) 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 (fn) 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 |fn(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 fn1S\N satisfy the assumptions everywhere on S.

Dominated convergence in Lp-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   (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]

See also



  1. ^ For the real case, see Evans, Lawrence C; Gariepy, Ronald F (2015). Measure Theory and Fine Properties of Functions. CRC Press. pp. Theorem 1.19.
  2. ^ Zitkovic 2013, Proposition 10.5.


  • Bartle, R.G. (1995). The Elements of Integration and Lebesgue Measure. Wiley Interscience. ISBN 9780471042228.
  • Royden, H.L. (1988). Real Analysis. Prentice Hall. ISBN 9780024041517.
  • Weir, Alan J. (1973). "The Convergence Theorems". Lebesgue Integration and Measure. Cambridge: Cambridge University Press. pp. 93–118. ISBN 0-521-08728-7.
  • Williams, D. (1991). Probability with martingales. Cambridge University Press. ISBN 0-521-40605-6.
  • Zitkovic, Gordan (Fall 2013). "Lecture10: Conditional Expectation" (PDF). Retrieved December 25, 2020.