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.
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.
Suppose that the sequence converges pointwise to a function
are integrable (in the Lebesgue sense) and In fact, we have the stronger statement
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
However, there is no harm in ignoring the limit inside the null set
, the condition that there is a dominating integrable function
can be relaxed to uniform integrability of the sequence (fn), see Vitali convergence theorem.
Remark 5 The stronger version of the dominated convergence theorem can be reformulated as: if a sequence of measurable complex functions
and almost everywhere bounded in absolute value by an integrable function then
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.
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 |f−fn| is bounded above by an integrable function) which implies that the limit exists and vanishes i.e.
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 x ∈ S \ N and by f(x) = 0 for x ∈ N, 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.
The assumption that the sequence is dominated by some integrable g cannot be dispensed with.
Observe that by the divergence of the harmonic series.
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.
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 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.
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. Let
Lp space), i.e., for every natural number
, i.e.: Idea of the proof: Apply the original theorem to the function sequence
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 dominated convergence theorem applies also to conditional expectations.