Dominated convergence theorem

Lebesgue's dominated convergence theorem.[1] Let ${\displaystyle (f_{n})}$ be a sequence of complex-valued measurable functions on a measure space ${\displaystyle (S,\Sigma ,\mu )}$. Suppose that the sequence converges pointwise to a function ${\displaystyle f}$ and is dominated by some integrable function ${\displaystyle g}$ in the sense that

${\displaystyle |f_{n}(x)|\leq g(x)}$

for all numbers n in the index set of the sequence and all points ${\displaystyle x\in S}$. Then f is integrable (in the Lebesgue sense) and

${\displaystyle \lim _{n\to \infty }\int _{S}|f_{n}-f|\,d\mu =0}$

which also implies

${\displaystyle \lim _{n\to \infty }\int _{S}f_{n}\,d\mu =\int _{S}f\,d\mu }$

Remark 1. The statement "g is integrable" means that measurable function ${\displaystyle g}$ is Lebesgue integrable; i.e.

${\displaystyle \int _{S}|g|\,d\mu <\infty .}$

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

Remark 3. If ${\displaystyle \mu (S)<\infty }$, the condition that there is a dominating integrable function ${\displaystyle g}$ can be relaxed to uniform integrability of the sequence (f_n), see Vitali convergence theorem.

Remark 4. While ${\displaystyle f}$ is Lebesgue integrable, it is not in general Riemann integrable. For example, take f_n to be defined in ${\displaystyle [0,1]}$ so that it is 1/n at rational numbers and zero everywhere else (on the irrationals). The series (f_n) converges pointwise to 0, so f is identically zero, but ${\displaystyle |f_{n}-f|=f_{n}}$ is not Riemann integrable, since its image in every finite interval is ${\displaystyle \{0,1/n\}}$ and thus the upper and lower Darboux integrals are 1/n and 0, respectively.

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),

${\displaystyle |f-f_{n}|\leq |f|+|f_{n}|\leq 2g}$

for all n and

${\displaystyle \limsup _{n\to \infty }|f-f_{n}|=0.}$

The second of these is trivially true (by the very definition of f). Using linearity and monotonicity of the Lebesgue integral,

${\displaystyle \left|\int _{S}{f\,d\mu }-\int _{S}{f_{n}\,d\mu }\right|=\left|\int _{S}{(f-f_{n})\,d\mu }\right|\leq \int _{S}{|f-f_{n}|\,d\mu }.}$

By the reverse Fatou lemma (it is here that we use the fact that |f−f_n| is bounded above by an integrable function)

${\displaystyle \limsup _{n\to \infty }\int _{S}|f-f_{n}|\,d\mu \leq \int _{S}\limsup _{n\to \infty }|f-f_{n}|\,d\mu =0,}$

which implies that the limit exists and vanishes i.e.

${\displaystyle \lim _{n\to \infty }\int _{S}|f-f_{n}|\,d\mu =0.}$

Finally, since

${\displaystyle \lim _{n\to \infty }\left|\int _{S}fd\mu -\int _{S}f_{n}d\mu \right|\leq \lim _{n\to \infty }\int _{S}|f-f_{n}|\,d\mu =0.}$

we have that

${\displaystyle \lim _{n\to \infty }\int _{S}f_{n}\,d\mu =\int _{S}f\,d\mu .}$

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 1_S \ N satisfy the assumptions everywhere on S. Then the function f(x) defined as the pointwise limit of f_n(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 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(x) = n for x in the interval (0, 1/n] and f_n(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = sup_n f_n. Observe that

${\displaystyle \int _{0}^{1}h(x)\,dx\geq \int _{\frac {1}{m}}^{1}{h(x)\,dx}=\sum _{n=1}^{m-1}\int _{\left({\frac {1}{n+1}},{\frac {1}{n}}\right]}{h(x)\,dx}\geq \sum _{n=1}^{m-1}\int _{\left({\frac {1}{n+1}},{\frac {1}{n}}\right]}{n\,dx}=\sum _{n=1}^{m-1}{\frac {1}{n+1}}\to \infty \qquad {\text{as }}m\to \infty }$

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:

${\displaystyle \int _{0}^{1}\lim _{n\to \infty }f_{n}(x)\,dx=0\neq 1=\lim _{n\to \infty }\int _{0}^{1}f_{n}(x)\,dx,}$

because the pointwise limit of the sequence is the zero function. Note that the sequence (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

${\displaystyle \lim _{n\to \infty }\int _{S}{f_{n}\,d\mu }=\int _{S}{f\,d\mu }.}$

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 f is chosen as a measurable function which agrees μ-almost everywhere with the μ-almost everywhere existing pointwise limit.

Let ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ be a measure space, ${\displaystyle 1\leq p<\infty }$ a real number and ${\displaystyle (f_{n})}$ a sequence of ${\displaystyle {\mathcal {A}}}$-measurable functions ${\displaystyle f_{n}:\Omega \to \mathbb {C} \cup \{\infty \}}$.

Assume the sequence ${\displaystyle (f_{n})}$ converges ${\displaystyle \mu }$-almost everywhere to an ${\displaystyle {\mathcal {A}}}$-measurable function ${\displaystyle f}$, and is dominated by a ${\displaystyle g\in L^{p}}$ (cf. Lp space), i.e., for every natural number ${\displaystyle n}$ we have: ${\displaystyle |f_{n}|\leq g}$, μ-almost everywhere.

Then all ${\displaystyle f_{n}}$ as well as ${\displaystyle f}$ are in ${\displaystyle L^{p}}$ and the sequence ${\displaystyle (f_{n})}$ converges to ${\displaystyle f}$ in the sense of ${\displaystyle L^{p}}$, i.e.:

${\displaystyle \lim _{n\to \infty }\|f_{n}-f\|_{p}=\lim _{n\to \infty }\left(\int _{\Omega }|f_{n}-f|^{p}\,d\mu \right)^{\frac {1}{p}}=0.}$

Idea of the proof: Apply the original theorem to the function sequence ${\displaystyle h_{n}=|f_{n}-f|^{p}}$ with the dominating function ${\displaystyle (2g)^{p}}$.

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]

• 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)

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• Zitkovic, Gordan (Fall 2013). "Lecture10: Conditional Expectation" (PDF). Retrieved December 25, 2020.