This section includes a list ofgeneral references, butit lacks sufficient correspondinginline citations. Please help toimprove this section byintroducing more precise citations.(September 2024) (Learn how and when to remove this message)

Inmeasure theory,Lebesgue'sdominated convergence theorem gives a mildsufficient condition under which limits and integrals of a sequence of functions can be interchanged. More technically it says that if asequence of functions is bounded in absolute value by an integrable function and isalmost everywhere point wiseconvergent to afunction then the sequence converges in${\displaystyle L_1}$ 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 ofLebesgue integration overRiemann integration.

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

Lebesgue's dominated convergence theorem.^[1] Let${\displaystyle (f_n)}$ be a sequence ofcomplex-valuedmeasurable functions on ameasure space${\displaystyle (S,\mathrm{\Sigma },\mu )}$. Suppose that the sequenceconverges pointwise to a function${\displaystyle f}$ i.e.

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f(x)}$

exists for every${\displaystyle x\in S}$. Assume moreover that the sequence${\displaystyle f_n}$ is dominated by some integrable function${\displaystyle g}$ in the sense that

${\displaystyle |f_n(x)|\le g(x)}$

for all points${\displaystyle x\in S}$ and all${\displaystyle n}$ in the index set. Then${\displaystyle f_n,f}$ are integrable (in theLebesgue sense) and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_{n}d\mu ={\int }_S\underset{n\to \mathrm{\infty }}{lim}{f}_{n}d\mu ={\int }_{S}fd\mu }$.

In fact, we have the stronger statement

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S|f_n-f|d\mu =0.}$

Remark 1. The statement "${\displaystyle g}$ is integrable" means that the measurable function${\displaystyle g}$ is Lebesgue integrable; i.e since${\displaystyle g\ge 0}$.

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

Remark 3. If, the condition that there is a dominating integrable function${\displaystyle g}$ can be relaxed touniform integrability of the sequence (f_n), seeVitali convergence theorem.

Remark 4. While${\displaystyle f}$ is Lebesgue integrable, it is not in generalRiemann integrable. For example, order the rationals in${\displaystyle [0,1]}$, and let${\displaystyle f_n}$ be defined on${\displaystyle [0,1]}$ to take the value 1 on the first n rationals and 0 otherwise. Then${\displaystyle f}$ is theDirichlet function on${\displaystyle [0,1]}$, 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${\displaystyle f_n}$ is almost everywhere pointwise convergent to a function${\displaystyle f}$ and almost everywhere bounded in absolute value by an integrable function then${\displaystyle f_n\to f}$ in theBanach space${\displaystyle L_1(S,\mu )}$

Without loss of generality, one can assume thatf is real, because one can splitf into its real and imaginary parts (remember that a sequence of complex numbers convergesif and only if both its real and imaginary counterparts converge) and apply thetriangle inequality at the end.

Lebesgue's dominated convergence theorem is a special case of theFatou–Lebesgue theorem. Below, however, is a direct proof that usesFatou’s lemma as the essential tool.

Sincef is the pointwise limit of the sequence (f_n) of measurable functions that are dominated byg, it is also measurable and dominated byg, hence it is integrable. Furthermore, (these will be needed later),

${\displaystyle |f-f_n|\le |f|+|f_n|\le 2g}$

for alln and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}|f-f_n|=0.}$

The second of these is trivially true (by the very definition off). Usinglinearity and monotonicity of the Lebesgue integral,

${\displaystyle \left|{\int }_{S}fd\mu -{\int }_S{f}_{n}d\mu \right|=\left|{\int }_S(f-f_n)d\mu \right|\le {\int }_S|f-f_n|d\mu .}$

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

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{\int }_S|f-f_n|d\mu \le {\int }_S\underset{n\to \mathrm{\infty }}{lim sup}|f-f_n|d\mu =0,}$

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

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S|f-f_n|d\mu =0.}$

Finally, since

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\int }_{S}fd\mu -{\int }_S{f}_{n}d\mu \right|\le \underset{n\to \mathrm{\infty }}{lim}{\int }_S|f-f_n|d\mu =0.}$

we have that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_{n}d\mu ={\int }_{S}fd\mu .}$

The theorem now follows.

If the assumptions hold onlyμ-almost everywhere, then there exists aμ-null setN ∈ Σ such that the functionsf_n1_S \ N satisfy the assumptions everywhere on S. Then the functionf(x) defined as the pointwise limit off_n(x) forx ∈S \ N and byf(x) = 0 forx ∈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 iff_n converges tof in measure (finite measure) and the dominating function is non-negative almost everywhere.

The assumption that the sequence is dominated by some integrableg cannot be dispensed with. This may be seen as follows: definef_n(x) =n forx in theinterval(0, 1/n] andf_n(x) = 0 otherwise. Anyg which dominates the sequence must also dominate the pointwisesupremumh = sup_nf_n. Observe that

${\displaystyle {\int }_0^{1}h(x)dx\ge {\int }_{\frac{1}{m}}^{1}h(x)dx=\sum _{n=1}^{m-1}{\int }_{\left(\frac{1}{n+1},\frac{1}{n}\right]}h(x)dx\ge \sum _{n=1}^{m-1}{\int }_{\left(\frac{1}{n+1},\frac{1}{n}\right]}ndx=\sum _{n=1}^{m-1}\frac{1}{n+1}\to \mathrm{\infty }\text{as }m\to \mathrm{\infty }}$

by the divergence of theharmonic 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:

${\displaystyle {\int }_0^1\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)dx=0\ne 1=\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1{f}_n(x)dx,}$

because the pointwise limit of the sequence is thezero function. Note that the sequence (f_n) is not evenuniformly integrable, hence also theVitali convergence theorem is not applicable.

One corollary to the dominated convergence theorem is thebounded convergence theorem, which states that if (f_n) is a sequence ofuniformly boundedcomplex-valuedmeasurable functions which converges pointwise on a boundedmeasure space(S, Σ, μ) (i.e. one in which μ(S) is finite) to a functionf, then the limitf is an integrable function and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\int }_S{f}_{n}d\mu ={\int }_{S}fd\mu .}$

Remark: The pointwise convergence and uniform boundedness of the sequence can be relaxed to hold onlyμ-almost everywhere, provided the measure space(S, Σ, μ) iscomplete orf is chosen as a measurable function which agrees μ-almost everywhere with theμ-almost everywhere existing pointwise limit.

Since the sequence is uniformly bounded, there is a real numberM such that|f_n(x)| ≤M for allx ∈S and for alln. Defineg(x) =M for allx ∈S. Then the sequence is dominated byg. 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 onlyμ-almost everywhere, then there exists aμ-null setN ∈ Σ such that the functionsf_n1_S\N satisfy the assumptions everywhere on S.

Let${\displaystyle (\mathrm{\Omega },\mathcal{A},\mu )}$ be ameasure space, a real number and${\displaystyle (f_n)}$ a sequence of${\displaystyle \mathcal{A}}$-measurable functions${\displaystyle f_n:\mathrm{\Omega }\to \mathbb{C}\cup \{\mathrm{\infty }\}}$.

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:${\displaystyle |f_n|\le g}$, 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${\displaystyle f}$ inthe sense of${\displaystyle L^p}$, i.e.:

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

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 aBanach space, with the dominating function still being non-negative and integrable as above. The assumption of convergence almost everywhere can be weakened to require onlyconvergence 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)

• 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). RetrievedDecember 25, 2020.

• Theorems in real analysis
• Theorems in measure theory
• Probability theorems

• Articles lacking in-text citations from September 2024
• All articles lacking in-text citations
• Articles with short description
• Short description is different from Wikidata
• Articles containing proofs