Dominated convergence theorem

Inmeasure theory,Lebesgue'sdominated convergence theoremgives a mildsufficient conditionunder which limits and integrals of a sequence of functions can be interchanged. More technically it says that if asequenceof functions is bounded in absolute value by an integrable function and isalmost everywherepoint wiseconvergentto afunctionthen the sequence converges in$L_{1}$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 integrationoverRiemann 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 valuesofrandom variables.

Statement

Lebesgue's dominated convergence theorem.Let$(f_{n})$be a sequence ofcomplex-valuedmeasurable functionson ameasure space$(S,\Sigma,\mu )$.Suppose that the sequenceconverges pointwiseto a function$f$i.e.

$\lim _{n\to \infty }f_{n}(x)=f(x)$

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

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

for all points$x\in S$and all$n$in the index set. Then$f_{n},f$are integrable (in theLebesguesense) and

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

In fact, we have the stronger,

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

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

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

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

Remark 3.If$\mu (S)<\infty$,the condition that there is a dominating integrable function$g$can be relaxed touniform integrabilityof the sequence (f_n), seeVitali convergence theorem.

Remark 4.While$f$is Lebesgue integrable, it is not in generalRiemann integrable.For example, order the rationals in$[0,1]$,and let$f_{n}$be defined on$[0,1]$to take the value 1 on the first n rationals and 0 otherwise. Then$f$is theDirichlet functionon$[0,1]$,which is not Riemann integrable but is Lebesgue integrable.

Remark 5The stronger version of the dominated convergence theorem can be reformulated as: if a sequence of measurable complex functions$f_{n}$is almost everywhere pointwise convergent to a function$f$and almost everywhere bounded in absolute value by an integrable function then$f_{n}\to f$in theBanach space$L_{1}(S,\mu )$

Proof

Without loss of generality,one can assume thatfis real, because one can splitfinto its real and imaginary parts (remember that a sequence of complex numbers convergesif and only ifboth its real and imaginary counterparts converge) and apply thetriangle inequalityat the end.

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

Sincefis 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),

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

for allnand

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

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

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

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

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

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

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

Finally, since

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

we have that

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

The theorem now follows.

If the assumptions hold onlyμ-almosteverywhere, then there exists aμ-nullsetN∈ Σsuch that the functionsf_n1_S \ Nsatisfy the assumptions everywhere onS.Then the functionf(x) defined as the pointwise limit off_n(x) forx∈S \ Nand byf(x) = 0forx∈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 setN,so the theorem continues to hold.

DCT holds even iff_nconverges tofin 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 integrablegcannot be dispensed with. This may be seen as follows: definef_n(x) =nforxin theinterval(0, 1/n]andf_n(x) = 0otherwise. Anygwhich dominates the sequence must also dominate the pointwisesupremumh= sup_nf_n.Observe that

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

by the divergence of 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:

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

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

Bounded convergence theorem

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

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

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

Proof

Since the sequence is uniformly bounded, there is a real numberMsuch that|f_n(x)| ≤Mfor allx∈Sand for alln.Defineg(x) =Mfor allx∈S.Then the sequence is dominated byg.Furthermore,gis 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-almosteverywhere, then there exists aμ-nullsetN∈ Σsuch that the functionsf_n1_S\Nsatisfy the assumptions everywhere onS.

Dominated convergence inL^p-spaces (corollary)

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

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

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

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

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

Extensions

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.

See also

• 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)

References

• Bartle, R.G. (1995).The Elements of Integration and Lebesgue Measure.Wiley Interscience.ISBN9780471042228.
• Royden, H.L. (1988).Real Analysis.Prentice Hall.ISBN9780024041517.
• Weir, Alan J. (1973). "The Convergence Theorems".Lebesgue Integration and Measure.Cambridge: Cambridge University Press. pp. 93–118.ISBN0-521-08728-7.
• Williams, D.(1991).Probability with martingales.Cambridge University Press.ISBN0-521-40605-6.
• Zitkovic, Gordan (Fall 2013)."Lecture10: Conditional Expectation"(PDF).RetrievedDecember 25,2020.