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| In [[measure theory]], [[Henri Lebesgue|Lebesgue]]'s '''dominated convergence theorem''' provides [[sufficient condition]]s under which two [[Limit (mathematics)|limit processes]] commute, namely [[Lebesgue integral|Lebesgue integration]] and [[almost everywhere]] [[convergence (mathematics)|convergence]] of a [[sequence]] of [[Function (mathematics)|functions]]. The dominated convergence theorem does not hold for the [[Riemann integral]] because the limit of a sequence of Riemann-integrable functions is in many cases not Riemann-integrable. Its power and utility are two of the primary theoretical advantages of Lebesgue integration over Riemann integration.
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| It is widely used in [[probability theory]], since it gives a sufficient condition for the convergence of [[expected value]]s of [[random variable]]s.
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| ==Statement of the theorem==
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| <blockquote>'''Lebesgue's Dominated Convergence Theorem.''' Let {''f<sub>n</sub>''} be a sequence of [[real number|real]]-valued [[measurable function]]s on a [[measure space]] {{nowrap|(''S'', Σ, μ)}}. Suppose that the sequence converges pointwise to a function ''f'' and is dominated by some integrable function ''g'' in the sense that
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| : <math> |f_n(x)| \le g(x)</math>
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| for all numbers ''n'' in the index set of the sequence and all points ''x'' ∈ ''S''.
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| Then ''f'' is integrable and
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| : <math> \lim_{n\to\infty} \int_S |f_n-f|\,d\mu = 0</math>
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| which also implies
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| :<math>\lim_{n\to\infty} \int_S f_n\,d\mu = \int_S f\,d\mu</math></blockquote>
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| '''Remark 1.''' The statement "''g'' is integrable" is meant in the sense of Lebesgue; that is
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| :<math>\int_Sg\,d\mu < \infty.</math>
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| '''Remark 2.''' The convergence of the sequence and domination by ''g'' can be relaxed to hold only {{nowrap|μ-}}[[almost everywhere]] provided the measure space {{nowrap|(''S'', Σ, μ)}} is [[measure (mathematics)#Completeness|complete]] or ''f'' is chosen as a measurable function which agrees {{nowrap|μ-almost}} everywhere with the {{nowrap|μ-almost}} everywhere existing pointwise limit. (These precautions are necessary, because otherwise there might exist a [[non-measurable set|non-measurable subset]] of a {{nowrap|μ-null}} set {{nowrap|''N'' ∈ Σ}}, hence ''f'' might not be measurable.)<br>
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| '''Remark 3.''' If μ(''S'') < ∞, the condition that there is a dominating integrable function ''g'' can be relaxed to [[uniformly integrable|uniform integrability]] of the sequence {''f<sub>n</sub>''}, see [[Vitali convergence theorem]].
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| ==Proof of the theorem==
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| 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.
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| Since ''f'' is the pointwise limit of the sequence ''(f<sub>n</sub>)'' of measurable functions that is dominated by ''g'', it is also measurable and dominated by ''g'', hence it is integrable. Furthermore (these will be needed later),
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| : <math> |f-f_n| \le |f| + |f_n| \leq 2g</math> | |
| for all ''n'' and
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| : <math> \limsup_{n\to\infty} |f-f_n| = 0.</math>
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| The second of these is trivially true (by the very definition of ''f''). Using [[Lebesgue integral#Basic theorems of the Lebesgue integral|linearity and monotonicity of the Lebesgue integral]],
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| : <math> \left | \int_S{f\,d\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}.</math>
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| By the [[reverse Fatou lemma]] (it is here that we use the fact that |''f''−''f<sub>n</sub>''| is bounded above by an integrable function)
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| : <math>\limsup_{n\to\infty} \int_S |f-f_n|\,d\mu \le \int_S \limsup_{n\to\infty} |f-f_n|\,d\mu = 0,</math>
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| which implies that the limit exists and vanishes i.e.
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| : <math>\lim_{n\to\infty} \int_S |f-f_n|\,d\mu= 0.</math>
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| The theorem now follows.
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| If the assumptions hold only {{nowrap|μ-almost}} everywhere, then there exists a {{nowrap|μ-null}} set {{nowrap|''N'' ∈ Σ}} such that the functions ''f<sub>n</sub>'' '''1'''<sub>''N''</sub> satisfy the assumptions everywhere on ''S''. Then ''f''(''x'') is the pointwise limit of ''f<sub>n</sub>''(''x'') for {{nowrap|''x'' ∈ ''S'' \ ''N''}} and {{nowrap|''f''(''x'') {{=}} 0}} for {{nowrap|''x'' ∈ ''N''}}, hence ''f'' is measurable. The values of the integrals are not influenced by this μ-null set ''N''.
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| ==Discussion of the assumptions==
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| The assumption that the sequence is dominated by some integrable ''g'' can '''not''' be dispensed with. This may be seen as follows: define {{nowrap|''f<sub>n</sub>''(''x'') {{=}} ''n''}} for ''x'' in the [[interval (mathematics)|interval]] {{nowrap|(0, 1/''n'']}} and {{nowrap|''f''<sub>''n''</sub>(''x'') {{=}} 0}} otherwise. Any ''g'' which dominates the sequence must also dominate the pointwise [[supremum]] {{nowrap|''h'' {{=}} sup<sub>''n''</sub> ''f<sub>n</sub>''}}. Observe that
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| : <math>\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]}{n\,dx}=\sum_{n=1}^{m-1} \frac{1}{n+1} \to \infty \qquad \text{as }m\to\infty </math>
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| by the divergence of the [[harmonic series (mathematics)|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:
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| : <math>\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,</math>
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| because the pointwise limit of the sequence is the [[zero function]]. Note that the sequence {''f<sub>n</sub>''} is not even [[uniformly integrable]], hence also the [[Vitali convergence theorem]] is not applicable.
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| == Bounded convergence theorem ==
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| One corollary to the dominated convergence theorem is the '''bounded convergence theorem''', which states that if {''f<sub>n</sub>''} is a sequence of [[uniform boundedness|uniformly bounded]] [[real number|real]]-valued [[measurable function]]s which converges pointwise on a bounded [[measure space]] {{nowrap|(''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
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| :<math>\lim_{n\to\infty} \int_S{f_n\,d\mu} = \int_S{f\,d\mu}.</math>
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| '''Remark:''' The pointwise convergence and uniform boundedness of the sequence can be relaxed to hold only {{nowrap|μ-}}[[almost everywhere]], provided the measure space {{nowrap|(''S'', Σ, μ)}} is [[measure (mathematics)#Completeness|complete]] or ''f'' is chosen as a measurable function which agrees μ-almost everywhere with the {{nowrap|μ-almost}} everywhere existing pointwise limit.
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| === Proof ===
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| Since the sequence is uniformly bounded, there is a real number ''M'' such that {{nowrap|{{!}}''f<sub>n</sub>''(''x''){{!}} ≤ ''M''}} for all {{nowrap|''x'' ∈ ''S''}} and for all ''n''. Define {{nowrap|''g''(''x'') {{=}} ''M''}} for all {{nowrap|''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. Therefore the result follows from the dominated convergence theorem.
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| If the assumptions hold only {{nowrap|μ-almost}} everywhere, then there exists a {{nowrap|μ-null}} set {{nowrap|''N'' ∈ Σ}} such that the functions ''f<sub>n</sub>'''''1'''<sub>''N''</sub> satisfy the assumptions everywhere on ''S''.
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| == Dominated convergence in ''L''<sup>''p''</sup>-spaces (corollary) ==
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| Let <math>(\Omega,\mathcal{A},\mu)</math> be a [[measure space]], {{nowrap| 1 ≤ ''p'' < ∞}} a real number and {''f<sub>n</sub>''} a sequence of <math>\mathcal{A}</math>-measurable functions <math>f_n:\Omega\to\R\cup\{\infty\}</math>.
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| Assume the sequence {''f<sub>n</sub>''} converges μ-almost everywhere to an <math>\mathcal{A}</math>-measurable function ''f'', and is dominated by a <math>g \in L^p</math>, i.e., for every natural number ''n'' we have: |''f<sub>n</sub>''| ≤ ''g'', μ-almost everywhere.
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| Then all ''f<sub>n</sub>'' as well as ''f'' are in <math>L^p</math> and the sequence {''f<sub>n</sub>''} converges to ''f'' in [[Lp-space|the sense of <math>L^p</math>]], i.e.:
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| :<math>\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.</math>
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| Idea of the proof: Apply the original theorem to the function sequence <math>h_n = |f_n-f|^p</math> with the dominating function <math>(2g)^p</math>.
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| ==Extensions==
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| 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]].
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| ==See also==
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| * [[Convergence of random variables]], [[Convergence in mean]]
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| * [[Monotone convergence theorem]] (does not require domination by an integrable function but assumes monotonicity of the sequence instead)
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| * [[Scheffé’s lemma]]
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| * [[Uniform integrability]]
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| * [[Vitali convergence theorem]] (a generalization of Lebesgue's dominated convergence theorem)
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| ==References==
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| {{refbegin}}
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| * {{cite book
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| | last = Bartle | first = R.G.
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| | title = The elements of integration and Lebesgue measure
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| | year = 1995
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| | publisher = Wiley Interscience
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| | ref = harv
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| }}
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| * {{cite book
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| | last = Royden | first = H.L.
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| | title = Real analysis
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| | year = 1988
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| | publisher = Prentice Hall
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| | ref = harv
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| }}
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| * {{cite book
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| | last = Williams | first = D. | authorlink = David Williams (mathematician)
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| | title = Probability with martingales
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| | year = 1991
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| | publisher = Cambridge University Press
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| | isbn = 0-521-40605-6
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| | ref = harv
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| }}
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| {{refend}}
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| [[Category:Theorems in real analysis]]
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| [[Category:Theorems in measure theory]]
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| [[Category:Probability theorems]]
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| [[Category:Articles containing proofs]]
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