Editing Dominated convergence theorem

{{More footnotes needed|section|date=September 2024}}

{{Short description|Theorem in measure theory}}
In [[measure theory]], [[Henri Lebesgue|Lebesgue]]'s '''dominated convergence theorem''' gives a mild [[sufficient condition]] under which limits and integrals of a sequence of functions can be interchanged. More technically it says that if a [[sequence]] of functions is bounded in absolute value by an integrable function and is [[almost everywhere]] pointwise [[convergence (mathematics)|convergent]] to a [[Function (mathematics)|function]] then the sequence converges in <math>L_1</math> to its pointwise 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 of [[Lebesgue integral|Lebesgue integration]] over [[Riemann integral|Riemann integration]].

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 value]]s of [[random variable]]s.

==Statement==
'''Lebesgue's dominated convergence theorem.'''<ref>For the real case, see {{cite book |last1=Evans |first1=Lawrence C |last2=Gariepy |first2=Ronald F |title=Measure Theory and Fine Properties of Functions |date=2015 |publisher=CRC Press |pages=Theorem 1.19}}</ref> Let <math>(f_n)</math> be a sequence of [[complex number|complex]]-valued [[measurable function]]s on a [[measure space]] {{nowrap|<math>(S,\Sigma,\mu)</math>}}. Suppose that the sequence [[Pointwise convergence|converges pointwise]] to a function <math>f</math> i.e. 
:<math> \lim_{n \to \infty} f_n(x) = f(x)</math>
exists for every <math>x \in S</math>. Assume moreover that the sequence <math>f_n</math> is dominated by some integrable function <math>g</math> in the sense that
: <math> |f_n(x)| \le g(x)</math>
for all points <math>x\in S</math> and all <math>n</math> in the index set. 
Then <math>f_n, f</math> are integrable (in the [[Lebesgue integration|Lebesgue]] sense) and
:<math>\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</math>.
In fact, we have the stronger statement
: <math> \lim_{n\to\infty} \int_S |f_n-f| \, d\mu = 0.</math>


'''Remark 1.''' The statement "<math>g</math> is integrable" means that the measurable function <math>g</math> is Lebesgue integrable; i.e since <math>g \ge 0</math>.
:<math>\int_S g\,d\mu < \infty.</math>

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

'''Remark 3.''' If <math>\mu (S) < \infty</math>, the condition that there is a dominating integrable function <math>g</math> can be relaxed to [[uniformly integrable|uniform integrability]] of the sequence (''f<sub>n</sub>''), see [[Vitali convergence theorem]].

'''Remark 4.''' While <math>f</math> is Lebesgue integrable, it is not in general [[Riemann integrable]]. For example, order the rationals in <math>[0,1]</math>, and let <math>f_n</math> be defined on <math>[0,1]</math> to take the value 1 on the first n rationals and 0 otherwise. Then <math>f</math> is the [[Dirichlet function]] on <math>[0,1]</math>, 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 <math>f_n</math> is almost everywhere pointwise convergent to a function <math>f</math> and almost everywhere bounded in absolute value by an integrable function then <math>f_n \to f</math> in the [[Banach space]] <math>L_1(S, \mu)</math>

==Proof==
[[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''<sub>''n''</sub>) 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),
: <math>    |f-f_n| \le |f| + |f_n| \leq 2g</math>
for all ''n'' and
: <math>    \limsup_{n\to\infty} |f-f_n| = 0.</math>
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]],
: <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>
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)
: <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>
which implies that the limit exists and vanishes i.e.
: <math>\lim_{n\to\infty} \int_S |f-f_n|\,d\mu= 0.</math>
Finally, since
: <math>\lim_{n\to\infty} \left|\int_S fd\mu-\int_S f_nd\mu\right| \leq\lim_{n\to\infty} \int_S |f-f_n|\,d\mu= 0.</math>
we have that
: <math>\lim_{n\to\infty} \int_S f_n\,d\mu= \int_S f\,d\mu.</math>
The theorem now follows.

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>''S'' \ ''N''</sub> satisfy the assumptions everywhere on&nbsp;''S''. Then the function ''f''(''x'') defined as the pointwise limit of ''f<sub>n</sub>''(''x'') for {{nowrap|''x'' ∈ ''S'' \ ''N''}} and by {{nowrap|''f''(''x'') {{=}} 0}} for {{nowrap|''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&nbsp;''N'', so the theorem continues to hold.

DCT holds even if ''f''<sub>''n''</sub> converges to ''f'' in 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 integrable ''g'' cannot 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
: <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>
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:
: <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>
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.

==Bounded convergence theorem==
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|complex]]-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

:<math>\lim_{n\to\infty} \int_S{f_n\,d\mu} = \int_S{f\,d\mu}.</math>

'''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.

===Proof===
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.

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>''S''\''N''</sub> satisfy the assumptions everywhere on&nbsp;''S''.

==Dominated convergence in ''L''<sup>''p''</sup>-spaces (corollary)==
Let <math>(\Omega,\mathcal{A},\mu)</math> be a [[measure space]], {{nowrap|<math> 1\leq p<\infty</math>}} a real number and <math>(f_n)</math> a sequence of <math>\mathcal{A}</math>-measurable functions <math>f_n:\Omega\to\Complex\cup\{\infty\}</math>.

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

Then all <math>f_n</math> as well as <math>f</math> are in <math>L^p</math> and the sequence <math>(f_n)</math> converges to <math>f</math> in [[Lp-space|the sense of <math>L^p</math>]], i.e.:

:<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>

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>.

==Extensions==
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.<ref>Zitkovic 2013, Proposition 10.5.</ref>

==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)

== Notes ==
{{Reflist}}

==References==
* {{cite book
  | last = Bartle | first = R.G.
  | title = The Elements of Integration and Lebesgue Measure
  | year = 1995
  | publisher = Wiley Interscience
  | isbn = 9780471042228
 | url = https://books.google.com/books?id=8oNGAAAAYAAJ
  }}
* {{cite book
  | last = Royden | first = H.L.
  | title = Real Analysis
  | year = 1988
  | publisher = Prentice Hall
  | isbn = 9780024041517
 | url = https://books.google.com/books?id=J4k_AQAAIAAJ
  }}
* {{cite book |first=Alan J. |last=Weir |title=Lebesgue Integration and Measure |location=Cambridge |publisher=Cambridge University Press |year=1973 |pages=93–118 |chapter=The Convergence Theorems |isbn=0-521-08728-7 }}
* {{cite book
  | last = Williams | first = D. | authorlink = David Williams (mathematician)
  | title = Probability with martingales
  | year = 1991
  | publisher = Cambridge University Press
  | isbn = 0-521-40605-6
  }}
*{{cite web
 | url = https://web.ma.utexas.edu/users/gordanz/notes/conditional_expectation.pdf
 | title = Lecture10: Conditional Expectation
 | last = Zitkovic
 | first = Gordan
 | date = Fall 2013
 | access-date = December 25, 2020
}}

{{Measure theory}}

[[Category:Theorems in real analysis]]
[[Category:Theorems in measure theory]]
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