Editing Dominated convergence theorem (section)

==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>