Editing Dominated convergence theorem (section)

==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''.