Editing Dominated convergence theorem (section)

===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''.