Editing Dominated convergence theorem (section)

==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.