(mathematics) If a sequence {ƒn} of Lebesgue measurable functions converges almost everywhere to ƒ and if the absolute value of each ƒn is dominated by the same integrable function, then ƒ is integrable and lim ∫ ƒndm = ∫ ƒdm.
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(mathematics) If a sequence {ƒn} of Lebesgue measurable functions converges almost everywhere to ƒ and if the absolute value of each ƒn is dominated by the same integrable function, then ƒ is integrable and lim ∫ ƒndm = ∫ ƒdm.
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In measure theory, a branch of mathematical analysis, Lebesgue's dominated convergence theorem provides sufficient conditions under which two limit processes commute, namely Lebesgue integration and almost everywhere convergence for a sequence of functions. The dominated convergence theorem does not hold for the Riemann integral, and its power and utility are one of the primary theoretical advantages of Lebesgue integration.
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Let f1, f2, f3, ... denote a sequence of real-valued measurable functions on a measure space (S,Σ,μ). Assume that the sequence converges almost everywhere and is dominated by some integrable function g. Then the limit is an integrable function and

To say that the sequence is "dominated" by g means that

for all natural numbers n and all points x in S. By integrable we mean

The convergence of the sequence and domination by g can be relaxed to hold only μ–almost everywhere.
Lebesgue's dominated convergence theorem is a special case of the Fatou–Lebesgue theorem. Below is a direct proof, using Fatou's lemma as the essential tool.
If f denotes the limit of the sequence, then f is also measurable and dominated by g, hence integrable. Furthermore,

for all n and

By the reverse Fatou lemma,

Using linearity and monotonicity of the Lebesgue integral,

and the theorem follows.
That the assumption that the sequence is dominated by some integrable g can not be dispensed with may be seen as follows: define fn(x) = n for x in the interval (0,1/n] and fn(x) = 0 otherwise. Any g which dominates the sequence must also dominate the pointwise supremum h = supn fn. Observe that
![\int_0^1 h(x)\,dx
\ge\int_{1/m}^1 h(x)\,dx
=\sum_{n=1}^{m-1}\int_{\left(\frac1{n+1},\frac1n\right]}n\,dx
=\sum_{n=1}^{m-1}\frac1{n+1}
\to\infty\quad\text{as }m\to\infty](http://wpcontent.answers.com/math/1/9/4/194ea08a721dd5b0f8591832a1d816f2.png)
by the divergence of the 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:

because the pointwise limit of the sequence is the zero function.
One corollary to the dominated convergence theorem is the bounded convergence theorem, which states that if f1, f2, f3, ... is a sequence of uniformly bounded real-valued measurable functions which converges almost everywhere on a bounded measure space (S,∑,μ) (i.e. one in which μ(S) is finite), then the limit is an integrable function and

(The convergence and uniform boundedness of the sequence can be relaxed to hold only μ–almost everywhere.)
To prove the bounded convergence theorem from the dominated convergence theorem, note that since the sequence is uniformly bounded, there is a real number M such that |fn(x)|<M for all (or at least μ-almost all) x in S and for all n. If a function g is defined so that g(x)=M for all x in S, then the sequence is dominated (at least μ-almost everywhere) by g. Furthermore, g is integrable since it is bounded on a set of finite measure. Therefore the result follows from the dominated convergence theorem.
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.
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