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Encyclopedia > Dominated convergence theorem

In mathematics, Lebesgue's dominated convergence theorem states that if a sequence { fn : n = 1, 2, 3, ... } of real-valued measurable functions on a measure space S converges almost everywhere, and is "dominated" (explained below) by some nonnegative function g in L1, then Euclid, Greek mathematician, 3rd century BC, as imagined by by Raphael in this detail from The School of Athens. ... Henri Lebesgue Henri Léon Lebesgue (June 28, 1875, Beauvais – July 26, 1941, Paris) was a French mathematician, most famous for his theory of integration. ... In mathematics, the real numbers may be described informally as numbers that can be given by an infinite decimal representation, such as 2. ... In mathematics, measurable functions are well-behaved functions between measurable spaces. ... In mathematics, a measure is a function that assigns a number, e. ... In measure theory (a branch of mathematical analysis), one says that a property holds almost everywhere if the set of elements for which the property does not hold is a null set, i. ...

int_Slim_{nrightarrowinfty} f_n=lim_{nrightarrowinfty}int_S f_n.

It is proven using Fatou's lemma. Fatous lemma establishes an inequality relating the integral (in the sense of Lebesgue) of the limit inferior of a sequence of functions to the limit inferior of the sequence of integrals of the functions. ...


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

|f_n(x)| leq g(x)

for every n and almost every x (i.e., the measure of the set of exceptional values of x is zero). By g in L1, we mean In measure theory (a branch of mathematical analysis), one says that a property holds almost everywhere if the set of elements for which the property does not hold is a null set, i. ... In mathematics, a measure is a function that assigns a number, e. ...

int_Sleft|gright|<infty.

So the theorem provides a sufficient condition under which integration and passing to the pointwise limit commute. The theorem applies also to measurable functions with values in a Banach space, with the dominating function still being non-negative and integrable as above.


That the assumption that the sequence is dominated by some g in L1 can not be dispensed with may be seen as follows: let fn(x) = n if 0 < x < 1/n and fn(x) = 0 otherwise. Any g which dominates the sequence must also dominate h given by h(x) = supn fn(x) for x > 0 (and 0 otherwise). Since

int_0^1 h(x),dx = infty

the order property of the Lebesgue integral tells us that there exists no function in L1 which dominates the sequence. A direct calculation shows integration and point-wise limit do not commute for this sequence:

int_0^1lim_{nrightarrowinfty} f_n(x),dx=0neq 1=lim_{nrightarrowinfty}int_0^1 f_n(x),dx.

In contrast, Lebesgue's monotone convergence theorem does not require a sequence being dominated by an integrable g and instead assumes the given sequence is monotone. It hence states:

lim_{ktoinfty} int f_k(x),dmu(x) = intlim_{ktoinfty} f_k(x),dmu(x)

whenever {fn} is a monotonically increasing sequence of non-negative measurable functions.


  Results from FactBites:
 
Dominated convergence theorem - Wikipedia, the free encyclopedia (257 words)
The theorem assumes that g is "integrable", i.e.,
By contrast, Lebesgue's monotone convergence theorem does not require such finite integral of g.
Either theorem, dominated convergence theorem or monotone convergence theorem, can be shown as a corollary of the Fatou-Lebesgue theorem.
  More results at FactBites »


 

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