# dominated convergence theorem

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## dominated convergence theorem

[′däm·ə‚nād·əd kən′vər·jəns ‚thir·əm]
(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 ∫ ƒ ndm = ∫ ƒ dm.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
thanks to Lebesgue's dominated convergence theorem.
By the dominated convergence theorem and Theorem 5 we have the following theorem.
Similarly, using the continuity and the growth assumptions in (A2) and (A3) and the fact that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] in E, P-a.s, once again it follows from Lebesgue dominated convergence theorem that
Then the dominated convergence theorem yields that [lim.sub.n [right arrow] [infinity]] ([C.sub.p]f)([x.sub.n]) = ([C.sub.p]f)(x).
From Lebesgue's dominated convergence theorem and the continuity of f and h we obtain, for all t [member of] I
If [g.sub.[alpha]] ([lambda]) is continuous in a then so is [r.sub.[alpha]] ([lambda]); from the Lebesgue dominated convergence theorem it follows that [psi] (., [y.sub.[delta]]) is continuous, thus (A4) holds.
Using Lebesque's dominated convergence theorem on time scales , F is completely continuous on [[0,T].sub.T].
Using Lebesgue's dominated convergence theorem and taking the limit in (3.5), (3.6) we deduce that x(.) is a solution of (1.1).
Now since [G.sub.D] is continuous outside the diagonal, we deduce by the dominated convergence theorem, ([H.sub.2]) and Proposition 1.5 (iii), that
Letting [delta] [right arrow] [0.sup.+], we have by the dominated convergence theorem
So, from Lebesgue's dominated convergence theorem, Lemma 1.2 and Corollary 1.3, it follows that [A.sub.n] strongly statistically converges to I - [E.sub.o] as n [right arrow] [infinity].
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