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.

(85) together with the

dominated convergence theorem leads to

Using Lebesque's

dominated convergence theorem on time scales [4], 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 theoremSo, 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].