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Tt < . . < tz = 1, + --f 28 2. Sample Calculus Approach where t; = k2-" for k = 0,1,2,. . ,2". We construct a sequence of ([0,1] x measurable functions as follows: a)- for t E [ti,ti+l], o E a, x,(t,o) = x ( t ; , o ) for k = 0,1,2,. . ,2". It is obvious that x ( t i )E R[Q,R"] and the sequence xn(t,o)is product-measurable on [0,1] x Q. O n the other hand, for IIX,(t,w) - x(t,o)(l = Ilx(t;,,o)- x(t,w)lJ t E [t"kti+l]. s. continuity, it then follows that for fixed t, { P w : lim [IIx,,(t,o) - x ( t , o ) l l ] # O n+m 1 = 0.

3 and the facts that f,(t, x,o)is sample continuous in x for fixed t and the solution process x(t, t o , xo, o)is sample continuous in (to,xo)for fixed t. 7. Assume also that A(t, o)is as. sample Lebesgueintegrable on J . 1) with the initial condition yo(w)= ek, k = 1, 2,. . ,n. Clearly, @(to,o)= unit matrix, and @(t,o)satisfies the random matrix differential equation @'(t,o) = A(t,co)@(t,w), @(t0,o) = unit matrix. 2) The following result is analogous to the corresponding deterministic result, whose proof can be formulated similarly.

1) with respect to the initial conditions ( t o 7 xo). 3. 1 hold. 1) through (to, uo) are sample continuous with respect to the initial conditions ( t o , u o ) . Then the solutions 2. s. unique and sample continuous with respect to the initial conditions ( t o ,xo). Proof. 1. 2. 13) where ( t o ,x,) is considered a random parameter. 2. 1)through (t,, x,) and ( t 2 ,x2), respectively. Without loss of generality, assume that t z I t,. Define m ( 4 4 = Ilx,(t,o) -x2(t,49 m(t,,w) = 11x1 - XZ(tl,w)11.

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