This needs to change, and while industry must be supported by promoting a growth-oriented environment, a clear and

measurable set of targets must be set for sectors that have a realistic potential to be globally competitive, such as information technology and pharmaceutical manufacturing.

for any

measurable set E [subset] R with the Lebesgue measure [absolute value of (E)] < [delta].

(ii) F is weakly measurable and there exists a

measurable set E [subset] [a, b] with [mu]([a, b] - E) = 0 such that F(E) is separable.

"If you're gonna create a section, be damn sure those products adhere to a

measurable set of standards," he says, adding that Patron has a companywide focus on educating trade mixologists about its environmentally conscious ways, which includes a state-of-the-art reverse-osmosis system of water treatment.

Assume that S [??] X x X is a

measurable set. Then the family of sections [S.sub.x] = {y [member of] X : (x, y) [member of] X} contains at most continuum of distinct sets and consequently the diagonal D = {(x, x) : x [member of] X} is no longer a

measurable set (see [2], p.

Next we prove that [[OMEGA].sub.P] is a

measurable set. For the latter, let [psi] : N [right arrow] [F.sub.0] be an enumeration of [F.sub.0].

Since Y is a separable Banach space, to use von Neumann selection theorem, it is enough to show that the graph of H, {(t, y) [member of] [OMEGA] x Y: y [member of] H(t)} = {(t,y) [member of] [OMEGA] x Y: [parallel]f (t) - y[parallel] = d(f (t), Y) and [parallel]y[parallel] [less than or equal to] d(0,[P.sub.Y] (f (t))) + [epsilon]} is a

measurable set in the product space.

Let G be a bounded

measurable set, then meas(G+(8)) and meas(G"(8)) are continuous functions of 8.

You need a

measurable set of variables to observe in order to show improvement over time.

where A [member of] F is an arbitrary

measurable set.

However, defining legacy is problematic especially if conceived as an entirely predictable or

measurable set of objectives.

if B is any [T.sup.-1][SIGMA]

measurable set for which [[integral].sub.B] fd[lambda] converges, we have [[integral].sub.B] fd[lambda] = [[integral].sub.B]E(f)d[lambda].