Sentence examples for measurable selection for from inspiring English sources

Since the multivalued operator is measurable (see [64, proposition III4]) there exists a measurable selection for.

Let v 2 be a measurable selection for U (which exists by Kuratowski-Ryll-Nardzewski's selection theorem [21, 22]).

Moreover, in [25] sufficient conditions are given to obtain the existence of a strongly measurable selection for the multimap F ( ⋅, x ) in not necessarily separable Banach spaces.

Since the multivalued operator (U t cap F t,bar{x}(t))) is measurable (Proposition III.4 [36]), there exists a function (v_{2}(t)) which is a measurable selection for U.

Equivalently, it is a measurable selection for the multivalued map (operatorname{Fix} F: Omegato {mathcal {P}}(X)) defined by operatorname{Fix} F omega)=bigl{ xin X: x=f omega,x bigr},quad omegain Omega.

Since the multivalued operator U ( t ) ∩ F ( t, x ̄ ( t ) ) is measurable ([[57], Proposition III.4])), there exists a function v2 t) which is a measurable selection for V.

We define levelwise the Δ-integral of f in ([0,t]_{ mathbb{T}}) (denoted by (int_{[0,t]_{mathbb{T}}}f(t)Delta t) or (int_{0}^{t} f(t)Delta t)) as the set of the integrals of the measurable selections for ([f]^{alpha}), for each (alphain 0,1]).

A consequence of various well-known selection theorems is that if r is a continuous affine map from E onto a metric space F, then there exists a Borel-measurable selection for r.

A measurable function is called a measurable selection of if for every Denote by (2.4).

Furthermore, according to the Pettis measurability theorem (see Theorem 2.4), it is possible to show that the maps t ↦ p i ( t, ⋅, I α φ ( t, ⋅ ), i = 1, 2, are measurable selection of F for every α ∈ L 2 ( Ω ; R ) ; hence condition (F1) is satisfied.

Then there exists a measurable selection of such that, for any, (21).