Implicit Vector Integral Equations Associated with Discontinuous Operators

and Applied Analysis 3 Then, there exist sets Q 1 , . . . , Q k , with Q i ∈ B(X i ) and ψ i (Q i ) = 0 for all i = 1, . . . , k, and a function φ : T×X → S such that (a) φ(t, x) ∈ F(t, x) for all (t, x) ∈ T × X; (b) for all x := (x 1 , x 2 , . . . , x k ) ∈ X \ [(⋃ k i=1 P −1 ∗,i (Q i )) ∪E], the function φ(⋅, x) isT μ -measurable over T; (c) for a.e. t ∈ T, one has {x := (x 1 , x 2 , . . . , x k ) ∈ X : φ (t, ⋅) is discontinuous at x} ⊆ E ∪ [ k

Of course, the main peculiarity of Theorem 1 resides in the kind of discontinuity that is allowed for .Indeed, it is easy to construct examples of functions , , and ℎ satisfying the assumptions of Theorem 1 and such that for all  ∈  the function (, ⋅) is discontinuous at all points  ∈ [0, ].
Theorem 1 extends a previous result (Theorem 1 of [6]), valid for the case where  does not depend on  explicitly.At this point, it is natural to ask if Theorem 1 above can be extended to the more general case where the function  takes its values in the space R  .In this direction, we note that some results exist for the vector explicit equation (see [7,8]), while for the implicit equation (1) the problem is still unsolved.The aim of this note is exactly to provide such an extension.In the following, if  ∈ N and  ∈ {1, . . ., }, we will denote by   : R  → R the projection over the th axis.Moreover, we will denote by   the -dimensional Lebesgue measure over R  .If  := ( 1 , . . .,   ) ∈ R  , with   > 0 for all  = 1, . . ., , we will put   := ∏  =1 [0,   ] .Finally, if  and  are as above, we will denote by F , the family of all subsets  ⊆   such that there exist sets  1 ,  2 , . . .,   ⊆ R  , with  1 (  (  )) = 0 for all  = 1 . . ., , such that  := ⋃  =1   .The following is our main result (where R +  denotes the positive open orthant of R  , and int  () is the interior of  in ).
Let  be a separable metric space, and let  :  ×  → 2  be a multifunction with nonempty complete values.Finally, let  ⊆  be a given set, and, for each  ∈ {1, . . ., }, let  * , :  →   be the projection over   .Assume that The proof of Theorem 2 will be given in Section 2. Further, we will point out some counterexamples to possible improvements of Theorem 2.

Proof of Theorem 2
Before giving the proof of Theorem 2, we fix some notations.If  ∈ N, the space R  will be considered with its Euclidean norm ‖ ⋅ ‖  .Moreover, if  ∈ R  and  > 0, we put If  ∈ [1, +∞], the space   (, R  ) will be considered with the usual norm As usual, we put   () :=   (, R).For the basic definitions and facts about multifunctions, we refer to [10].
Proof of Theorem 2. Without loss of generality, we can assume that ( 8) and ( 10) hold for all  ∈ .Moreover, we can suppose that  < +∞.Firstly, we prove that the functions V and  are measurable.Observe that, by assumption (ii), for all  ∈ , one has which is absurd.Therefore, the second equality in ( 18) is proved.The first one can be checked in analogous way.Since  is closed, it can be easily checked that the set  ∩ (  \ ) is nonempty, countable, and dense in (  \ ).Consequently, by Lemma at page 198 of [11], the function  * | ×(  \) is L() ⊗ B(  \ )-measurable (where L() denotes the family of all Lebesgue-measurable subsets of ).By (18) and Lemma III.39 of [12], it follows that the functions V and  are measurable over , as claimed.
Observe that  is well-defined since for all (, ) ∈  ×   one has Moreover, by the lower semicontinuity of  and by ( 25), for all  ∈ , we get { ∈   :  (, ⋅) is not lower semicontinuous at } ⊆ .