Without loss of generality we can assume that assumptions (ii) and (iv) are satisfied for all \(t\in[a,b]\). Moreover, let
$$E:=\bigcup_{i=1}^{2n} E_{i} . $$
Now, observe that by assumption (iii) and Theorem 2.4 of [4] the function h is inductively open. That is, there exists a set \(Y\subseteq X\) such that the function
is open and \(h(Y)=h(X)\). It follows that the multifunction \(T:h(X)\to2^{Y}\) defined by putting, for each \(s\in h(X)\),
$$T(s)=h^{1}(s)\cap Y $$
is lower semicontinuous in \(h(X)\) with nonempty values. To see this, fix any set \(\Omega_{1}\subseteq Y\), with \(\Omega_{1}\) open in the relative topology of Y. We get
$$\begin{aligned} T^{}(\Omega_{1})&:=\bigl\{ s\in h(X): T(s)\cap\Omega_{1}\ne \emptyset\bigr\} \\ &=\bigl\{ s\in h(X): h^{1}(s)\cap Y\cap\Omega_{1}\ne \emptyset\bigr\} \\ &=\bigl\{ s\in h(X): h^{1}(s)\cap\Omega_{1}\ne\emptyset\bigr\} \\ &=h(\Omega_{1}). \end{aligned}$$
Since the function \(h_{Y}:Y\to h(X)\) is open, the set \(h(\Omega_{1})\) is open in \(h(X)\). It follows that the set \(T^{}(\Omega_{1})\) is open in \(h(X)\), hence T is lower semicontinuous in \(h(X)\), as claimed.
Let \(\Psi:[a,b]\times\mathbf{R}^{n}\times \mathbf{R}^{n}\to2^{Y}\) be defined by
$$\Psi(t,x,y):=T\bigl(G(t,x,y)\bigr)=h^{1}\bigl(G(t,x,y)\bigr)\cap Y $$
(note that Ψ is well defined since \(G(t,x,y)\subseteq h(X)\) for all \((t,x,y)\in[a,b]\times\mathbf{R}^{n}\times\mathbf{R}^{n}\)). We observe the following facts:

(a)
the multifunction Ψ is \(\mathcal{L}([a,b])\otimes\mathcal {B}(\mathbf{R} ^{n})\otimes\mathcal{B}(\mathbf{R}^{n})\)weakly measurable; that is, for each set \(\Omega\subseteq Y\), with Ω open in the relative topology of Y, the set
$$\Psi^{}(\Omega):=\bigl\{ (t,x,y)\in[a,b]\times\mathbf{R}^{n}\times \mathbf{R}^{n}: \Psi(t,x)\cap\Omega\ne\emptyset\bigr\} $$
belongs to \(\mathcal{L}([a,b])\otimes\mathcal{B}(\mathbf {R}^{n})\otimes\mathcal{B}(\mathbf{R}^{n})\) (this follows from Proposition 13.2.1 of [3], since G is \(\mathcal{L}([a,b])\otimes\mathcal{B}(\mathbf{R}^{n})\otimes \mathcal{B}(\mathbf{R}^{n})\)measurable and T is lower semicontinuous);

(b)
Ψ has nonempty values and for all \(t\in[a,b]\) one has
$$\bigl\{ (x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}:\Psi(t, \cdot, \cdot) \mbox{ is not lower semicontinuous at }(x,y) \bigr\} \subseteq E. $$
Indeed, if \(t\in[a,b]\) and \((\overline{x},\overline{y})\in (\mathbf{R}^{n}\times\mathbf{R}^{n})\setminus E\) are fixed, then the multifunction \(G(t,\cdot,\cdot)\) is lower semicontinuous at \((\overline{x},\overline{y})\), hence (by the lower semicontinuity of T) the multifunction
$$(x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}\to\Psi(t,x,y)=T \bigl(G(t,x,y)\bigr) $$
is lower semicontinuous at \((\overline{x},\overline{y})\), as claimed.
Now, let \(\overline{\Psi}:[a,b]\times\mathbf{R}^{n}\times\mathbf {R}^{n}\to2^{\mathbf{R}^{n}}\) be defined by
$$\overline{\Psi}(t,x,y):=\overline{\Psi(t,x,y)}. $$
It follows by assumption (iv) and by construction that
$$ \overline{\Psi}(t,x,y)\subseteq \prod_{i=1}^{n} \bigl[\alpha_{i}(t),\beta _{i}(t)\bigr] \quad \mbox{for all } (t,x,y)\in[a,b]\times\mathbf{R}^{n}\times\mathbf{R}^{n}. $$
(4)
By Proposition 2.6 and Theorem 3.5 of [5], the multifunction \(\overline{\Psi}\) is \(\mathcal{L}([a,b])\otimes \mathcal{B}(\mathbf{R} ^{n})\otimes\mathcal{B}(\mathbf{R}^{n})\)measurable with nonempty values. Moreover, for all \(t\in[a,b]\) one has
$$\bigl\{ (x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}: \overline{\Psi}(t,\cdot ,\cdot)\mbox{ is not l.s.c. at }(x,y) \bigr\} \subseteq E. $$
At this point, let us apply Theorem 2.2 of [1] to the multifunction \(\overline{\Psi}\), choosing \(k=2n\), \(T=[a,b]\), \(X_{i}=\mathbf{R}\) for all \(i=1,\ldots, 2n\), \(S=\mathbf{R}^{n}\) (all the spaces are considered with their Euclidean distance and with the usual Lebesgue measure over their Borel family), and, as above,
$$E:=\bigcup_{i=1}^{2n} E_{i} \subseteq \mathbf{R}^{2n}. $$
We find that there exist \(Q_{1},\ldots,Q_{2n}\subseteq \mathbf{R}\), with \(Q_{i}\in \mathcal{B}(\mathbf{R})\) and \(m_{1}(Q_{i})=0\) for all \(i=1,\ldots,2n\), and a function \(\phi:[a,b]\times\mathbf{R}^{n}\times\mathbf{R}^{n}\to\mathbf {R}^{n}\), such that
 (i)′:

\(\phi(t,x,y)\in\overline{\Psi}(t,x,y)\) for all \((t,x,y)\in [a,b]\times\mathbf{R} ^{n}\times \mathbf{R}^{n}\);
 (ii)′:

for all \((x,y)\in(\mathbf{R}^{n}\times\mathbf{R}^{n}) \setminus [\bigcup_{i=1}^{2n}(P_{i}^{1}(Q_{i})\cup E_{i}) ]\), the function \(\phi(\cdot,x,y)\) is \(\mathcal{L}([a,b])\)measurable;
 (iii)′:

for a.e. \(t\in [a,b]\), one has
$$\bigl\{ (x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}: \phi( t, \cdot ,\cdot )\mbox{ is discontinuous at } (x,y) \bigr\} \subseteq \bigcup _{i=1}^{2n} \bigl(E_{i}\cup P_{i}^{1}(Q_{i})\bigr). $$
Of course, for all \(i=1,\ldots,2n\), one has \(m_{1}(P_{i}(E_{i}\cup P_{i}^{1}(Q_{i})))=0\).
Now, let us apply Theorem 2.1 of [1] with \(f=g=\phi\), taking into account that for a.e. \(t\in[a,b]\) and for all \((x,y)\in\mathbf{R}^{n}\times\mathbf{R}^{n}\) one has
$$\phi(t,x,y)\in\overline{\Psi}(t,x,y)\subseteq \prod _{i=1}^{n}\bigl[\alpha _{i}(t), \beta_{i}(t)\bigr]. $$
We find that there exist \(u\in W^{2,p}([a,b],\mathbf{R}^{n})\) and a set \(U_{0}\in\mathcal{L}([a,b])\), with \(m_{1}(U_{0})=0\), such that
$$ \textstyle\begin{cases} u^{\prime\prime}(t)=\phi(t,u(t),u^{\prime}(t))& \mbox{for all } t\in [a,b]\setminus U_{0}, \\ u(a)=u(b)=0_{\mathbf{R}^{n}}, \end{cases} $$
and also
$$\bigl(u(t),u^{\prime}(t)\bigr)\notin\bigcup_{i=1}^{2n} \bigl(E_{i}\cup P_{i}^{1}(Q_{i}) \bigr) \quad \mbox{for all } t\in[a,b]\setminus U_{0}. $$
In particular, by assumption (ii) we get
$$G\bigl(t,u(t),u^{\prime}(t)\bigr)\subseteq F\bigl(t,u(t),u^{\prime}(t) \bigr) \quad \mbox{for all } t\in[a,b]\setminus U_{0}. $$
Consequently, taking into account the continuity of h and the closedness of X, for all \(t\in[a,b]\setminus U_{0}\) we get
$$u^{\prime\prime}(t)=\phi\bigl(t,u(t),u^{\prime}(t)\bigr) \in\overline{\Psi} \bigl(t,u(t),u^{\prime}(t)\bigr) \subseteq h^{1}\bigl(G \bigl(t,u(t),u^{\prime}(t)\bigr)\bigr). $$
In particular, we get
$$h\bigl(u^{\prime\prime}(t)\bigr)\in G\bigl(t,u(t),u^{\prime}(t)\bigr) \subseteq F\bigl(t,u(t),u^{\prime}(t)\bigr) $$
for all \(t\in[a,b]\setminus U_{0}\) and hence the function u satisfies our conclusion. The proof is now complete.
Remark
We now give two simple examples of application of Theorem 1.1. The first example concerns with the scalar singlevalued case, while the second one deals with the vector multivalued case.
Example 2.1
Let \(n=1\) and let \([a,b]\) be any compact interval. Let us consider the problem
$$ \textstyle\begin{cases} \cos(u^{\prime\prime}(t))= F(t,u(t),u^{\prime}(t))& \mbox{for a.e. } t\in[a,b], \\ u(a)=u(b)=0, \end{cases} $$
(5)
where \(F:[a,b]\times\mathbf{R}\times\mathbf{R}\to\mathbf{R}\) is the (singlevalued) function defined by putting, for each \((x,y)\in\mathbf{R}^{2}\),
$$F(t,x,y)= \textstyle\begin{cases} 0&\mbox{if }x\in\mathbf{Q}\mbox{ or }y\in\mathbf{Q}, \\ 1&\mbox{otherwise}. \end{cases} $$
Of course, such a function F (which does not depend on t explicitly) is discontinuous at all points
\((x,y)\in\mathbf{R}^{2}\). We observe that Theorem 1.1 can easily be applied by choosing, for any \(p\in [1,+\infty[\),
$$G:[a,b]\times\mathbf{R}\times\mathbf{R}\to\mathbf{R},\qquad G(t,x,y)\equiv1, $$
\(E_{1}=\mathbf{Q}\times\mathbf{R}\), \(E_{2}=\mathbf{R}\times\mathbf {Q}\), \(X=[2\pi,4\pi]\), \(\alpha(t)\equiv2\pi\), \(\beta(t)\equiv4\pi\), \(h(t)=\cos(t)\) (restricted to the interval \([2\pi,4\pi]\)). Indeed, such a function G is \(\mathcal{L}([a,b])\otimes\mathcal {B}(\mathbf{R} )\otimes\mathcal{B}(\mathbf{R})\)measurable, \(P_{1}(E_{1})=P_{2}(E_{2})=\mathbf{Q}\), and for all \(t\in[a,b]\) one has
$$\begin{aligned}& \bigl\{ (x,y)\in\mathbf{R}^{2}: G(t, \cdot ,\cdot ) \mbox{ is discontinuous at }(x,y) \bigr\} \\& \quad {}\cup \bigl\{ (x,y)\in\mathbf{R}^{2}:G(t,x,y)\ne F(t,x,y) \bigr\} = E_{1}\cup E_{2}. \end{aligned}$$
Moreover, for all \((t,x,y)\in[a,b]\times\mathbf{R}\times\mathbf {R}\) we have
$$G(t,x,y)=1\in h(X)\quad \mbox{and}\quad h^{1}\bigl(G(t,x,y) \bigr)=h^{1}(1)\subseteq [2\pi,4\pi]=\bigl[\alpha(t),\beta(t)\bigr]. $$
Finally, observe that assumption (iii) is satisfied since for all \(t\in\operatorname {int}_{\mathbf{R}}(h(X))= \, ]1,1[ \) the set \(h^{1}(t)\) contains only two points.
Consequently, by Theorem 1.1, problem (5) has a solution in \(u\in W^{2,p}([a,b],\mathbf{R})\). Such a solution u also satisfies the condition \(u^{\prime\prime}(t)\in[2\pi,4\pi]\) for a.e. \(t\in [a,b]\), hence we get \(u\in W^{2,\infty}([a,b],\mathbf{R})\).
Moreover, observe that Theorem 1.1 can be applied in analogous way also by taking \(X=[2\pi+4k\pi,4\pi+4k\pi]\) and \(\alpha(t)\equiv 2\pi +4k\pi\), \(\beta(t)\equiv4\pi+4k\pi\), with \(k\in\mathbf{N}\). Then, for each \(k\in\mathbf{N}\), we get the existence of a solution \(u_{k}\in W^{2,\infty}([a,b],\mathbf{R})\) such that \(u_{k}^{\prime\prime}(t)\in[2\pi+4k\pi,4\pi+4k\pi]\) for a.e. \(t\in [a,b]\). Hence, problem (5) has infinitely many solutions. Finally, we note that problem (5) does not admit the trivial solution \(u(t)\equiv0\), since \(F(t, x,0)=0\) for all \((t,x)\in[a,b]\times\mathbf{R}\) and \(\cos(0)=1\).
Example 2.2
Let \(n=2\), and let \(\psi:\mathbf{R}^{2}\to\mathbf {R}\) be defined by \(\psi(v,z)=v+z\). In what follows, we denote vectors of \(\mathbf{R}^{2}\) by the notations \(x:=(x_{1},x_{2})\) and \(y:=(y_{1},y_{2})\). Let \(E\subseteq \mathbf{R}^{4}\) be the set of vectors such that at least one component is rational, that is,
$$E:=\bigl\{ (x,y)\in\mathbf{R}^{4} : \mbox{at least one of } x_{1},x_{2},y_{1},y_{2} \mbox{ is rational}\bigr\} . $$
Let \(F:[0,1]\times\mathbf{R}^{2}\times\mathbf{R}^{2}\to2^{\mathbf{R}}\) be defined by
$$F(t,x,y)= \textstyle\begin{cases} {1} &\mbox{if }(x,y)\in E, \\ {[2+t,3+t]}&\mbox{if }(x,y)\notin E\mbox{ and }x_{1}< 1, \\ {[4+t,5+t]}&\mbox{if }(x,y)\notin E\mbox{ and }x_{1}>1. \end{cases} $$
Of course, for any fixed \(t\in[0,1]\), the multifunction \(F(t,\cdot ,\cdot )\)
is not lower semicontinuous at any point \((x,y)\in\mathbf{R}^{4}\). By Theorem 1.1, it is easily seen that for any fixed \(p\in[1,+\infty[\) the problem
$$ \textstyle\begin{cases} \psi(u^{\prime\prime}(t))\in F(t,u(t),u^{\prime}(t))& \mbox{a.e. in } [0,1], \\ u(0)=u(1)=0_{\mathbf{R}^{2}} \end{cases} $$
(6)
has a solution \(u\in W^{2,p}([0,1],\mathbf{R}^{2})\). To this aim, choose \(X=[1,5]\times[1,5]\), \(h:=\psi_{X}\), \(\alpha(t)\equiv(1,1)\), \(\beta(t)\equiv(5,5)\), \(G:[0,1]\times\mathbf{R}^{2}\times\mathbf{R}^{2}\to2^{\mathbf{R}}\), with
$$G(t,x,y)= \textstyle\begin{cases} {[2+t,3+t]}&\mbox{if }x_{1}\le1, \\ {[4+t,5+t]}&\mbox{if }x_{1}>1 \end{cases} $$
and
$$\begin{aligned}& E_{1}:=\mathbf{Q}\times\mathbf{R}\times\mathbf{R}\times\mathbf {R}, \qquad E_{2}:=\mathbf{R}\times\mathbf{Q} \times\mathbf{R}\times \mathbf{R}, \\& E_{3}:=\mathbf{R}\times\mathbf{R}\times\mathbf{Q} \times \mathbf{R}, \qquad E_{4}:=\mathbf{R}\times \mathbf{R}\times \mathbf{R}\times\mathbf{Q}. \end{aligned}$$
Of course, such a multifunction G is \(\mathcal{L}([0,1])\otimes\mathcal{B}(\mathbf{R}^{2})\otimes \mathcal{B}(\mathbf{R}^{2})\)measurable with nonempty closed values. Moreover, if one fix any \(t\in[0,1]\), then the multifunction \(G(t,\cdot,\cdot)\) is lower semicontinuous at each point \((x,y)\in\mathbf{R}^{4}\), with \(x_{1}\ne1\). Consequently, for all \(t\in[0,1]\) we have
$$\begin{aligned}& \bigl\{ (x,y)\in\mathbf{R}^{2}\times\mathbf{R}^{2}: G(t, \cdot ,\cdot )\mbox{ is not lower semicontinuous at }(x,y) \bigr\} \\& \quad {}\cup \bigl\{ (x,y)\in\mathbf{R}^{2}\times\mathbf {R}^{2}:G(t,x,y)\nsubseteq F(t,x,y) \bigr\} = \bigcup _{i=1}^{4} E_{i}=E. \end{aligned}$$
Clearly, for all \(i=1,2,3,4\), one has \(P_{i}(E_{i})=\mathbf{Q}\). Now, observe that for all \((t,x,y)\in[0,1]\times\mathbf{R}^{2}\times \mathbf{R}^{2}\) we have
$$G(t,x,y)\subseteq [2,6]\subseteq h(X)=[2,10] $$
and
$$h^{1}\bigl(G(t,x,y)\bigr)\subseteq X=[1,5]\times[1,5]=\bigl[ \alpha_{1}(t),\beta_{1}(t)\bigr] \times\bigl[ \alpha_{2}(t),\beta_{2}(t)\bigr]. $$
Finally, let us show that for all \(s\in\operatorname{int}_{\mathbf{R}}(h(X))=\, ]2,10[\), we have \(\operatorname{int}_{X} (h^{1}(s))=\emptyset\) (though this fact is quite intuitive  since h is never locally constant  we shall provide an explicit proof). To this aim, fix \(s\in\operatorname{int}_{\mathbf{R}}(h(X))\), and let \((v_{0},z_{0})\in h^{1}(s)\). Therefore, \((v_{0},z_{0})\in X\) and \(h(v_{0},z_{0})=v_{0}+z_{0}=s\). Let Ω be an open set in X such that \((v_{0},z_{0})\in\Omega\). Of course, one can find \(r>0\) such that
$$\bigl([v_{0}r,v_{0}+r]\times[z_{0}r,z_{0}+r] \bigr)\cap X\subseteq \Omega. $$
Let \(v_{1}\) be any point in \([v_{0}r,v_{0}+r]\cap[1,5]\), with \(v_{1}\ne v_{0}\). We have
$$(v_{1},z_{0})\in\Omega $$
and
$$h(v_{1},z_{0})=v_{1}+z_{0}\ne v_{0}+z_{0}=s. $$
Hence, the set \(h^{1}(s)\) has empty interior in X, as claimed. Thus, all the assumptions of Theorem 1.1 are satisfied. Consequently, problem (6) has at least a solution \(u\in W^{2,p}([0,1],\mathbf{R}^{2})\).
As a matter of fact, since \(u^{\prime\prime}(t)\in X\) for a.e. \(t\in[0,1]\), we get \(u\in W^{2,\infty}([0,1],\mathbf{R}^{2})\). As before, we note that problem (6) does not admit the trivial solution \(u(t)\equiv0_{\mathbf{R}^{2}}\).