= \vp(v(t_0 + t; t_0, a))$ for all $t\in [0, T(t_0, a))$. Because this is true for every $\vp\in\Phi$ and the family $\Phi$ separates the points of $X$, we obtain $u(t_0 + t; t_0, a) = v(t_0 + t; t_0, a)$ for all $t\in [0, T(t_0, a))$, i.e., $u = v$ as paths.

For the unique path in the funnel $S^\infty(t_0, a)$, let us use the notation ${\frak u}(t; t_0, a)$. If $t_0 < t_1 < T(t_0, a)$, then $(\sigma_{t_1 - t_0} {\frak u})(t) = {\frak u}(t; t_1, {\frak u}(t_1; t_0, a))$ for $t_1\le t < T(t_0, a)$, by property [**LS3**]{}. Thus, if $t_2 > t_1$, $(\sigma_{t_1 - t_0} {\frak u})(t_2) = {\frak u}(t_2; t_1, {\frak u}(t_1; t_0, a))$. On the other hand, $(\sigma_{t_1 - t_0} {\frak u})(t_2) = {\frak u}(t_2; t_0, a)$ by the definition of the shift operator $\sigma_\tau$. This establishes the semigroup property of the measurable selection ${\frak u}(t; t_0, a)$. The fact that the maps $X\owns a \mapsto {\frak u}(\cdot; t_0, a)\in C([t_0, T(t_0, a))\to X$ are measurable is established by an argument with the distance function similar to the one in the end of the proof of Theorem \[thm:main\]. This completes the proof.

There are 