Let $\alpha_n:\,(0,T)\rightarrow A$, where $A\subset\mathbb{R}^d$ compact, a sequence of measurable functions. Since $A$ is compact, for almost every $t\in (0,T)$, there exists a subsequence $\alpha_{n_k}(t)$ and $\alpha_t\in A$ such that $\alpha_{n_k}(t)\rightarrow \alpha_t$ as $k\rightarrow\infty$. Now, we define $\alpha:(0,T)\rightarrow A$ via: \begin{align*} \alpha(t):=\alpha_t. \end{align*} Is $\alpha$ also measurable?
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The limit superior of a sequence of measurable functions is measurable (e.g. see Rudin's Real and Complex Analysis Thm 1.14.). In your case, $a$ is the limit (and hence the limit superior) of the sequence of measurable functions $a_{n_k}$, hence measurable.