Recall that the inner Hausdorff measure of noncompactness in a Banach space $Y$ is defined as $$ \chi_{i}(B):=\inf\{\varepsilon>0: B\subset \cup_{i=1}^{n} B(y_{i},\varepsilon), \textrm{for some } y_{1},\ldots,y_{n}\in B\} $$ for each bounded $B\subset Y$, where $ B(y_{i},\varepsilon)$ stands for the closed ball centered at $x_{i}$ and radius $\varepsilon$.
Now, let $X$ be a Banach space and $C([a,b],X)$ the (also Banach) space of the continuous functions $x:[a,b]\longrightarrow X$ endowed the usual supremum norm. For a given non-empty and bounded $B\subset C([a,b],X)$ and $t\in [a,b]$, denote $B(t):=\{x(t):x\in B\}$, and $v(t):=\chi_{i}(B(t))$.
Then, is true that the function $v(t)$ is integrable in the Lebesgue sense?
Many thanks in advance for your comments.