Let $A\subset\mathbb{R}^n$ be compact, $\alpha\in L^\infty([0,T];A)$ and $\alpha_n\subset L^\infty([0,T];A)$ with $\alpha_n\rightarrow \alpha$ in $\|\cdot\|_\infty$. Let also $f:A\rightarrow\mathbb{R}^d$ be continuous.
Does $\int_0^T\|f(\alpha_n(s))-f(\alpha(s))\|ds \rightarrow 0$ as $n\rightarrow \infty$?
If $\alpha_n$ would be continuous this wouldn't be a problem, because $\alpha_n$ would converge pointwise and, hence, $\|f(\alpha_n(s))-f(\alpha(s))\|\rightarrow 0$ for all $s\in[0,T]$. This certaintly does not work here, since we have only the measurability of $\alpha_n$. If we can show that $\text{ess}\sup_{s\in[0,T]}\|f(\alpha_n(s))-f(\alpha(s))\|\rightarrow 0$, also the integral converges to zero. Does anyone have an idea how to do that?