Given a measurable function $f \in L^{\infty}([0,1],\mathbb{R}^d)$, we know that for all $\epsilon > 0$ there exists a continuous function $f_{\epsilon}$ such that $$ \|f - f_{\epsilon}\|_{L^1} \leq \epsilon.$$
But, I wonder if $f$ satisfies, in addition, a constraint of the form: $f(t) \in A$ for a.e. $t \in [0,1]$, where $A$ is a compact set in $\mathbb{R}^d$. Can we prove that for all $\epsilon > 0$ there exists a continuous function $f_{\epsilon}$ such that $$ \|f - f_{\epsilon}\|_{L^1} \leq \epsilon, \quad f_{\epsilon}(t) \in A, \quad \forall t \in [0,1].$$
I think the idea is to use the first density result: so for all $\epsilon>0$ there is a $f_{\epsilon}$ that is continuous satisfying: $$ \|f-f_{\epsilon}\|_{L^1} \leq \epsilon,$$ so can we construct/move/rescale the function $f_{\epsilon}$ such that we get a function $g_{\epsilon}$ that satisfies: $$ \|f - g_{\epsilon}\|_{L^1} \leq \epsilon, \quad g_{\epsilon}(t) \in A, \quad \forall t \in [0,1],$$ but I do not know how to construct it.