TITLE: Is the set of the absolutely continuous functions a Borel set of the space of the continuous functions?
QUESTION [6 upvotes]: Does anyone knows whether the set of the absolutely continous functions $F :[0,1]\to \mathbb{R}^d$ of the form $$F(t)= a + \int_0^tf(s) ds$$ where $f$ is an integrable function is a Borel set of the Banach space $C$ of the continuous funtions $$F : t\in [0,1] \to F(t)\in \mathbb{R}^d$$ with the norm of the uniform convergence ?

REPLY [2 votes]: With a little more machinery we can give a very short proof.  A theorem of Lusin and Souslin states that if $X,Y$ are Polish, $\Phi : X \to Y$ is continuous, $A \subset X$ is Borel and $\Phi|_A$ is injective, then $\Phi(A)$ is Borel.  (See for instance Theorem 15.1 of Kechris's Classical Descriptive Set Theory.)  Taking $A = X = \mathbb{R} \times L^1([0,1])$, $Y = C([0,1])$, and considering the map $(a,f) \mapsto a + \int_0^\cdot f$ which is continuous and injective and whose image is the absolutely continuous functions, we have the result.