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 ?

4$\begingroup$ Let $E_{n,m}$ be the set of all $f$ in $C[0,1]$ s.t. whenever $xy \le 1/m$ we have $f(x)f(y)\le 1/n$ and consider $\cap_n\cup_m E_{n,m}$. $$ $$ In other words, just use the definition of absolute continuity. $\endgroup$ – Bill Johnson Mar 19 '13 at 19:16

1$\begingroup$ Indeed, any function of $C$ is uniformly continuous but what's the link with the absolute continuity ? $\endgroup$ – Theluze Mar 21 '13 at 19:56

$\begingroup$ en.wikipedia.org/wiki/Absolute_continuity $\endgroup$ – Bill Johnson Mar 22 '13 at 15:49
Let $\phi:C\to[0,\infty]$ be defined for $F\in C$ as the norm of $F$ in $W^{1,1}$ if $F$ is absolutely continuous, and $+\infty$ otherwise. Then $\phi$ is lower semicontinuous for the topology of uniform convergence and $W^{1,1}=\{\phi<\infty\}$ is Borel measurable.

1$\begingroup$ Thank you for your elegant proof. However, i don't know how to prove that $\phi$ is lower semicontinuous. Indeed if, for $F(t)= a +\int_0^t f(t) dt$ i write $$F_{1,1} = G(F) + H(F) $$ where $$G(F) =\int_0^1 F(t)dt $$ and $$H(F) = \int_0^1 f(t)dt $$ then $F\to G(F)$ clearly continuous for the norm of the uniform convergence, but i can't prove that $H$ is lower semicontinuous. Moreover $W^{1,1}$ does not seem to be closed in $C$. Maybe i don't take the proof in the good way. Could someone enlighten me on the good way to prove the lower semicontinuity of $\phi$ ? $\endgroup$ – Theluze Mar 19 '13 at 17:17
I think Bill Johnson's answer ``just use the definition" is correct, but that he did not write what he wanted to write. A function $f\in\mathcal C([0,1],\mathbb R^d)$ is absolutely continuous if and only if the following holds: $$\forall p\in\mathbb N\; \exists q\in\mathbb N\; \forall x_1,\dots ,x_N,y_1,\dots ,y_N\in\mathbb Q\cap[0,1]$$
$$\sum_{i=1}^N\vert y_ix_i\vert <\frac 1q\;\implies\;\sum_{i=1}^N\Vert f(y_i)f(x_i)\Vert<\frac 1p\cdot $$
For fixed $x_1,\dots ,x_N,y_1,\dots y_N$, the set of all $f\in\mathcal C([0,1],\mathbb R^d)$ satisfying the condition written in the second displayed line is obviously open in $\mathcal C([0,1],\mathbb R^d)$. This shows that $AC([0,1],\mathbb R^d)$ is indeed Borel in $\mathcal C([0,1],\mathbb R^d)$.
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.