MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Let $E_{n,m}$ be the set of all $f$ in $C[0,1]$ s.t. whenever $|x-y| \le 1/m$ 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. – Bill Johnson Mar 19 '13 at 19:16
Indeed, any function of $C$ is uniformly continuous but what's the link with the absolute continuity ? – Theluze Mar 21 '13 at 19:56
up vote 1 down vote accepted

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 semi-continuous for the topology of uniform convergence and $W^{1,1}=\{\phi<\infty\}$ is Borel measurable.

share|cite|improve this answer
Thank you for your elegant proof. However, i don't know how to prove that $\phi$ is lower semi-continuous. 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 semi-continuous. 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 $\phi$ ? – 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: $$\forall p\in\mathbb N\; \exists q\in\mathbb N\; \forall x_1,\dots ,x_N,y_1,\dots ,y_N\in\mathbb Q\cap[0,1]$$

$$\sum_{i=1}^N\vert y_i-x_i\vert <\frac 1q\;\implies\;\sum_{i=1}^N\Vert f(y_i)-f(x_i)\Vert<\frac 1p\cdot $$

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 $\mathcal C([0,1],\mathbb R^d)$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.