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Let $\Omega:=(0,1)^n$ and define $ACL_i(\Omega)$ as the set of all Borel functions $u:\Omega\to\mathbb{R}$ such that $$ t\mapsto u(x_1,\dots,x_{i-1},t,x_{i+1},\dots,x_n) $$ is $AC$ for a.e. $(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)$. Let $ACL(\Omega):=\cap_i ACL_i(\Omega)$ and denote by $ACL^p(\Omega)$ the set of functions $u\in ACL(\Omega)$ having all partial derivatives in $L^p(\Omega)$.

It seems to be well-known that any element of $W^{1,p}$ has a representative in $ACL^p(\Omega)$, but all proofs I have found actually show that, for any $i$, one can find a representative in $ACL_i(\Omega)$.

How can one deduce that there exists a representative in $ACL(\Omega)$? Notice that, given $u\in W^{1,p}(\Omega)$, it is not (a priori) sufficient to subsequently modify $u$ on the negligible sets $\Sigma_1,\dots,\Sigma_n$ provided by the above weaker assertion.

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    $\begingroup$ The procedure to make $u$ AC on lines of a given direction is to pass to $\int_a^x \partial_t u \, dt$, and it seems applying this repeatedly doesn't destroy previous achievements. $\endgroup$ – Christian Remling Jul 21 '16 at 0:02
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This is addressed in section 4.9.2 of the Evans and Gariepy's book on measure theory. The key is to show that the representative in $ACL_i$ is the precise representative of the function (defined in the book), so it does not depend on $i$.

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