In the book Optimal Transport for Applied Mathematicians, the author states that the condition that a measure $\mu \in \mathcal P(\mathbb R^d)$ is absolutely continuous relative to the Lebesgue measure is a stronger than "$\mu(A) = 0$ for any $A \subset \mathbb R^d$ such that $\mathcal H^{d-1}(A)<+\infty$".
Now, I can intuitively understand what absolutely continuity of a measure implies, for example, but I do not have any intuition when it comes to the "$\mu(A) = 0$ for any $A \subset \mathbb R^d$ such that $\mathcal H^{d-1}(A)<+\infty$".
What would be the intuition behind such condition? Can anyone provide examples where $\mu$ is not absolutely continuous, but satisfies this second condition?
Yes, in this context you can think about the Lebesgue measure as a sort of characterization of the (let me use this word a bit informally) dimension of the sets. All the Borel sets $A\subset\mathbb{R}^d$ of dimension $d$ verify $\mathcal{L}^d(A)>0$, while if $\mathcal{L}^d(A)=0$ then dimension of $A$ is strictly smaller than $d$. Thus, for $\mu$ to be absolutely continuous w.r.t. $\mathcal{L}^d$ it should vanish also in sets of dimension strictly smaller than $d$. On the other hand, the condition $\mu(A)=0$ for any $A\subset\mathbb{R}^d$ such that $\mathcal{H}^{d-1}(A)<+\infty$ leaves room to define $\mu$ so that it does not vanish for Borel sets $A$ of dimension $d-1$ but such that $\mathcal{H}^{d-1}(A)=+\infty$.
Example: A line in $\mathbb{R}^2$. Consider $d=2$ and $B=\{(x,y)\in\mathbb{R}^2:y=x\}$, hence $\mathcal{L}^2(B)=0$ but $\mathcal{H}^{1}(B)=+\infty$. Then you can define $\mu$ so that $\mu(A)=0$ for any $A\subset\mathbb{R}^d$ such that $\mathcal{H}^{d-1}(A)<+\infty$ but $\mu(B)>0$. Therefore it is not absolutely continuous w.r.t. $\mathcal{L}^2$.