Is this outer measure the same as the absolutely continuous part? (related to Lebesgue differentiation on more general spaces)

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In this paper about Lebesgue differentiation: https://arxiv.org/pdf/1802.02069.pdf they define an outer measure $\mu_{\lambda}(A) = \inf \{ \mu(E): E \in \mathcal{B}(X) \text{ and } \lambda(A \setminus E) = 0\}$ where $\lambda$ and $\mu$ are measures on the Borel sets of a space $X$.

Theorem 2.2 says that $\mu_{\lambda}$ is essentially a restriction of $\mu$. I couldn't really follow the proof (i.e. the one in Federer's book) but since $\mu_{\lambda} \ll \lambda$, I suspect that $\mu_{\lambda}$ is related to the absolutely continuous part of $\mu$.

I showed that if $\mu$ has Lebesgue decomposition $\mu = \rho + f d \lambda$, then $f d\mu \leq \mu_{\lambda}$ on Borel sets by using the proof of the decomposition theorem in Folland's book.

Are they generally different? When $\mu$ and $\lambda$ are Radon or regular, can we show they are the same?