= Solution
Suppose $\mu=\alpha+\sigma=\beta+\tau$ are two such decompositions. Choose measurable supports $S,T$ of the singular parts with $\kappa(S)=\kappa(T)=0$, and put $N=S\cup T$. <Absolute continuity> makes both $\alpha$ and $\beta$ vanish on $N$. On $X\setminus N$ the two singular parts vanish, so for every measurable $A$,
$$
\alpha(A)=\alpha(A\setminus N)=\mu(A\setminus N)=\beta(A\setminus N)=\beta(A).
$$
Subtracting from the <finite measure> $\mu$ gives $\sigma=\tau$. This proves \b[uniqueness of the <Lebesgue decomposition>] without subtracting infinite masses.
For the specified $M$, the restriction to $M$ is singular because $\kappa(M)=0$, and the complementary restriction is absolutely continuous by assumption. Their sum is $\mu$. Uniqueness therefore gives
$$
\boxed{\mu^s=\mu|_M,\qquad\mu^a=\mu|_{X\setminus M}.}
$$
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