Decomposition of a function/measure into a continuous and discontinuous function/measure. I'm reading a book where the author states

In Chapter 1 we have seen that, to every increasing function α(t)
defined on [0, ∞[, there corresponds a measure dα(t). In the
terminology of measure theory, α is continuous if and only if dα is a
diffuse measure (that is, the support of dα has no atoms). If dα is
purely atomic (that is, the support of dα consists of a countable
number of atoms), then α is said to be purely discontinuous. Every
σ-finite measure on [0, ∞[ can be written in a unique way as the sum
of a diffuse and a purely atomic measure. This corresponds to the
unique decomposition of an increasing function into the sum of a
continuous and purely discontinuous function.

In Chapter 1 of that book, the only decomposition theorem that was alluded to was the Lebesgue decomposition theorem.
How, from that theorem, do we get «Every σ-finite measure on [0, ∞[ can be written in a unique way as the sum of a diffuse and a purely atomic measure. This corresponds to the unique decomposition of an increasing function into the sum of a continuous and purely discontinuous function.» ? In the theorem, the decomposition is with respect to a second different measure. There's no reference to it in this sentence, or in the paragraph above...
 A: First you have to prove that a $\sigma-$finite measure on $\mathbb{R}^+$ has only points for atoms (this also depends on the definition you have of atom, the one I am using here is: a set $A$ such that $B\subsetneq A$, $B$ measurable implies $\mu(B)\in\{0,\mu(A)\}$). This is a nice exercise but if you need I can be more explicit.
Now that we have this result, let $S=\{x\in\mathbb{R}^+:d\alpha(\{x\})>0\}$. This set must be countable. One proves this result with a standard trick: suppose that $S$ is uncountable. Then, since $d\alpha$ is $\sigma-$finite, $\mathbb{R}^+=\cup B_n$ with $\mu(B_n)<+\infty$. Since $S=\cup_n (S\cap B_n)$, there must be a $j:$ $T:=S\cap B_j$ is uncountable (actually, it suffices for $d\alpha$ to be semi-finite but this is not relevant here). Now, let $T_n=\{x\in T:d\alpha(x)>\frac1n\}$. Since $T=\cup T_n$ and $T$ is uncountable, there must be a $k:T_k$ is uncountable. Let $\{x_1,\dots,x_n,\dots,\}$ be a countable sequence in $T_k$. This implies $d\alpha(B_j)\ge \sum d\alpha(x_l)\ge \frac{1}{k}\sum_l 1=+\infty$, absurd.
Since $S$ must be countable, let $\mu:=d\alpha_{|S}$ and $\nu=d\alpha-\mu$. Can you see how $\mu$ is purely discontinuous and $\nu$ is diffuse?
