TITLE: Refinement of Lebesgue Decomposition Theorem
QUESTION [3 upvotes]: On Wikipedia, a "refinement" of the Lebesgue decomposition theorem is given, and it is also given as problems in Stein and Shakarchi and Bruckner and Thomson. Can someone provide a comprehensive proof of it (I've written it below), as I'm having trouble.
Suppose $F$ is an increasing function on [a,b]. Prove we can write $F = A + B + C$, where $A, B, C$, are increasing functions and: A is absolutely continuous; $B$ is continuous, but $B'(x)=0$ for almost everywhere x; and C is a jump function. Moreover, prove $A, B$, and $C$ are uniquely determined up to an additive constant.

REPLY [2 votes]: First we prove the existence of such a decomposition:

It's easy to observe that any increasing function on a bounded interval is bounded (in fact, all value of $F$ lie between $F(a)$ and $F(b)$. Note that I exclude the situation of infinite value and it's rather worthless of discussing), hence, $F$ could be expressed as the sum of a increasing continuous function and a jump function (see Lemma 3.13, Ch3). Denote that $F=F_J +f$, where $F_J$ is a jump function and $f$ is continuous and increasing;
Since $f$ is increasing, it is obviously of bounded variation. Thus, $F'$ exists almost everywhere and according to Corollary 3.7, it is locally integrable on $[a,b]$. Hence, it is reasonable to write $f=F_C + F_A$, where $F_A (x) =\int_{a}^{x} F'(x) dx$ and $F_C (x) =f(x) - F_A (x)$. Due to the absolute continuity of L-intregration, it is easy to see that $F_A$ is absolutely continuous and $F_C$ is continuous. Besides, according to the Lebesgue differentiation theorem, $F_C ' (x)=0$ almost everywhere. Here we conclude the proof of existence.

The proof of uniqueness proceeds as follows. We assume that there exists two different decompositions, i.e. $F=F_A^{1}+F_C^{1}+F_J^{1}=F_A^{2}+F_C^{2}+F_J^{2}$.

Since $\Delta_J = F_J ^{1} - F_J ^{2}=F_A^{2}+F_C^{2} -F_A^{1}-F_C^{1}$ is also a jump function and the RHS of the equation is a continuous function, the jump points of $F_J ^{1}$ must meet the ones of $F_J ^{2}$. Thus, they differ from each other with a constant, namely $F_J ^{1} - F_J ^{2} =C_J$;
Since absolutely continuous function is differentiable almost everywhere, we have $F_A^{1'} -F_A^{2'}=F_C^{2'} -F_C^{1'} - C_J ' =0 $ (a.e.). Let $\Delta_A = F_A^{1} -F_A^{2}$ and it is easy to observe that $\Delta_A$ is absolutely continuous. Due to Lebesgue differentiation theorem, we have $\Delta_A (x) - \Delta_A (a) = \int_{a}^x (F_A^{1'} -F_A^{2'}) dx =0$. Let $C_A =\Delta_A (a)$, we have $F_A^{1} -F_A^{2} =C_A$;
Let $C_C =-(C_A +C_J)$, we have $F_C^{1} -F_C^{2} =C_C$. Here we conclude the proof.
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