Two questions on Lebesgue Decomposition of an increasing function? I come up with the question in doing Stein's Real analysis, Chap3. Ex. 24, which assert that any increasing function $f$ on $[a,b]$ can be decomposed as $$F=F_A+F_C+F_J,$$ with $F_A$ is absolutely continuous, $F_J$ is a jump function, and $F_C$ is a singular function, i.e., it is continuous and  $F'_C=0$, a.e.. Both $F_A$, $F_J$, $F_C$ are increasing on $[a,b]$. Moreover, the decomposition is unique upto a constant.
My questions are:


*

*Did we need to add the boundedness of $F$? such that it is a BV function on $[a,b]$?(since only in this case I have proved the existence.)

*How to show the uniqueness?


Any help or suggestion will be welcome.
 A: Any increasing function on a closed interval is bounded. This is because everything lies between $F(a)$ and $F(b)$.
For uniqueness, it suffices to show that two out of three are unique. But the AC part is uniquely determined by the derivative of the function (which exists a.e.) and the jump function is uniquely determined by the jumps of the original function. Thus, the decomposition is unique.
A: The question 1 is answered by Brian Rushton. My idea for question 2 is listed below. We assume that there are two different decomposoitions, 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.

