Lebesgue decomposition of a function $F(x)=\begin{cases} x^3+5, & x\ge 1\\ x^3+2, & 0\leq x<1\\ x^3, & x<0 \end{cases}$
Let $\mu_{F}$ be the Lebesgue-Stieltjes measure associated with $F$. Find the lebesgue decomposition of $\mu_{F}$ with respect to the Lebesgue measure $m$.
$\frac{dF(x)}{dx}=3x^2$. It has jump of height 2 at $x=0$ and has jump of height 3 at $x=1$. 
My answer is $\mu_{F}=\mu_{a}(E)+\mu_{d}(E)=\int_{E} 3x^2dm(x)+2\delta_{0}+3\delta_{1}$, where $\mu_{a}<<m$, and $\mu_{a}\perp\mu_{d}$.
I am asking because I do not quite understand the concept of Lebesgue decomposition. Are there any flaws in my analysis? I asked similar questions here before, but no one answers.
 A: You first try to subtract off $3x^2 dm(x)$ from $\mu_F$, since that's definitely an absolutely continuous measure that's in there.
What you're left with is now $\mu_G$ where $G(x)=\begin{cases} 0 & x < 0 \\ 2 & 0 \leq x < 1 \\ 5 & x \geq 1 \end{cases}$. At this point you see that $\mu_G$ and $m$ are mutually singular, by decomposing $\mathbb{R}$ into the disjoint union $\{ 0,1 \} \cup (\mathbb{R} \setminus \{ 0,1 \})$. So $\mu_F=3x^2 dm(x) + \mu_G$ is a Lebesgue decomposition of $\mu_F$.
A: This function is not absolutely continuous,to be precise.For one thing, absolutely continuous functions need to be continuous,which is clearly not the case here.Secondly,you can think of Lebesgue decomposion theorem in terms of projections on cones.You call two measures mutually orthogonal if either of them is zero on the support of the other,so in a sense,they don't 'overlap'.You can see for yourself that the space of all positive measures(take finite,if you wish) on a measurable space forms a convex cone over $\mathbb{R}_{\geq 0}$.The collection of measures absolutely continuous with respect to a given measure $m$ forms a cone too.So you think of Lebesgue decomposition theorem in terms of an orthogonal decomposition,a concept you might be familiar with from geometry and linear algebra.The absolutely continuous part can be thought of as the projection of the measure on the cone of measures absolutely continuous with respect to it.
A: F is an increasing function, so you have the Lebesgue decomposition for the positive Lebesgue Stieltjes measure. See for example Theorem 8 of Complex Measure, Dual Space of L
p Space, Radon-Nikodym Theorem and Riesz Representation Theorems. (https://my-calculus-web.firebaseapp.com/MA3110/Complex%20Measure_Dual_space_Riesz_Thm.pdf). The proof there is a general existence proof. For Lebesgue Stieltjes measure, specifically we have a more detail decomposition: for the function F here, we can write it as a sum of an increasing continuous function x3 and an obvious Saltus function. Note that x3 is not absolutely continuous on the set of real numbers but is locally absolutely continuous. Taking the domain as a bounded interval, from Theorem 19 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,
Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals (https://my-calculus-web.firebaseapp.com/MA3110/Lebesgue%20Stieltjes%20Measure%20and%20de%20La%20Vallee%20Poussin%20Decomposition.pdf), appropriate interpreted, we have a Lebesgue decomposition on any bounded interval. By Passing the domain to the whole of the reals we have the Lebesgue decomposition. The explicit form of the decomposition is given by Theorem 22 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition, Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals, where we can ignore the contributions from the points having infinite derivative, since there are none. Note that the absolute part is given by the integral of the derivative of x3.  The contribution from the saltus part of the function F is given by the sum of the jumps at 0 and 1 which is the same as the sum of the jumps at all points of a Borel set and so are the sum of the function given by you, 2 times the characteristic function of {0} plus 3 times the characteristic function at {1}. Note that Theorem 19 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,
Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals says that these two parts are mutually singular.  Proposition 12 of Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals affirms that the absolute part of the Lebesgue Stieltjes measure of F, is absolutely continuous with respect to the Lebesgue measure on the reals since x3 is a locally absolutely continuous function on the reals. Hope the references cited here would be useful.
