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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:

  1. 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.)
  2. How to show the uniqueness?

Any help or suggestion will be welcome.

Brian Rushton
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van abel
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2 Answers2

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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. $F=F_A^{1}+F_C^{1}+F_J^{1}=F_A^{2}+F_C^{2}+F_J^{2}$.

  1. 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 $F_J ^{1} - F_J ^{2} =C_J$;
  2. 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 $F_A^{1} -F_A^{2} =C_A$;
  3. Let $C_C =-(C_A +C_J)$, we have $F_C^{1} -F_C^{2} =C_C$. Here we conclude the proof.
Jiayu QIU
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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.

Brian Rushton
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  • Add a constant to any one, subtract it from any other. This doesn't change the derivative or the jumps. Perhaps postulate that they all vanish at $a$ to get uniqueness. – GEdgar Jul 05 '13 at 18:37
  • Yes, I think you're right. The two parts I specified are unique with your condition. – Brian Rushton Jul 05 '13 at 18:50
  • It seems that the boundness is not followed from what you have side, since in the real analysis case, we can consider the function of value $\infty$, also you can see that almost every theorem or lemma in chap3 in stein's book will say something like:"let $f$ be bounded and increasing on $[a,b]$"... – van abel Jul 06 '13 at 11:33
  • Why the jump function is uniquely determined by the jumps of the original function? It seems that we can add any continuous increasing function to the jump function such that it is also a jump function ... – van abel Jul 06 '13 at 12:52
  • I assumed that a jump function was a function that is constant on the complement of the jump points. – Brian Rushton Jul 06 '13 at 13:14
  • @BrianRushton It seem here we did not assume it, from http://www.encyclopediaofmath.org/index.php/Lebesgue_decomposition we can find another definiton of $F_J$: "It can be characterized as the function $g$ with smallest variation such that $F-g$ is continuous." And it seems that we should first show that $F_C$ is unique... – van abel Jul 06 '13 at 13:36
  • Sorry to make a mistake, that the jump function is defined as $f^+-f^-$, thus it is constant on the complement of the jump. – van abel Jul 06 '13 at 14:25