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From Lemma $62$ of Math $245A$ note $5$:

Continuous-singular decomposition for monotone functions: Let ${F: {\bf R} \rightarrow {\bf R}}$ be a bounded, monotone non-decreasing function. Then ${F}$ can be expressed as the sum of a continuous monotone non-decreasing function ${F_c}$ and a jump function ${F_{pp}}$.

It can be shown that the decomposition is given by setting $F_{pp} := \sum_{x \in A} c_xJ_x$, where the jump $${c_x := F_+(x) - F_-(x) > 0}$$ and the fraction for the basic jump function $J_x$ $${\theta_x := \frac{F(x)-F_-(x)}{F_+(x)-F_-(x)} \in [0,1]}$$

are defined for points $x$ in $A$ - the set of discontinuities of $F$; and then setting $F_c := F - F_{pp}$.

Problem: Find a suitable generalisation of the notion of a jump function that allows one to extend the above decomposition to unbounded monotone functions, and then prove this extension. (Hint: the notion to shoot for here is that of a “locally jump function”.)

Attempt: The restriction of any monotone function to a compact interval $[a,b]$ is bounded, so if one extends $F \downharpoonright_{[a,b]}$ to ${\bf R}$ by setting $F(x) := F(a)$ for $x < a$ and $F(x) := F(b)$ for $x > b$, then $F$ is bounded, and the Lemma applies. As such, we define a locally jump function to be a function that agrees with a jump function on any compact interval, and claim that $F$ can be expressed as the sum of a continuous monotone non-decreasing function $F_c$ and a locally jump function $F_{pp}$. For any interval $[n,n+1]$, denote the above extension by $F_n$, with \begin{align*} F_n := F_{c}^n + F_{pp}^n \end{align*} as given by the Lemma. I was trying to show that $$F_c := \sum_{n \in {\bf Z}} F_c^n \downharpoonright_{(n,n+1]},\ F_{pp} := \sum_{n \in {\bf Z}} F_{pp}^n \downharpoonright_{(n,n+1]}$$ is the desired decomposition, but ran into issues showing that the continuous $F_c^n$ agree at the endpoints. Namely, that $F_c^n(n+1) = F_c^{n+1}(n+1)$ for all $n$. Is this argument on the right track?

For context, the definition of a jump function is given here: Uniqueness of a special case of Lebesgue decomposition.

Edit: With the generalization of the notion of a jump function to locally jump function, it seems that no further changes on the original decomposition are needed.

shark
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  • If $F_c^n(n+1) \ne F_c^{n+1}(n+1)$ can we not make them equal by subtracting the difference from the continuous component and adding it to the jump function (add a jump at $n+1$)? – Jayanth R Varma Jun 25 '25 at 12:17
  • @JayanthRVarma: Then the $F_c^n$ is no longer continuous at the right endpoint. – shark Jun 25 '25 at 17:15
  • You can add it to $F_c^{n+1}$ where this will be the left end point which does not impact continuity. – Jayanth R Varma Jun 26 '25 at 08:44
  • @JayanthRVarma: $F_c^{n+1}$ will then need to be glued to $F_c^{n+2}$ and so on. In general I don’t think this is a valid operation. – shark Jun 26 '25 at 16:14
  • I look at it this way. For the decomposition over a compact interval, the jump at the left end point is arbitrary as it can be compensated by adding an offsetting constant to the continuous component. Therefore, for the jump component over the interval $[n, n+1]$, we are free to choose the jump at the left end point to be equal to the jump in the original function $F$ at the same point if any. That will ensure that the continuous components are glued together correctly. – Jayanth R Varma Jun 27 '25 at 07:11
  • @JayanthRVarma: As stated before, one would need countably many such minor adjustments (add and subtract a constant), one for each interval $[n-1, n]$, which is a not a valid operation. Actually, with the generalized notion of a "locally jump function", the original construction of the proof would suffice, which I added below for completeness. – shark Jun 27 '25 at 15:47

1 Answers1

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Let $F: {\bf R} \rightarrow {\bf R}$ be a monotone non-decreasing function, and $A$ the set of discontinuities of $F$, which is at most countable. We define a locally jump function to be a function that is a jump function on any compact interval $[a,b]$, and claim that $F$ can be expressed as the sum of a continuous monotone non-decreasing function $F_c$ and a locally jump function $F_{pp}$.

For each $x \in A$, we define the jump $c_x := F_+(x) - F_-(x) > 0$, and the fraction $\theta_x := \frac{F(x)-F_-(x)}{F_+(x)-F_-(x)} \in [0,1]$. Thus \begin{align*} F_+(x) = F_-(x) + c_x \hbox{ and } F(x) = F_-(x) + \theta_x c_x. \end{align*} Note that $c_x$ is the measure of the interval $(F_-(x),F_+(x))$. By monotonicity, these intervals are disjoint. Since $F$ is bounded on $[a,b]$, the union of these intervals for $x \in A \cap [a,b]$ is bounded. By countable additivity, we thus have $\sum_{x \in A \cap [a,b]} c_x < \infty$, and so if we let $J_x$ be the basic jump function with point of discontinuity $x$ and fraction $\theta_x$, then the function \begin{align*} F_{pp} := \sum_{x \in A} c_xJ_x \end{align*} is a locally jump function. $F$ is discontinuous only at $A$, and for each $x \in A$ one easily checks that \begin{align*} (F_{pp})_+(x) = (F_{pp})_-(x) + c_x \hbox{ and } F_{pp}(x) = (F_{pp})_-(x) + \theta_x c_x \end{align*} where $(F_{pp})_-(x) := \lim_{y \rightarrow x^-} F_{pp}(y)$, and $(F_{pp})_+(x) := \lim_{y\rightarrow x^+} F_{pp}(y)$. We thus see that the difference $F_c := F-F_{pp}$ is continuous. Finally, we verify that $F_c$ is monotone non-decreasing. By continuity it suffices to verify this away from the (countably many) jump discontinuities, thus we need \begin{align*} F_{pp}(b)-F_{pp}(a) \leq F(b)-F(a) \end{align*} for all $a < b$ that are not jump discontinuities. But the LHS can be rewritten as $\sum_{x \in A \cap [a,b]} c_x$, while the RHS is $F_-(b) - F_+(a)$. As each $c_x$ is the measure of the interval $(F_-(x), F_+(x))$, and these intervals for $x \in A \cap [a,b]$ are disjoint and lie in $(F_+(a),F_-(b))$, the claim follows from countable additivity.

shark
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