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This question is related to the post: Lebesgue decomposition of increasing function on an unbounded interval?

Let ${F: {\bf R} \rightarrow {\bf R}}$ be a monotone non-decreasing function. If ${F}$ is bounded, then ${F}$ can be expressed as the sum of a continuous monotone non-decreasing function ${F_c}$ and a jump function ${F_{pp}}$. Here a jump function is any absolutely convergent combination of basic jump functions, i.e. a function of the form ${F = \sum_n c_n J_n}$, where ${n}$ ranges over an at most countable set, each ${J_n}$ is a basic jump function, meaning

$\displaystyle J_n(x) := \left\{ \begin{array}{ll} 0 & \hbox{ when } x < x_0 \\ \theta & \hbox{ when } x = x_0 \\ 1 & \hbox{ when } x > x_0 \end{array} \right.$

for some real numbers ${x_0 \in {\bf R}}$ and ${0 \leq \theta \leq 1}$.

Question: Show that the decomposition is unique. As noted elsewhere, it can be shown that the decomposition is unique up to a constant by showing that any other decomposition $F= F_{c}' + F_{pp}'$ into continuous $F_{c}'$ and jump component $F_{pp}'$ will be such that $F_{pp} - F_{pp}' = c$ for some constant $c$. Is there any feature of this special case that dictate the value of $c$ to be zero here?

shark
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1 Answers1

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In a bounded interval $[a,b]$, the constant function is both a continuous function (obvious) and also a jump function (the jump is at $a$). So you can add a constant (as a jump function) to the $F_{pp}$ and subtract the same constant (as a continuous function) from $F_c$. This will give you a different decomposition of $F$. But if the interval is unbounded ($\mathbb R$), the constant is not a jump function and you cannot "add" it to $F_{pp}$.

Jayanth R Varma
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  • Fair point. The only constant jump function on the real line consistent with the definition given here is the zero function. – shark Jun 24 '25 at 07:04