Lebesgue decomposition of increasing function on an unbounded interval?

Question

The Lebesgue decomposition theorem for functions of bounded variation states: If $f : I \to \mathbb{R}$ is a right-continuous function of bounded variation on an interval $I$ then $f$ has a decomposition $f = f_A + f_C + f_J$ that is unique up to constants where:

• $f_A$ is absolutely continuous.

• $f_C$ is singular (it is continuous, of bounded variation, and has zero derivative almost everywhere).

• $f_J$ is a jump function.

Meanwhile Stein et al set an exercise (Chapter 3, Exercise 24) to show: If $f : [a,b] \to \mathbb{R}$ is increasing on $[a,b]$ then $f = f_A + f_C + f_J$ that is unique up to constants where each of the functions $f_A$, $f_C$, $f_J$ is increasing and:

• $f_A$ is absolutely continuous.

• $f_C$ is continuous but $f_C'(x) = 0$ for almost everywhere $x$

• $f_J$ is a jump function.

(Stein et al note the existence of the corresponding decomposition for functions of bounded variation.) See this question, this question, and this question.

My question: If $f : I \to \mathbb{R}$ is increasing on an unbounded interval $I$, does it necessarily have a decomposition $f = f_A + f_C + f_J$ that is unique up to constants where each of the functions $f_A$, $f_C$, $f_J$ is increasing and $f_A$ is absolutely continuous, $f_C$ is continuous with zero derivative almost everywhere, and $f_J$ is a jump function? (Without imposing an extra condition of $f$ being of bounded variation on $I$?)

It seems to me that I can't just use the Lebesgue decomposition theorem for functions of bounded variation because $f$ is not necessarily of bounded variation on $I$. But my intuition is that I can use the 'Stein et al' decomposition theorem for increasing functions as follows:

1. Construct a countable, mutually disjoint set of intervals $\{ I_k \}_k$ such that $I = \bigcup_k I_k $:

   • If $I$ is unbounded on both sides then for each $k \in \mathbb{Z}$ let $I_k = [k-1,k]$.

   • If $I$ is a one-sided unbounded interval then by a change of variables, we can write $I = [0,\infty)$ without loss of generality. For each $k \in \mathbb{Z}$ let $I_k = [2^{k-1}, 2^k]$.

2. Define $f_k : I_k \to \mathbb{R}$ as $f_k(x) = f(x)$. Apply the 'Stein et al' decomposition theorem to $f_k$ to generate a decomposition $f_k = f_{A,k} + f_{C,k} + f_{J,k}$.

3. Stitch the neighbouring decompositions together: construct $f_A$ by adding constants to align $f_{A,k-1}$ with $f_{A,k}$ and likewise construct $f_C$ by adding constants to align $f_{C,k-1}$ with $f_{C,k}$. Then mod out those constants by introducing a jump into $f_J$ at the interface from $I_{k-1}$ to $I_k$.

Is that right? Many thanks in advance for any thoughts.

Answer 1

I think this is false in general unless $f$ is bounded as you mentioned. For instance, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x) = x^{2}$ for $x\geq 0$ and $f(x) = 0$ for $x\leq 0$. Suppose a decomposition $f = f_{A}+f_{C}$ existed on $\mathbb{R}$ (with $f_{J} \equiv 0$ obviously). Since a unique Lebesgue decomposition is valid for each bounded interval, up to constants we have $f_{A}|_{[-M, M]} = f|_{[-M, M]}$ (noting that the restriction of $f$ to any bounded interval is absolutely continuous) and $f_{C}|_{[-M, M]}\equiv 0$ for any $M > 0$. From this simple observation, we get $f_{A} = f$ on all of $\mathbb{R}$ but $f$ is not absolutely continuous on $\mathbb{R}$ (since it is not even uniformly continuous).