I forgot to add details on my attempt in previous post. Since that one is closed, here is new post: To be honest, I don't have any idea on this problem. Because, original definition of bounded variation concerns bounded intervals whereas we are said to try whole real numbers in this problem. Any help is strongly appreciated!
Suppose that $f$ is of bounded variation on $\mathbb{R}$. Let $\mathrm{c}$ be any point in $\mathbb{R}$. Prove that $f$ is of bounded variation on $(-\infty, c]$ and on $[c, \infty)$. That is, prove that $$ V(f ;-\infty, c)=\sup (V(f: a, b):[a, b] \subset(-\infty, c]) $$ and $$ V(f ; c, \infty)=\sup (V(f ; a, b):[a, b] \subset[c, \infty) \mid $$ are finite. Further, prove that $$ V(f ; \mathbb{R})=V(f ;-\infty, c)+V(f ; c, \infty) $$


1 Answer 1


I assume $V(f; \mathbb{R})$ is defined by $\sup(V(f; a,b) : [a,b] \subset \mathbb{R})$, and that "$f$ is of bounded variation of $\mathbb{R}$" means that this quantity is finite.

Let $A := \{V(f; a,b) : [a,b] \subset (-\infty, c]\}$ and $B := \{V(f; a,b) : [a,b] \subset [c, \infty)\}$ and $C := \{V(f; a,b) : [a,b] \subset \mathbb{R}\}$. You know $\sup C < \infty$ and want to show

  1. $\sup A < \infty$, $\sup B < \infty$
  2. $\sup C = \sup A + \sup B$

The fact that $A \cup B \subset C$ proves (1), as well as $\sup C \ge \sup A + \sup B$, so the only remaining step is to show $\sup C \le \sup A + \sup B$. To do this, we will show that for any $v_1 \in C$, we can produce $v_2 \in A$ and $v_3 \in B$ such that $v_1 \le v_2 + v_3$.

Take any $[a,b] \subset \mathbb{R}$.

  • If $c \le a$, then $V(f; [a,b])$ is in $C$ as well as in $A$
  • If $c \ge b$, then $V(f; [a,b])$ is in $C$ as well as in $B$
  • If $a < c < b$, then check that $V(f;[a,b]) \le V(f;[a,c]) + V(f;[c, b])$ (take any partition of $[a,b]$, and split the interval containing $c$ into two intervals), and note that $V(f; [a,b]) \in C$, $V(f; [a,c]) \in A$, and $V(f; [c,b]) \in B$.

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