I read this question: Prove that $\sin x$ is not of bounded variation, $\sin x\not\in BV$ and the answer by Robert Z is:
Hint. Let $x_n = \frac{\pi(2n + 1)}{2}$ for $n\in\mathbb{N}$. Then $\{x_n\}$ is a strictly increasing sequence and $\sin(x_n)=(-1)^n$. Now consider the sum $$\sum_{n=1}^N \left| \sin(x_n) - \sin(x_{n-1})\right|=\sum_{n=1}^N 2=2N.$$ What may we conclude? Recall the definition of bounded variation function and take a look at Functions of bounded variation on $\Bbb R$
However, I don't understand how I can conclude that in light of the following definition:
A function $f\colon\mathbb{R}\to\mathbb{R}$ is said a bounded variation on $\mathbb{R}$ if it is in $BV([a,b])$ for all $-\infty<a<b<+\infty$ and results $$V(f):=\sup_{-\infty<a<b <+\infty} V_a^b(f)<\infty.$$
In this case $V_a^b(f):=\sup_{\mathcal{P}\in \Pi} V_a^b(f,\mathcal{P})$, where $\Pi$ is the collection of all partition of $[a,b]$.
Could you give me a further suggestion? Thanks!
