Let $f$ be absolutely continuous on $[a,b]$. I want to prove that $f$ is of bounded variation. I am reading Royden and Fitzpatrick so they use the following notations: Let $P=\{x_0=a,x_1,\ldots,x_{n-1},x_n=b\}$ be a partition of $[a,b]$. Then $$ V(f,P)=\sum_{i=1}^n\lvert f(x_i)-f(x_{i-1})\rvert $$ is the variation of $f$ over $P$ and $$ TV(f_{[a,b]})=\sup_{P}\{V(f,P)\,:\,P\text{ a partition of }[a,b]\} $$ is the total variation of $f$ over $[a,b]$. For any subinterval $[c,d]\subset[a,b]$ we have $TV(f_{[c,d]})$ denoting the total variation of $f$ when restricted to $[c,d]$.
Then they start the proof that if $f$ is absolutely continuous then it is of bounded variation in the following way:
Let $\delta$ respond to the $\varepsilon=1$ challenge regarding the absolute continuity of $f$. Let $P$ be a partition of $[a,b]$ into $N$ closed intervals $\{[c_k,d_k]\}_{k=1}^N$ each of length less than $\delta$. Then by the definition of $\delta$ in relation to the absolute continuity of $f$, it is clear that $TV(f_{[c_k,d_k]})\le 1$, for $1\le k\le N$.
This is not clear to me at all.
I understand that by the absolute continuity of $f$ we have $$ (d_k-c_k)<\delta\,\,\Longrightarrow\,\,\lvert f(d_k)-f(c_k)\rvert<1 $$ for all $1\le k\le N$. But in general $|f(d_k)-f(c_k)|\le TV(f_{[c_k,d_k]})$ and we get equality when $f$ is nondecreasing on $[c_k,d_k]$. So how is it clear that that $TV(f_{[c_k,d_k]})\le1$ as well, as stated in the text?
I must be missing something mundane here, but could someone shed some light on this for me?
UPDATE: I fixed it. Since every $d_k-c_k<\delta$, every partition of $[c_k,d_k]$ will yield $V(f_{[c_k,d_k]},P)<1$ so that $TV(f_{[c_k,d_k]})=\sup_P V(f_{[c_k,d_k]})\le 1$.
