$BV[a,b] \subset B[a,b]$

Question

I am a little confused about this statement. Can't we take $\tan(x)$ where $x \in [0,\pi/2]$, then it is monotonic hence of bounded variation but obviously it is not bounded as a counterexample?

In the proof, Carothers uses for $x \in [a,b]$ and $P = \{a,x,b\}$: then $$ \lvert f(x)-f(a)\rvert \leq V(f,P) \leq V_a^bf $$ which implies $$\lvert f(x) \rvert \leq \lvert f(a) \rvert + V_a^bf. $$ I am not quite sure how he gets the last inequality (reverse triangular inequality?)

Answer 1

By definition $|f(x)-f(a)| + |f(b)-f(x)| \leq V_a^{b}f$. This implies $|f(x)-f(a)| \leq V_a^{b}f$. Hence $|f(x)| \leq |f(a)|+|f(x)-f(a)| \leq |f(a)|+ V_a^{b}f$.

As far as your example is converned, $tan \, x $ is not a real valued function on $[0,\pi/2]$