Does bounded variation imply boundedness

That $\|f\|_{TV}<\infty$ implies that $f$ is bounded is quite straightforward: $|f(x)|\le|f(0)|+|f(x)-f(0)|\le |f(0)|+\|f\|_{TV}$ holds for all $x$.

We can also do it by contradiction: assuming $f$ is not bounded, then using definition of unbounded functions, choose such a partition and hence prove that our function can not be of bounded variation.

If $f$ is a real valued function of bounded variation on an interval [a,b] then $f$ is bounded because

$$ |f(x)| \leq |f(a)| + |f(x) - f(a)| \leq |f(a)| + \sup_{x_0 < \cdots < x_n}\sum_{i = 1}^{n} |f(x_i) - f(x_{i - 1})|. $$ for all $x \in [a,b]$. I believe the argument by Hagen was for the closed interval $[0,1]$.