Assume $f$ is of bounded variation on $[a,b]$. Show that there is a sequence of partitions $\{P_n\}$ of $[a,b]$ for which the sequence $\{TV(f,P_n)\}$is increasing and converges to $TV(f)$? Remark:-$TV(f)$ stand for total variation .
2026-03-28 09:54:31.1774691671
Problem on Bounded Variation
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By definition $$ \mathrm{TV}(f)=\sup_P\mathrm{V}(f,P). $$ So (from the definition of the supremum) for every $\varepsilon>0$, there exists a partition $P$ of $[a,b]$, such that $$ \mathrm{TV}(f)-\varepsilon\le\mathrm{V}(f,P) \le \mathrm{TV}(f). $$ In particular for $\varepsilon=1/n$, there exists a partition $Q_n$ of $[a,b]$, such that $$ \mathrm{TV}(f)-\frac{1}{n}\le\mathrm{V}(f,Q_n) \le \mathrm{TV}(f). $$ Let now a new sequence of partitions be defined by $$ P_n=Q_1\cup Q_2\cup\cdots\cup Q_n. $$ Clearly $P_n\subset P_{n+1}$ and thus $$ \mathrm{V}(f,P_n)\le \mathrm{V}(f,P_{n+1}), $$ i.e., $\{\mathrm{V}(f,P_n)\}$ is increasing, and also $$ \mathrm{TV}(f)-\frac{1}{n}\le\mathrm{V}(f,Q_n) \le \mathrm{V}(f,P_n) \le \mathrm{TV}(f), $$ and thus $$ \lim_{n\to\infty}\mathrm{V}(f,P_n)=\mathrm{TV}(f). $$