Does there exist a sequence of partitions $\{P_n\}$ of $[a,b]$ such that $\{V_a^b(f,P_n) \}$ is increasing and converges to $V_a^b(f)$ ?

Question

LEt $f:[a,b] \to \mathbb R$ be of bounded variation . Then does there exist a sequence of partitions $\{P_n\}$ of $[a,b]$ such that $\{V_a^b(f,P_n) \}$ is increasing and converges to $V_a^b(f)$ ?

Answer 1

Let $\mathcal{P}$ be the set of partitions of $[a, b]$. Then $V = V_a^b (f, \cdot)$ is a function from $\mathcal{P}$ to $[0, \infty)$. We observe that

1. $\mathcal{P}$ is equipped with the ordering $\subseteq$. Also, if $P, Q \in \mathcal{P}$, then $P\cup Q \in \mathcal{P}$.

2. $V : \mathcal{P} \to [0, \infty)$ is an non-decreasing function. That is, $P \subseteq Q$ implies $V(P) \leq V(Q)$.

3. We can choose a sequence $(\tilde{P}_n)$ of partitions such that $V(\tilde{P}_n) \to V_a^b(f)$.

Now all the ingredients are at our hand. Let

$$ P_n = \tilde{P}_1 \cup \cdots \cup \tilde{P}_n. $$

Then $(P_n)$ is an increasing sequence of partitions. This implies that $(V(P_n))$ is also increasing. Moreover, we have the following inequality:

$$ V(\tilde{P}_n) \leq V(P_n) \leq V_a^b (f). $$

Taking $n\to\infty$, squeezing lemma shows that $V(P_n)$ also converges to $V_a^b (f)$ as well. So we are done.