Why is this function of bounded variation?

Question

Let $g=[0,1]\rightarrow \mathbb R$ be a function such that for any sequence $(f_n)$ of left-continuous step functions on $[0,1]$ decreasing to $0$, we have that $\int_0^1 f_n(x) dg(x)$ converges to $0$.

I read that this implies that $g$ is of bounded variation. How to show that? I imagine this has something to do with Carathéodory extension theorem?

Answer 1

Preliminaries on Riemann-Stieltjes integration.

Proposition 1. If $f,g \colon [0,1] \to \mathbb{R}$ are functions sharing a point of discontinuity $x_0 \in [0,1]$, then the R-S integral $\int_0^1 f(x) \, dg(x)$ does not exist.

Therefore, for the hypothesis of the question to make sense (with respect to R-S integration), we require that $g \colon [0,1] \to \mathbb{R}$ is continuous on $[0,1)$.

Proof of Proposition 1. We split into three (overlapping) cases:

• $x_0$ is a shared point of right-discontinuity;
• $x_0$ is a shared point of left-discontinuity;
• $x_0 \in (0,1)$ and $g(x)$ does not converge as $x \to x_0$.

For the first case: Let $\varepsilon_f>0$ be such that for all $x \in (x_0,1]$, the range of $f$ over $[x_0,x]$ is greater than $\varepsilon_f$. Let $x_n$ be a sequence in $(x_0,1]$ converging to $x_0$ such that for some value $\varepsilon_g>0$, every $n$ has $|g(x_n) - g(x_0)| \geq \varepsilon_g$. Let $\mathcal{P}_n$ be a sequence of partitions of mesh tending to $0$ such that for each $n$, $[x_0,x_n]$ is one of the intervals defined by the partition $\mathcal{P}_n$. Using the sequence of partitions $\mathcal{P}_n$, define two corresponding sequences of Riemann-Stieltjes sums of $f$ against $g$: for each $n$, for every partition-interval except $[x_0,x_n]$, choose the same point in that interval for both of the Riemann-Stieltjes sums; but for the interval $[x_0,x_n]$, choose the points $\zeta_n,\xi_n \in [x_0,x_n]$ for the two respective Riemann-Stieltjes sums such that $|f(\zeta_n)-f(\xi_n)| \geq \varepsilon_f$. Then for every $n$, the difference between the two Riemann-Stieltjes sums is at least $\varepsilon_{f\,} \varepsilon_g$.

The second case is proved the same way as the first case.

For the third case: Let $\varepsilon_f>0$ be such that for all $x \in [0,x_0)$ and $y \in (x_0,1]$, the range of $f$ over $[x,y]$ is greater than $\varepsilon_f$. Let $x_n$ and $y_n$ be sequences in $[0,x_0)$ and $(x_0,1]$ respectively, both converging to $x_0$, such that for some value $\varepsilon_g>0$, every $n$ has $|g(y_n)-g(x_n)| \geq \varepsilon_g$. The rest of the proof proceeds exactly as in the first case, except with $[x_n,y_n]$ in place of $[x_0,x_n]$. $\ \square$

Proposition 2. Given a left-continuous step function \begin{align*} &f \colon [0,1] \to \mathbb{R} \\ &f = c_{-1}\mathbb{1}_{\{0\}} + \sum_{i=0}^N c_i \mathbb{1}_{(a_i,\,a_{i+1}]}, \quad a_0=0, \ a_{N+1}=1, \ N \geq 0, \end{align*} and a function $g \colon [0,1] \to \mathbb{R}$ that is continuous on $[0,1)$, we have $$ \int_0^1 f(x) \, dg(x) = \sum_{i=0}^N c_i (g(a_{i+1})-g(a_i)). $$

Proof. Fix $\varepsilon>0$, and take $0<\delta<\min(a_{i+1}-a_i : i \in \{0,\ldots,N\})$ such that for all $i \in \{0,\ldots,N\}$ and $x \in [a_i-\delta,a_i+\delta] \cap [0,1]$, $$ |c_i-c_{i-1}||g(x)-g(a_i)| < \tfrac{\varepsilon}{N+1}. $$ For any partition $\mathcal{P}=\{0=a_0'< \ldots < a_{N'+1}'=1\}$ of mesh less than $\delta$, for any choice of $\xi_i \in [a_i',a_{i+1}']$ across $i \in \{0,\ldots,N'\}$, letting \begin{align*} Q^- &= \{i \in \{0,\ldots,N'\} : a_{j(i)} \in [a_i',a_{i+1}') \text{ and } \xi_i \in [a_i',a_{j(i)}] \text{ for some } j(i) \in \{0,\ldots,N\} \} \\ Q^+ &= \{i \in \{0,\ldots,N'\} : a_{j(i)} \in [a_i',a_{i+1}') \text{ and } \xi_i \in [a_{j(i)},a_{i+1}') \text{ for some } j(i) \in \{0,\ldots,N\} \} \end{align*} the Riemann-Stieltjes sum is given by $$ \text{(claimed result)} + \left( \sum_{i \in Q^-} (c_{i-1}-c_i)(g(a_{i+1}')-g(a_{j(i)})) \right) + \left( \sum_{i \in Q^+} (c_i-c_{i-1})(g(a_{j(i)})-g(a_i')) \right). $$ Hence the difference between the Riemann-Stieltjes sum and the claimed result is less than $\varepsilon$. $\ \square$

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Answering the original question.

Assume that $g$ is continuous on $[0,1)$. We argue by contrapositive: Suppose $g$ does not have bounded variation; we need to construct a sequence $(f_n)$ of left-continuous step functions decreasing to $0$ such that $\int_0^1 f_n(x) \, dg(x) \not\to 0$ as $n \to \infty$.

For any partition $\mathcal{P}=\{0=a_0 < \ldots < a_{N+1} = 1\}$ with $N \geq 0$, define \begin{align*} \mathcal{P}^+(g) &= \{i \in \{0,\ldots,N\} : g(a_{i+1}) \geq g(a_i) \} \\ \mathcal{P}^-(g) &= \{i \in \{0,\ldots,N\} : g(a_{i+1}) < g(a_i) \} = \{0,\ldots,N\} \setminus \mathcal{P}^+(g) \\ \mathcal{V}^+(\mathcal{P},g) &= \sum_{i \in \mathcal{P}^+(g)} g(a_{i+1}) - g(a_i) \\ \mathcal{V}^-(\mathcal{P},g) &= \sum_{i \in \mathcal{P}^-(g)} g(a_i) - g(a_{i+1}) \\ \mathcal{V}^\vee(\mathcal{P},g) &= \max( \, \mathcal{V}^+(\mathcal{P},g) \, , \, \mathcal{V}^-(\mathcal{P},g) \,) \\ \mathcal{V}^\wedge(\mathcal{P},g) &= \min( \, \mathcal{V}^+(\mathcal{P},g) \, , \, \mathcal{V}^-(\mathcal{P},g) \,) \\ \mathcal{P}^\vee(g) &= \Bigg\{ \begin{array}{c l} \mathcal{P}^+(g) & \text{if } \mathcal{V}^+(\mathcal{P},g) \geq \mathcal{V}^-(\mathcal{P},g) \\ \mathcal{P}^-(g) & \text{if } \mathcal{V}^+(\mathcal{P},g) < \mathcal{V}^-(\mathcal{P},g) \end{array} \\ \mathcal{P}^\wedge(g) &= \{0,\ldots,N\} \setminus \mathcal{P}^\vee(g) \end{align*}

Since $g$ does not have bounded variation, one can find partitions $\mathcal{P}$ for which $\mathcal{V}^\vee(\mathcal{P},g)$ is arbitrarily large. [I'll leave this as an exercise :)] Consequently, I can recursively construct a sequence of partitions $\mathcal{P}_n$ and sequence of positive numbers $\alpha_n>0$ as follows:

• take $\mathcal{P}_1$ to be an arbitrary partition with $\mathcal{V}^\vee(\mathcal{P}_1,g) \geq 1$, and take $\alpha_1 = \max(\mathcal{V}^\vee(\mathcal{P}_1,g),2\mathcal{V}^\wedge(\mathcal{P}_1,g))$;
• given $\mathcal{P}_n$ and $\alpha_n$ for some $n \geq 1$, take $\mathcal{P}_{n+1}$ such that $\mathcal{V}^\vee(\mathcal{P}_{n+1},g) \geq \max(n+1,\alpha_n)$, and take $\alpha_{n+1} = \max(\,\mathcal{V}^\vee(\mathcal{P}_{n+1},g)\, , \, 2\mathcal{V}^\wedge(\mathcal{P}_{n+1},g) \,)$.

Now writing $\mathcal{P}_n$ as $$ \mathcal{P}_n =\{0=a_0^{(n)} < \ldots < a_{N_n+1}^{(n)} = 1\}, $$ define $f_n$ by $$ f_n = \left( \frac{1}{\mathcal{V}^\vee(\mathcal{P}_n,g)} \sum_{i \in \mathcal{P}_n^\vee(g)} \mathbb{1}_{(a_i^{(n)},\,a_{i+1}^{(n)}]} \right) + \left( \frac{1}{\alpha_n} \sum_{i \in \mathcal{P}_n^\wedge(g)} \mathbb{1}_{(a_i^{(n)},\,a_{i+1}^{(n)}]} \right). $$

Claim: $f_n$ is decreasing to $0$, and $\int_0^1 f_n(x) \, dg(x) \geq \frac{1}{2}$ for all $n \geq 1$.

Proof of Claim.

• For the fact that $f_n \to 0$: We have $f_n \geq 0$. We also have that $\alpha_n \geq \mathcal{V}^\vee(\mathcal{P}_n,g) \geq n$, and so $f_n \leq \frac{1}{n}$.
• For the fact that $f_n$ is decreasing: By construction, $f_n(0)=0$ for all $n$, and $f_n(x)$ is equal to either $\frac{1}{\mathcal{V}^\vee(\mathcal{P}_n,g)}$ or $\frac{1}{\alpha_n}$ for each $x \in (0,1]$. So then, the fact that $$ \alpha_{n+1} \geq \mathcal{V}^\vee(\mathcal{P}_{n+1},g) \geq \alpha_n \geq \mathcal{V}^\vee(\mathcal{P}_n,g) $$ gives the result.
• For the fact that $\int_0^1 f_n(x) \, dg(x) \geq \frac{1}{2}$: By Proposition 2, we have that $\int_0^1 f_n(x) \, dg(x)$ is equal to $$ \underbrace{\left( \sum_{i \in \mathcal{P}_n^\vee(g)} \frac{1}{\mathcal{V}^\vee(\mathcal{P}_n,g)} (g(a_{i+1}^{(n)})-g(a_i^{(n)})) \right)}_{=: \, A_n} + \underbrace{\left( \sum_{i \in \mathcal{P}_n^\wedge(g)} \frac{1}{\alpha_n} (g(a_{i+1}^{(n)})-g(a_i^{(n)})) \right)}_{=: \, B_n}. $$ Now $$ A_n = \frac{1}{\mathcal{V}^\vee(\mathcal{P}_n,g)} \sum_{i \in \mathcal{P}_n^\vee(g)} g(a_{i+1}^{(n)})-g(a_i^{(n)}) = \frac{1}{\mathcal{V}^\vee(\mathcal{P}_n,g)} \mathcal{V}^\vee(\mathcal{P}_n,g) = 1 $$ and $$ B_n = \frac{1}{\alpha_n} \sum_{i \in \mathcal{P}_n^\wedge(g)} g(a_{i+1}^{(n)})-g(a_i^{(n)}) = -\frac{1}{\alpha_n} \mathcal{V}^\wedge(\mathcal{P}_n,g) \geq -\tfrac{1}{2}, $$ and so $A_n + B_n \geq \frac{1}{2}$.

Answer 2

Let $\mathcal{S}$ be the set of functions of the following type $$f(x) = \sum_{k=1}^{n}{ c_k 1_{(t_{k-1},t_k]} }$$ where $0 = t_0 < t_1 < \dots < t_n = 1$ and all the $c_k$ are either $+1$ or $-1$

Claim 1 : $$V(g) = \sup_{ f \in \mathcal{S}}{\int_{0}^{1}{f(x)dg(x)}}$$

Then if by contradicition $V(g) = \infty$ you find a sequence $\{ f_n \} \subset \mathcal{S} $ such that $$\int_{0}^{1}{ f_n dg(x) } > n^2$$

Now just take $$F_n := \frac{1}{n}f_n$$clearly the $F_n$ are simple, left continuos and $F_n \to 0$ but $$\int_{0}^{1}{ F_n(x)dg(x) } = \frac{1}{n}\int_{0}^{1}{ f_n(x)dg(x) } \geq n \not\to 0$$

The proof is complete, now I prove the Claim.

The claim follows from the equality

$$\sum_{k=1}^{n}{ |g(t_k) - g(t_{k-1}+)| } = \int_{0}^{1}{f(x)dg(x)}$$

Where $f(x)$ is defined so that if $t_{k-1} < x \leq t_{k}$ one has $f(x) = +1$ when $g(t_k) - g(t_{k-1} +) \geq 0$ and $f(x) = -1$ when $g(t_k) - g(t_{k-1}) < 0$.