Riemann–Stieltjes integral with itself.

Question

Suppose that $f:[a,b] \to \mathbb R$ is continuous everywhere and non-decreasing. I am wondering that whether $$\int_a^b f(x)^{n-1}df(x) = \frac{1}{n}[f(b)^n-f(a)^n]$$ for $n=1,2,\ldots$. Notice that $f$ is not necessarily absolutely continuous.

Answer 1

• When $f$ is non-decreasing and continuous it has bounded variation and therefore gives rise to a Riemann-Lebesgue-Stieltjes integral that satisfies the integration by parts formula. In your case this can be written as $$\tag{1} \int_a^bf^{n-1}(x)\,df(x)=f^n(b)-f^n(a)-\int_a^bf(x)\,df^{n-1}(x)\,. $$ This establishes your formula for $n=2\,.$ For $n>2$ the formula follows from the

Claim. $$\tag{2} \int_a^bf(x)\,df^{n-1}(x)=(n-1)\int_a^b f^{n-1}(x)\,df(x)\,. $$ Proof. Since $f$ is continuous and non-decreasing there is at most one $c\in[a,b]$ such that $f(c)=0$ and $f\le 0$ ond $[a,c]$ and $f\ge 0$ on $[c,b]\,.$ By the additivity of the integral we can consider each subinterval separately. Let's first consider the case $f\ge 0\,.$

We fix a partition $x_0<\dots<x_N$ of the interval and write \begin{align} f^k_i&:=f^k(x_i)\,,& \Delta f^k_i=f^k_i-f^k_{i-1}\,. \end{align} Then, \begin{align} \Delta f^{n-1}_i&= f^{n-2}_{i-1}\,\Delta f^1_i+f^1_i\,\Delta f^{n-2}_i\,. \end{align} This leads easily to \begin{align}\tag{3} \Delta f^{n-1}_i &=\sum_{j=0}^{n-2}f^j_i\,f^{n-2-j}_{i-1}\,\Delta f^1_i\,. \end{align} Since $f$ is non-decreasing and $f\ge 0$ we have $$ f^{n-2}_{i-1}\le f^j_i\,f^{n-2-j}_{i-1} \le f^{n-2}_i\,. $$ Therefore, $$\tag{4} \sum_{i=1}^Nf^{n-1}_{i-1}\,\Delta f^1_i\le \sum_{i=1}^Nf^{j+1}_i\,f^{n-2-j}_{i-1}\,\Delta f^1_i \le \sum_{i=1}^Nf^{n-1}_i\,\Delta f^1_i\,. $$ Both outer terms of (4) are Riemann-Stieltjes sums of $$\tag{5} \int_a^bf^{n-1}(x)\,df(x)\, $$ and therefore converge to the same limit (5) (since the integrand is continuous). So does the middle term. Looking at (3) it follows that the Riemann-Stieltjes sum $$ \sum_{i=1}^Nf_i\Delta f^{n-1}_i $$ of $\int_a^bf(x)\,df^{n-1}(x)$ converges to $$ (n-1)\int_a^b f(x)^{n-1}\,df(x)\,. $$ This proves the claim for $f\ge 0\,.$ The case $f\le 0$ is handled analogously.

• When we drop the assumption that $f$ is non-decreasing the claim is generally wrong.

The prime example of a continuous but not absolutely continuous function is a Brownian path since it has infinite variation. We know from Ito's formula that $$\int_a^b W_s\,dW_s=\frac{W_b^2-W_a^2-b+a}{2}\,$$ which is not of the form $\frac{W_b^2-W_a^2}{2}\,.$

If you want that this integral obeys standard calculus rules you should define it as Stratonovich integral $$ \int_a^tW_s\circ\,dW_s $$ which is obtained when the Riemann-Stieljes sums are defined as $$ \sum_{i=1}^n\frac{W_{t_{i+1}}+W_{t_i}}{2}(W_{t_{i+1}}-W_{t_i})\, $$ or equivalently as $$ \sum_{i=1}^nW_{\textstyle\frac{t_{i+1}+t_i}{2}}(W_{t_{i+1}}-W_{t_i})\,. $$

• When we assume that $f$ is non-decreasing but not absolutely continuous then the prime example is a Poisson process which is piecewise constant except when it has jumps of size one: $$ \int_0^bN_{s-}\,dN_s=\sum_{n=1}^{N_b}(n-1)=\frac{N_b(N_b-1)}{2}\,. $$ This shows that $$ \int_a^bN_{s-}\,dN_s=\frac{N_b^2-N_a^2-N_b+N_a}{2} $$ which is not of the form $$ \int_a^bN_{s-}\,dN_s=\frac{N_b^2-N_a^2}{2}\,. $$ Then again $N_t$ is not continuous as OP has assumed. Nonetheless, I find these examples important to know about.