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I have two functions $f$ and $g$ and I need to show that $f$ is Riemann-Stieltjes integrable with respect to $g$. I was able to calculate the integral, but I'm not sure how to actually prove why it is Riemann-Stieltjes integrable.

Let \begin{align*} f(x) &=x^2 \qquad x \in [0,5]\\ \\ g(x) &=\left\{ \begin{array}{ll} 0 & \textrm{if }0 \leq x<2 \\ p & \textrm{if } 2 \leq x<4 \\ 1 & \textrm{if } 4 \leq x \leq 5 \end{array} \right. \end{align*} After calculating the integral I got it equal to $16-12p$. Now how do I go about actually proving this? Or have I already done so?

Another User
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samsonite
  • 67

2 Answers2

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Note that $f$ is Riemann-Stieltjes integrable with respect to $g$ if for every $\epsilon > 0$ there exists a partition $P_\epsilon$ with

$$U(P_\epsilon,f,g) - L(P_\epsilon,f,g) < \epsilon,$$

where $U$ and $P$ are upper and lower sums, respectively:

$$U(P_\epsilon,f,g)= \sum_{k=1}^{n}\sup_{x \in [x_{k-1},x_k]}f(x)[g(x_k)-g(x_{k-1})]\\=\sum_{k=1}^{n}x_k^2[g(x_k)-g(x_{k-1})],\\L(P_\epsilon,f,g)= \sum_{k=1}^{n}\inf_{x \in [x_{k-1},x_k]}f(x)[g(x_k)-g(x_{k-1})]\\=\sum_{k=1}^{n}x_{k-1}^2[g(x_k)-g(x_{k-1})].$$

Choose any partition that includes subintervals $[2-\delta,2]$ and $[4-\delta,4]$ with $0<\delta < 2.$

Then

$$U(P_\epsilon,f,g) - L(P_\epsilon,f,g) = 2^2(p-0) + (4)^2(1-p) - (2-\delta)^2(p-0) - (4-\delta)^2(1-p)\\=16-12p- (2-\delta)^2p-(4-\delta)^2(1-p).$$

The RHS can be made smaller than $\epsilon$ by choosing $\delta$ sufficiently small.

Using the same partition, the upper sum and lower sum each approximate the integral to any desired accuracy by choosing $\delta$ sufficiently small. In this case the upper sum is $16 - 12p$, the exact value of the integral, regardless of $\delta$.

RRL
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Because $f$ is continuous, it is sufficient to show that $g$ is of bounded variation (equivalently: $g$ is the difference of two monotone functions).

BaronVT
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