Use the Riemann condition to show that \(f\in\mathcal R_\alpha[0,4]\) where \(f(x)=e^{2x}\) and \[\alpha(x)=\begin{cases} x+1,&0\le x\le2,\\ 3x-2,&2<x\le4. \end{cases}\] Compute the value of the Riemann–Stieltjes integral \[\int_0^4 f(x)\,d\alpha.\]
Solution
The key point is that the integrator is not continuous at \(2\): \[ \alpha(2)=3, \qquad \alpha(2+)=4. \] Thus there is a jump of size \(1\) immediately to the right of \(2\).
1. Verify the Riemann–Stieltjes condition.
The function \(\alpha\) is increasing on \([0,4]\). Let \(P\) be a partition containing \(2\). On intervals not adjacent to \(2\), the increment of \(\alpha\) is either \(\Delta x\) or \(3\Delta x\). On the interval \([2,x_i]\) immediately to the right of \(2\), \[ \alpha(x_i)-\alpha(2)=1+3(x_i-2). \] Hence \[ U(P,f,\alpha)-L(P,f,\alpha) \le 3\sum_i (M_i-m_i)\Delta x_i +(M_*-m_*), \] where \(M_*-m_*\) is the oscillation of \(f\) on the interval immediately to the right of \(2\).
Since \(f(x)=e^{2x}\) is uniformly continuous on \([0,4]\), both terms can be made arbitrarily small by taking the mesh of \(P\) sufficiently small. Therefore the Riemann condition holds and \[ f\in\mathcal R_\alpha[0,4]. \]
2. Separate the jump from the absolutely continuous pieces.
Define \[ \beta(x)= \begin{cases} x+1,&0\le x\le2,\\ 3x-3,&2<x\le4, \end{cases} \qquad H(x)=\mathbf1_{(2,4]}(x). \] Then \(\beta\) is continuous, piecewise \(C^1\), with \[ \beta'(x)=1\quad(0<x<2), \qquad \beta'(x)=3\quad(2<x<4), \] and \[ \alpha=\beta+H. \] The Riemann–Stieltjes integral against \(H\) is the jump contribution \[ \int_0^4 f\,dH=f(2). \] Therefore \[ \int_0^4 f\,d\alpha =\int_0^2 f(x)\,dx+f(2)+3\int_2^4 f(x)\,dx. \]
3. Compute the value.
With \(f(x)=e^{2x}\), \[ \begin{aligned} \int_0^4 e^{2x}\,d\alpha &=\frac{e^4-1}{2}+e^4+\frac{3(e^8-e^4)}2\\ &=\boxed{\frac{3e^8-1}{2}}. \end{aligned} \]