(June 2014 1)Define \(\alpha \colon [-1,1] \to \mathbb{R}\) by \[\alpha(x) := \begin{cases} -1 & x \in [-1,0] \\ 1 & x \in (0,1]. \end{cases}\] Let \(f \colon [-1,1] \to \mathbb{R}\) be a function that is uniformly bounded on \([-1,1]\) and continuous at \(x=0\), but not necessarily continuous for \(x \neq 0\). Prove that \(f\) is Riemann-Stieltjes integrable with respect to \(\alpha\) over \([-1,1]\) and that \[\int_{-1}^1 f(x)d\alpha(x) = 2f(0).\]
Solution
1. \(\alpha\) has bounded variation and is constant on each side of its single jump at \(0\): \(\alpha \equiv -1\) on \([-1,0]\) and \(\alpha \equiv 1\) on \((0,1]\), with jump \(\alpha(0+) - \alpha(0-) = 2\).
definition of \(\alpha\); the value \(\alpha(0)\) is irrelevant to the Stieltjes integral.
2. \(f\) is bounded and continuous at \(0\); let \(|f| \le M\).
hypothesis.
3. For any partition \(P\) with \(0\) as a partition point, the Stieltjes sum depends only on the interval containing \(0\).
on intervals where \(\alpha\) is constant, the \(\alpha\)-increment is \(0\), so those terms contribute nothing to \(\sum f(t_i)\Delta\alpha_i\).
4. Given \(\eps > 0\), choose \(\delta > 0\) with \(|f(x) - f(0)| < \eps\) for \(|x - 0| < \delta\) (continuity at \(0\)); then for any partition containing \(0\) whose interval around \(0\) has length \(< \delta\):
4.1. The upper and lower Stieltjes sums satisfy \(U(P) - L(P) \le 2\eps\) (the jump is \(2\)).
the only contributing interval \([x_j, x_{j+1}] \ni 0\) has \(\Delta\alpha = 2\) and oscillation of \(f\) on it \(\le 2\eps\); every other interval has \(\Delta\alpha = 0\).
4.2. Hence \(f\) is Riemann–Stieltjes integrable with respect to \(\alpha\).
the Riemann–Stieltjes criterion: \(U(P) - L(P)\) can be made arbitrarily small by refining the partition.
5. \(\int_{-1}^1 f\,d\alpha = 2f(0)\).
for partitions as in 4, \(\sum f(t_i)\Delta\alpha_i = f(t)(\alpha(0+) - \alpha(0-)) = 2f(t)\) for some \(t\) in the interval around \(0\), and \(|f(t) - f(0)| < \eps\), so the sums converge to \(2f(0)\) as the mesh tends to \(0\).