Prove that $\displaystyle \int_{0}^{1} t^p\ d\mu \lt \infty.$

Question

Let $f : [0,1] \longrightarrow \Bbb R$ be a monotonically increasing function (not necessarily continuous) with $f(0) = 0$ and $f(1) = 1.$ Suppose $\mu$ denotes the Borel measure on $[0,1]$ such that $\mu \left (\left (a,b \right ] \right )$ is the cardinality of the set $\displaystyle \left \{x \in [0,1]\ \bigg |\ a \lt \lim\limits_{h \to 0^+} \left [f(x+h) - f(x) \right ] \leq b \right \}$ for all $0 \leq a \lt b \leq 1.$ Prove that $\displaystyle \int_{0}^{1} t^p\ d\mu \lt \infty$ for $p > 1.$

How do I solve this question? Any help will be highly appreciated.

Thanks in advance.

Answer 1

We know that $f$ has countably many right discontinuities at $x_n$, and their 'sizes' $s_n = f(x_n+) - f(x_n)$ is a countable sequence such that $\sum_n s_n \le 1$.

Then $\mu((a,b])$ is the cardinality of those $n$ such that $s_n \in (a,b]$. That is, $$ \int h \, d\mu = \sum_n h(s_n) .$$