Hence(a) By Theorems 6.12(c) and 6.20, $$\lim_{c\rightarrow0}\int_c^1f(x)\,dx=\int_0^1f(x)\,dx-\lim_{c\rightarrow0}\int_0^cf(x)\,dx=\int_0^1f(x)\,dx.$$(By analambanomenos) For any positive integer $n$ define $g_1(x)=f(n)$ and $g_2(x)=f(n+1)$ for $x\in(n,n+1]$. The total length of the intervals of $E_n$ is $(2/3)^n$ and so $\rightarrow 0$ as $n\rightarrow\infty$. There exists a partition $P$ such that(d) The result follows from parts (a) – (c) and the fact that if $f$ is continuous at $0$, then $f(0-) = f(0) = f(0+)$. Then we haveNow suppose that $f \in \mathscr R (\beta_1)$ and let $\varepsilon >0$ be given. <> Then there exists a $\delta^* > 0$ such that $\left| x \right| < 0$ implies $\left| f(x) – f(0) \right| < \varepsilon$. Let $\delta = \min\{1,\delta^*/2\}$, and $P = \{-1,-\delta,\delta,1\}$.
Since $f$ decreases monotonically, $g_1(x)\ge f(x)\ge g_2(x)$ for all $x\in[1,\infty)$. Then $f$ is uniformly continuous on $K$ and there exists $\delta>0$ such that $\big|f(s)-f(t)\big|