Suppose $f$ is real function on $(0,1]$ and $f\in \mathcal{R}(\alpha)$ on $[c,1]$ for every $c>0$. If $f\in \mathcal{R}(\alpha)$ on $[0,1]$, show that $\displaystyle\int_{0}^{1}fd\alpha=\displaystyle \lim_{c\to 0}\int_{c}^{1}fd\alpha.$
attempt: Let $\varepsilon>0$, there exist $\delta>0$ such that:
\begin{align} \left|\int_{0}^{1}fd\alpha-\int_{c}^{1}fd\alpha\right| &=\left|\int_{0}^{1}fd\alpha +\int_{1}^{c}fd\alpha \right|\\ &=\left|\int_{0}^{c}fd\alpha\right|\\ &\leq\int_{0}^{c}|f|d\alpha\\ &\leq \left\|f\right\|(\alpha(c)-\alpha(0)) \end{align} Where $\left\|f \right\|=\sup \left\{|f(x)|:x\in[0,c]\right\}=K$
I tried take $ \displaystyle|c| < \delta=\frac{\varepsilon}{Kn}$ for $n\in \mathbb{Z}^{+}$, use the Archimedian property.
I could not continue with the proof, to guarantee that it is $<\varepsilon$.
Any suggestion is welcome.
Edit: Suppose that $f\in \mathcal{R}(\alpha)$ on $[0,1]$ and $\alpha$ is bounded and continuous at $0$, then:
$\displaystyle\lim_{c\to 0}\int_{c}^{1}fd \alpha=\int_{0}^{1}fd\alpha$.
Proof Since $\alpha$ is continuous at $0$, let $\varepsilon>0$ $|\alpha(c)-\alpha(0)|<\frac{\varepsilon}{M}$, if $|c|<\delta$, then:
$\displaystyle\left|\int_{1}^{c}fd\alpha-\int_{0}^{1}fd\alpha\right|=\left|\int_{0}^{1}fd\alpha+\int_{1}^{c}fd\alpha\right|=\left|\int_{0}^{c}fd\alpha\right|\leq\left\|f\right\|(\alpha(c)-\alpha(0))<M\frac{\varepsilon}{M}=\varepsilon$,
where $\left\|f\right\|=Sup_{c\in[0,1]}f(c)=M $, hence