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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.

Another User
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hJulian
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  • What properties of $\alpha$ are assumed? – Daniel Fischer Jun 08 '20 at 15:23
  • I think you need $f$ and $\alpha$ bounded. – Paramanand Singh Jun 08 '20 at 18:21
  • yeah, $\alpha$ be a monotonically increasing function on [0, c] , and bounded. – hJulian Jun 08 '20 at 18:28
  • See this thread https://math.stackexchange.com/q/3241005/72031 you will need continuity of $\alpha $ at $0$. – Paramanand Singh Jun 08 '20 at 18:31
  • Also remember this: If $f, \alpha $ are bounded with no further conditions and $f\in\mathcal{R} (\alpha) $ over some interval $[a, b] $ then $f\in\mathcal{R} (\alpha) $ over any subinterval $[c, d] $ contained in $[a, b] $. The proof is easy when $\alpha$ is monotone, but difficult somewhat for general $\alpha$. Thus your hypothesis that $f\in\mathcal{R} (\alpha) $ over $[c, 1]$ is redundant. – Paramanand Singh Jun 08 '20 at 18:35
  • @ParamanandSingh , I don't know if I am understanding, it is redundant to limit myself to the interval $[c,1]$ – hJulian Jun 08 '20 at 18:56
  • @ParamanandSingh, Why should $\alpha$ be continuous at zero? – hJulian Jun 08 '20 at 18:57
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    Otherwise the limit of $\int_c^1$ needn't be equal to $\int_0^1$. The difference is $f(0)\cdot \bigl(\alpha(0^+) - \alpha(0)\bigr)$ if $f$ has an extension that is continuous at $0$. – Daniel Fischer Jun 08 '20 at 19:32
  • Your question has the hypothesis that $f\in\mathcal {R} (\alpha) $ on $[0,1]$. This automatically implies that $f\in\mathcal {R} (\alpha) $ on $[c, 1]$ so this part of hypotheses is redundant and need not be mentioned explicitly. – Paramanand Singh Jun 08 '20 at 20:08

1 Answers1

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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

hJulian
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