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I'm working on the following problem:

Let $c\in(a,b)$ and $$f(x)=\begin{cases} 0 & a\leq x\leq c\\ 1 & c<x\leq b \end{cases}$$ $$\alpha (x)=\begin{cases} 0 & a\leq x< c\\ 1 & c\leq x\leq b \end{cases}$$

Show that $f\in R(\alpha)$

Unfortunately, what I seem to be proving is that $f\notin R(\alpha)$. My strategy was to show that given epsilon, we could find a partition $P_\epsilon $ such that the difference between the upper and lower sums is less than $\epsilon$.

So start by observing that for some $k$, $c\in[x_{k-1},x_k]$. Let $M_i=\sup_{x\in [x_{i-1},x_i]}f(x)$ and $m_i=\inf_{x\in [x_{i-1},x_i]}f(x)$. Fix $\epsilon>0$. Then

$$U\left(P,f,\alpha\right)-L\left(P,f,\alpha\right) = \sum_{i=1}^{N}\left(M_{i}-m_{i}\right)\left[\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)\right] = \sum_{i=1}^{k-1}\left(M_{i}-m_{i}\right)\left[\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)\right]+\left(M_{k}-m_{k}\right)\left[\alpha\left(x_{k}\right)-\alpha\left(x_{k-1}\right)\right]+\sum_{i=k+1}^{N}\left(M_{i}-m_{i}\right)\left[\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)\right] = (0-0)(0-0)+(1-0)(1-0)+(1-1)(1-1) = 1 $$

And as far as I know, 1 is not less than $\epsilon$ for many values of $\epsilon$. What am I doing wrong?

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shouldn't $\inf f(x) = 1$ in the interval $ c < x \le d $ – Santosh Linkha Jan 16 '13 at 7:46
@experimentX yes I'll fix that. Thanks. – crf Jan 16 '13 at 7:48
up vote 3 down vote accepted

In fact, you should want $c$ to be in your partition. So let $P = \{a = x_0 < x_1 < \ldots < x_k = c < \ldots < x_n = b\}$ be a partition. Then, letting $M_i = \sup_{x\in[x_{i-1}, x_i]} f(x)$ and $m_i = \inf_{x\in[x_{i-1},x_i]} f(x),$ we have:

$\begin{align*} U(P, f, \alpha) - L(P, f, \alpha) &= \sum_{i=1}^{n} (M_i - m_i)[\alpha(x_{i})-\alpha(x_{i-1})]\\ &= \sum_{i=1}^{k} (M_i - m_i)[\alpha(x_i) - \alpha(x_{i-1})] + \sum_{i=k+1}^{n} (M_i - m_i)[\alpha(x_i)-\alpha(x_{i-1})]\\ &= \sum_{i=1}^{k} (0-0)[\alpha(x_i)-\alpha(x_{i-1})] + \sum_{i=k+1}^{n} (M_i - m_i)[1-1]=0 < \epsilon \end{align*}$ for any $\epsilon.$ Hence, $f \in \mathscr{R}(\alpha).$

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