Prove that the function is Riemann-Integrable.

Question

This is an incorrect attempt! Look below for @Laithy 's advice and my second attempt.

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I tried this problem but I'm not sure whether I am getting it right. I am having trouble with the very last step where I am supposed to show that $I^-(f) = I^+(f)$. Here is my attempt. Please point out any errors and help me justify why $I^-(f)$ and $I^+(f)$ are equal. Thank you!

$ f(x) =\begin{cases} 1 & 0\leq x < \frac{1}{2} \\ 0 & x = \frac{1}{2} \\ 1 & \frac{1}{2} < x \leq 1 \end{cases} $

Let $P$ be the following partition $P = \{[0, \frac{1}{2}], [\frac{1}{2}, 1]\}$ of $[0, 1]$.

Consider $I_1 = [0, \frac{1}{2}]$. It follows that $m_1 = 0$ and $M_1 = 1$.

This implies $m_1(x_1 - x_0) = 0$ and $M_1(x_1 - x_0) = 1(\frac{1}{2} - 0) = \frac{1}{2}$.

Now consider $I_2 = [\frac{1}{2}, 1]$. It follows that $m_2 = 0$ and $M_2 = 1$.

This implies $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 1(1 - \frac{1}{2}) = \frac{1}{2}$.

All this gives the following:

$S^- = 0$ and $S^+ = 1$.

Therefore...

$I^-(f) = \sup\{S^-, \sigma\in G[0, 1]\} = \sup\{0, \sigma\in G[0, 1]\}$

and

$I^+(f) = \inf\{s^+, \sigma\in G[0, 1]\} = \inf\{1, \sigma\in G[0, 1]\}$

How do I decide whether $I^-(f)$ equals $I^+(f)$?

Answer 1

I don't quite understand your definition of $I_-$ and $I^+$. This should be the supremum of the lower sum over all partitions and the infimum of the upper sum over all partitions respectively. The partition you defined is not a good choice. You want to choose a partition that makes the upper sum and lower sum arbitrarily close.

Let $\varepsilon >0$. Consider the partition $P = \left\{ [0,\frac{1}{2} - \delta], [\frac{1}{2} - \delta, \frac{1}{2} + \delta], [\frac{1}{2} + \delta, 1] \right\} $ for some $\delta>0$.

Choose an appropriate $\delta >0$ so that $U(f,P) - L(f,P) < \varepsilon$

And so $0 \leq I^+ - I_- < \varepsilon$. Note that $I^+$ and $I_-$ do not depend on $\varepsilon$ and that this inequality holds for all $\varepsilon>0$. This implies that they are equal.

Answer 2

Okay, I tried this again and I hope @Laithy can help me figure out whether I got it correct this time. Feel free to chime in give your opinions, thanks!

$ f(x) =\begin{cases} 1 & 0\leq x < \frac{1}{2} \\ 0 & x = \frac{1}{2} \\ 1 & \frac{1}{2} < x \leq 1 \end{cases} $

Let $\epsilon\in\mathbb{R}^+$, choose $\delta = \frac{\epsilon}{4}$ and let $P$ be the following partition of $[0, 1]$... $P = \{[0, \frac{1}{2}-\delta], [\frac{1}{2} - \delta, \frac{1}{2} + \delta], [\frac{1}{2} + \delta, 1]\}$.

Consider $[0, \frac{1}{2} - \delta]$. It follows that $m_1 = 1$ and $M_1 = 1$.

This implies: $m_1(x_1 - x_0) = \frac{1}{2} - \delta$ and $M_1(x_1 - x_0) = \frac{1}{2} - \delta$.

Now consider $[\frac{1}{2} - \delta, \frac{1}{2} + \delta]$. It follows that $m_2 = 0$ and $M_2 = 1$.

This implies: $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 2\delta$.

Consider the final sub-division $[\frac{1}{2} + \delta, 1]$. It follows that $m_3 = 1$ and $M_3 = 1$.

This implies: $m_3(x_3 - x_2) = \frac{1}{2} - \delta$ and $M_3(x_3 - x_2) = \frac{1}{2} - \delta$.

All this gives the following results:

$U(f, P) = \frac{1}{2} - \delta + 2\delta + \frac{1}{2} - \delta = 1$

and

$L(f, P) = \frac{1}{2} - \delta + 0 + \frac{1}{2} - \delta = 1 - 2\delta$

Therefore...

$U(f, P) - L(f, P) = 1 - (1 - 2\delta) = 2\delta = 2\big(\frac{\epsilon}{4}\big) = \frac{\epsilon}{2} < \epsilon$

$f$ is Riemann-Integrable.