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Consider the task

Let $$ f(x) = \begin{cases} 1 &\text{if $0 \le x \le 1$} \\ 0 &\text{if $1 < x \le 2$} \end{cases} $$ Show that $f$ is Riemann integrable on $[0, 2]$ and find $\int_0^2f(x)\space dx$

This is my attempt to prove it, but I am not sure if its valid:

Let $\epsilon > 0$. Let $P$ be a partition of the interval $[0,2]$ where $P = \{0, 1 - \epsilon, 1 + \epsilon, 2\}$. We have that $L(f, P) = (1 - \epsilon) \cdot 1 + (1 + \epsilon) \cdot 0 + 2\cdot0 = 1 - \epsilon = U(f, P)$. Since $U(f,P) - L(f,P) = 0 < \epsilon$, $f$ is Riemann integrable on $[0, 2]$. Since $\epsilon$ is a small quantity, it follows that $\int_0^2 f(x) \space dx = 1$.

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1 Answer 1

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I think you made a slight mistake here. Your partition is fine but you should have:

$$ L(f,P) = (2 - (1 + \epsilon)) \cdot 1 + (2\epsilon) \cdot 0 + (1 - \epsilon) \cdot 0 $$

$$ U(f,P) = (2 - (1 + \epsilon)) \cdot 1 + (2\epsilon) \cdot 1 + (1 - \epsilon) \cdot 0 $$

Your conclusion is:

$$ U(f,P) - L(f,P) = 2\epsilon $$

This is sufficient to prove that $f$ is Riemann integrable on $[0,2]$. And correct, since

$$L(f,P) = 1 - \epsilon,$$

letting $\epsilon \to 0$ you get the integral is 1.

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  • $\begingroup$ Naturally, slight blunder. $\endgroup$ Feb 7, 2014 at 18:56
  • $\begingroup$ On $[1+\epsilon,2]$, in the $L(f, P)$ part, why is the $\inf$ $1$? $\endgroup$
    – EllaW
    Sep 15, 2022 at 5:43

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