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The function is $f(x) = 1$ for $ 0 \le x \lt 1 $ and $f(x) = 2$ for $x = 1$

I calculate the upper sum

$$U(P,f) = \sum_{i=1}^n M_i \Delta x_i = \sum_{i=1}^{n-1} 1\,\Delta x_i + 2 \,\Delta x_n = 1(x_{n-1} - x_0) + 2(x_n - x_{n-1}) = 2 - x_{n-1}$$

$$L(P,f) = \sum_{i=1}^n m_i \,\Delta x_i = \sum_{i=1}^n 1 \,\Delta x_i = 1 - 0 = 1$$

So $U(P,f) - L(P,f) = 1 - x_{n-1}$

Then for some arbitrary $\epsilon > 0$ we choose $1 - x_{n-1} \lt \epsilon$ so the function is Riemann integrable in $[0,1]$ because $U(P,f) - L(P,f) < \epsilon$.

Furthermore, the lower Riemann integral $\int_0^1 f = 1$

Just wondering if I did this correctly. Thanks for any help.

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    $\begingroup$ Yes, it is. In particular if you change the value of a Riemann integrable function at finitely many points, it would not change the integral. $\endgroup$ – Bombyx mori Jan 20 '15 at 20:48
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    $\begingroup$ (and sometimes even in infinitely many points) $\endgroup$ – user2345215 Jan 20 '15 at 20:49
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Of course because the discontinuities are finite.

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  • $\begingroup$ Thanks for the reply. Is the way that I did it enough to show that it is Riemann integrable? $\endgroup$ – user3251256 Jan 20 '15 at 20:53
  • $\begingroup$ Yes, you only need use Riemann-Lebesgue's Theorem that says: "A function defined on $[a,b]$ is Riemann integrable if and only if it's bounded and continious almost everywhere" $\endgroup$ – Heracles Jan 20 '15 at 21:01

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