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Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points then $\int^1_0f(x)dx$ exists.

I know that I can break up the interval $[a,b]$ into subintervals where each subinterval has one discontinuity point. Then if $f$ is integrable on each subinterval then it is integrable on $[a,b]$. So if I let $c$ be my point of discontinuity in the subinterval $[a_1,b_1]\subset [a,b]$, where $c\in[a_1,b_1]$. Then how can I show this subinterval is integrable?

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$c$ is your point of discontinuity. Let me assume that $c \in (a,b)$. Take the subinterval $(c- \frac{\delta}{2}, c+ \frac{\delta}{2})$. Let $M_k = \max\{f(x): x \in (c- \frac{\delta}{2}, c+ \frac{\delta}{2})\}$ and $m_k = \min\{f(x): x \in (c- \frac{\delta}{2}, c+ \frac{\delta}{2})\}$

Consider a pertition $P$ on $[a,b]$. $P = \{a, x_1, \dots , c- \frac{\delta}{2}, c+ \frac{\delta}{2}, \dots ,b\}$. For the subinterval $[c- \frac{\delta}{2}, c+ \frac{\delta}{2}]$ you shall get $(M_k - m_k)\delta$ when you shall try to find out the $U(P,f) - L(P,f)$. $U(P,f)$ and $L(P,f)$ are upper and lower Riemann sums respectively. To show integrability show $U(P,f) - L(P,f) < \epsilon$ for sutaible $P$. You can consider $P$ in such a manner that $(M_k - m_k)\delta$ be arbitrarily small.

Hope you can do the remaining patrs yourself.

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  • $\begingroup$ How did you get $(M_k - m_k)\delta$? $\endgroup$ – user104235 Feb 9 '14 at 6:13
  • $\begingroup$ Try to find out $U(P,f) - L(P,f)$. $\endgroup$ – Dutta Feb 9 '14 at 6:21

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