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Let $f: [a,b] \rightarrow \mathbb{R}$ be bounded, and let $ c\in (a,b) $. Suppose$ f$ is Riemann integrable on $[a, c]$ and $[b, c]$.

Prove that $f \in \mathcal{R} ([a,b]) $ and that $$\int_{a}^{b} f dx =\int_{c}^{a} f dx + \int_{b}^{c} f dx .$$ Could someone explain why this proves that any bounded function on $[a, b]$ with finitely many discontinuities is integrable?

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  • $\begingroup$ if $f(x)$ is bounded and piecewise continuous, a countably infinite number of discontinuities will work too, since the obtained series $\int_a^b f(x) dx = \sum_{k=1}^\infty \int_{x_k}^{x_{k+1}} f(x) dx = \sum_{k=1}^\infty c_k$ is absolutely convergent $\endgroup$ – reuns Jun 6 '16 at 3:12
  • $\begingroup$ First prove the case of one discontinuity, then use induction. $\endgroup$ – Vim Jun 6 '16 at 9:56
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Let $a=a_{1}<\cdots<a_{n}=b$ denote all the discontinuities of $f$ on $[a,b]$. Now use induction as follows: $$ \int_{a}^{b}f=\int_{a_{1}}^{a_{2}}f+\int_{a_{2}}^{a_{n}}f=\int_{a_{1}}^{a_{2}}f+\int_{a_{2}}^{a_{3}}f+\int_{a_{3}}^{a_{n}}f=\cdots $$

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Suppose the discontinuities are at $x_1,...,x_n \in (a,b)$, with $x_i < x_{i+1}$. Then $f$ is bounded and continuous on $(a,x_1), (x_1,x_2),...,(x_n,b)$, hence $f$ is Riemann integrable on $[a,x_1], [x_1,x_2],...,[x_n,b]$.

The stated result shows that $f$ is Riemann integrable on $[a,x_2]$, and so by repeating we see that $f$ is Riemann integrable on $[a,b]$.

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