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Here is Theorem 6.12 (c) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If $f \in \mathscr{R}(\alpha)$ on $[a, b]$ and if $a < c < b$, then $f \in \mathscr{R}(\alpha)$ on $[a, c]$ and on $[c, b]$, and $$ \int_a^c f d \alpha + \int_c^b f d \alpha = \int_a^b f d \alpha. $$

Here is my proof:

Let $\varepsilon > 0$ be given. As $f \in \mathscr{R}(\alpha)$ on $[a, b]$, so we can find a partition $P$ of $[a, b]$ such that $$ U(P, f, \alpha ) - L(P, f, \alpha) < \varepsilon. $$ Let $Q$ be any refinement of $P$ such that $Q$ also contains the point $c$. Then (by Theorem 6.4 in Baby Rudin, 3rd edition) we have $$ L(P, f, \alpha ) \leq L(Q, f, \alpha) \leq U(Q, f, \alpha) \leq U(P, f, \alpha), $$ and so $$ U(Q, f, \alpha) - L(Q, f, \alpha) \leq U(P, f, \alpha ) - L(P, f, \alpha) < \varepsilon. \tag{1} $$ Let $$ Q = \left\{ x_0, \ldots, x_{k-1}, c, x_k, \ldots, x_n \ \right\},$$ where $$ a = x_0 < \cdots < x_{k-1} < c < x_k < \cdots < x_n = b.$$ Let $$Q_1 \colon= \left\{ \ x_0, \ldots, x_{k-1}, c \ \right\}, \qquad Q_2 \colon= \left\{\ c, x_k, \ldots, x_n \ \right\}.$$ Then $Q_1$ and $Q_2$ are partitions, respectively, of $[a, c]$ and $[c, b]$, and $$ Q = Q_1 \cup Q_2. $$ Also $$ L(Q, f, \alpha) = L\left( Q_1, f, \alpha \right) + L\left( Q_2, f, \alpha \right), \tag{2}$$ and $$ U(Q, f, \alpha) = U\left( Q_1, f, \alpha \right) + U\left( Q_2, f, \alpha \right), \tag{3}$$ where $$ L\left( Q_1, f, \alpha \right) \colon= \sum_{i=1}^{k-1} \left( \inf_{x_{i-1}\leq x \leq x_i} f(x) \right) \left( \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right) \right) + \left( \inf_{x_{k-1}\leq x\leq c} f(x) \right) \left( \alpha (c) - \alpha \left( x_{k-1} \right) \right), $$ and $$L\left( Q_2, f, \alpha \right) \colon= \left( \inf_{c\leq x\leq x_k} f(x) \right) \left( \alpha \left(x_k \right) - \alpha(c) \right) + \sum_{i=k+1}^n \left( \inf_{x_{i-1}\leq x \leq x_i} f(x) \right) \left( \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right) \right), $$ and similarly for $U\left( Q_1, f, \alpha \right)$ and $U\left( Q_2, f, \alpha \right)$.

Moreover, for each $j= 1, 2$, $$ U \left( Q_j, f, \alpha \right) - L \left( Q_j, f, \alpha \right) \geq 0, $$ which together with (1) implies that, for each $j = 1, 2$, $$ U \left( Q_j, f, \alpha \right) - L \left( Q_j, f, \alpha \right) \leq U (Q, f, \alpha ) - L(Q, f, \alpha) < \varepsilon, $$ from which it follows that $f$ is Riemann-integrable with respect to $\alpha$ on $[a, c]$ and on $[c, b]$.

And, from (1)and (2) above we obtain \begin{align} \int_a^b f d \alpha &\leq U(Q, f, \alpha ) \\ &< L(Q, f, \alpha) + \varepsilon \qquad \mbox{ [ by (1) above ] } \\ &= L\left(Q_1, f, \alpha \right) + L \left( Q_2, f, \alpha \right) + \varepsilon \qquad \mbox{ [ by (2) above ] } \\ &\leq \int_a^c f d \alpha + \int_c^b f d \alpha + \varepsilon \end{align} for every real number $\varepsilon > 0$, which implies that $$ \int_a^b f d\alpha \leq \int_a^c f d\alpha + \int_c^b f d \alpha. \tag{A}$$

Now from (1) and (3) above, we obtain \begin{align} \int_a^c f d \alpha + \int_c^b f d \alpha &\leq U \left( Q_1, f, \alpha \right) + U \left( Q_2, f, \alpha \right) \\ &= U(Q, f, \alpha ) \qquad \mbox{ [ by (3) above ] } \\ &< L(Q, f, \alpha ) + \varepsilon \qquad \mbox{ [ by (1) above ] } \\ &\leq \int_a^b f d \alpha + \varepsilon \end{align} for every real number $\varepsilon > 0$, which implies that $$ \int_a^c f d \alpha + \int_c^b f d \alpha \leq \int_a^b f d \alpha. \tag{B}$$ From (A) and (B), we conclude that $$ \int_a^c f d \alpha + \int_c^b f d \alpha = \int_a^b f d \alpha, $$ as required.

Is the above proof correct (and as required by Rudin)? If so, then is my presentation good enough too? If not, then where lie the pitfalls?

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It seems good.

Maybe you could show as a lemma that, if $\mathcal{P}$ denotes the set of partitions of $[a,b]$ and $\mathcal{P}_c$ is the set of partitions which $c$ is a member of, then $$ \sup\{L(P,f,\alpha):P\in\mathcal{P}\}= \sup\{L(P,f,\alpha):P\in\mathcal{P}_c\} $$ and $$ \inf\{U(P,f,\alpha):P\in\mathcal{P}\}= \inf\{U(P,f,\alpha):P\in\mathcal{P}_c\} $$ using the fact that every partition can be refined to one that contains $c$. This would shorten the presentation, I believe.

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  • $\begingroup$ can you please elaborate on your answer by supplying the full details? $\endgroup$ – Saaqib Mahmood Jun 18 '17 at 13:42
  • $\begingroup$ @SaaqibMahmuud They're already in your answer. $\endgroup$ – egreg Jun 18 '17 at 13:47

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