TITLE: Theorem 6.12 (c) in Baby Rudin: If $f\in\mathscr{R}(\alpha)$ on $[a, b]$ and $a<c<b$, then $f\in\mathscr{R}(\alpha)$ on $[a, c]$ and $[c, b]$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.