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?


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.

  • $\begingroup$ 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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