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Here is Theorem 6.15 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If $a<s<b$, $f$ is bounded on $[a,b]$, $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then $$ \int_a^b f d \alpha = f(s). $$

Here is Definition 6.14:

The unit step function $I$ is defined by $$ I(x) = \begin{cases} 0 \qquad & (x \leq 0), \\ 1 \qquad & (x > 0). \end{cases} $$

And, here is Rudin's proof:

Consider partitions $P = \left\{ \ x_0, x_1, x_2, x_3 \ \right\}$, where $x_0 = a$, and $ x_1 = s < x_2 < x_3 = b$. Then $$ U(P, f, \alpha) = M_2, \qquad L(P, f, \alpha) = m_2. $$ Since $f$ is continuous at $s$, we see that $M_2$ and $m_2$ converge to $f(s)$ as $x_2 \to s$.

Now here is my reading of Rudin's proof:

First of all, here are Definitions 6.1 and 6.2 in Baby Rudin, 3rd edition:

Definition 6.1:

Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, \ldots, x_n$, where $$ a = x_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq x_n = b.$$ We write $$ \Delta x_i = x_i - x_{i-1} \qquad (i = 1, \ldots, n). $$ Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put $$ \begin{align} M_i &= \sup f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ m_i &= \inf f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ U(P, f) &= \sum_{i=1}^n M_i \Delta x_i, \\ L(P, f) &= \sum_{i=1}^n m_i \Delta x_i, \end{align} $$ and finally $$ \begin{align} \tag{1} \overline{\int}_a^b f dx &= \inf U(P, f), \\ \tag{2} \underline{\int}_a^b f dx &= \sup L(P, f), \end{align} $$ where the $\inf$ and the $\sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.

If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$, we write $f \in \mathscr{R}$ (that is, $\mathscr{R}$ denotes the set of Riemann-integrable functions), and we denote the common value of (1) and (2) by $$ \tag{3} \int_a^b f dx, $$ or by $$ \tag{4} \int_a^b f(x) dx. $$ This is the Riemann integral of $f$ over $[a, b]$. Since $f$ is bounded, there exist two numbers, $m$ and $M$, such that $$ m \leq f(x) \leq M \qquad (a \leq x \leq b). $$ Hence, for every $P$, $$ m(b-a) \leq L(P, f) \leq U(P, f) \leq M (b-a), $$ so that the numbers $L(P, f)$ and $U(P, f)$ form a bounded set. This shows that the upper and lower integrals are defined for every bounded function $f$. . . .

Definition 6.2:

Let $\alpha$ be a monotonically increasing function on $[a, b]$ (since $\alpha(a)$ and $\alpha(b)$ are finite, it follows that $\alpha$ is bounded on $[a, b]$). Corresponding to each partition $P$ of $[a, b]$, we write $$ \Delta \alpha_i = \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right). $$ It is clear that $\Delta \alpha_i \geq 0$. For any real function $f$ which is bounded on $[a, b]$ we put $$ \begin{align} U(P, f, \alpha) &= \sum_{i=1}^n M_i \Delta \alpha_i, \\ L(P, f, \alpha) &= \sum_{i=1}^n m_i \Delta \alpha_i, \end{align} $$ where $M_i$, $m_i$ have the same meaning as in Definition 6.1, and we define $$ \begin{align} \tag{5} \overline{\int}_a^b f d \alpha = \inf U(P, f, \alpha), \\ \tag{6} \underline{\int}_a^b f d \alpha = \sup L(P, f, \alpha), \end{align} $$ the $\inf$ and $\sup$ again being taken over all partitions. If the left members of (5) and (6) are equal, we denote their common value by $$ \tag{7} \int_a^b f d \alpha $$ or sometimes by $$ \tag{8} \int_a^b f(x) d \alpha(x). $$ This is the Riemann-Stieltjes integral (or simply the Stieltjes integral) of $f$ with respect to $\alpha$, over $[a, b]$.

If (7) exists, i.e., if (5) and (6) are equal, we say that $f$ is integrable with respect to $\alpha$, in the Riemann sense, and write $f \in \mathscr{R}(\alpha)$.

Now for the proof of Theorem 6.15:

For any partition $P = \left\{ \ x_0, x_1, x_2 , x_3 \ \right\}$ of the closed interval $[a, b]$, where $$a = x_0 < x_1 = s < x_2 < x_3 = b,$$ we note that $$ \begin{align} U(P, f, \alpha) &= M_1 \left[ \alpha \left( x_1 \right) - \alpha \left( x_0 \right) \right] + M_2 \left[ \alpha \left( x_2 \right) - \alpha \left( x_1 \right) \right] + M_3 \left[ \alpha \left( x_3 \right) - \alpha \left( x_2 \right) \right] \\ &= M_1 ( 0 - 0) + M_2 ( 1 - 0) + M_3 ( 1 - 1 ) \\ &= M_2, \end{align} $$ and similarly, $L(P, f, \alpha) = m_2$; thus $$ U(P, f, \alpha) = M_2, \qquad L(P, f, \alpha) = m_2, \tag{A} $$ and so $$U(P, f, \alpha) - L(P, f, \alpha) = M_2 - m_2, \tag{B} $$

Let $\varepsilon > 0$ be given. As $f$ is continuous at the point $s \in (a, b)$, so we can find a real number $\delta > 0$ such that $$ a < s-\delta < s < s + \delta < b,$$ and $$ \lvert f(x) - f(s) \rvert < { \varepsilon \over 4 } $$ for all $x$ which satisfy $\lvert x-s \rvert < \delta$.

Thus, if $s < x_2 < s+\delta$, then we must have $$ \lvert f(x) - f(s) \rvert < { \varepsilon \over 4 } $$ for all $x \in \left[ s, x_2 \right] = \left[ x_1 , x_2 \right]$. That is, $$ f(s) - { \varepsilon \over 4 } < f(x) < f(s) + { \varepsilon \over 4 }$$ for all $x \in \left[ x_1, x_2 \right]$. Therefore, we must have $$ f(s) - { \varepsilon \over 4 } \leq m_2 \leq M_2 \leq f(s) + { \varepsilon \over 4 }, \tag{C} $$ and so, by (B) above, $$ U(P, f, \alpha) - L(P, f, \alpha) = M_2 - m_2 \leq { \varepsilon \over 2 } < \varepsilon,$$ from which it follows (by Theorem 6.6 in Baby Rudin, 3rd edition) that $f \in \mathscr{R}(\alpha)$ on $[a, b]$.

Now as $$ m_2 = L(P, f, \alpha) \leq \int_a^b f d \alpha \leq U(P, f, \alpha) = M_2, \qquad \mbox{ [ using (A) ] } $$ so (C) implies that $$ f(s) - { \varepsilon \over 4 } \leq m_2 \leq \int_a^b f d \alpha \leq M_2 \leq f(s) + { \varepsilon \over 4 }, $$ and so $$ f(s) - { \varepsilon \over 4 } \leq \int_a^b f d \alpha \leq f(s) + { \varepsilon \over 4 }, $$ which in turn implies that $$ \left\lvert \int_a^b f d \alpha - f(s) \right\rvert \leq { \varepsilon \over 4 } < \varepsilon$$ for every real number $\varepsilon > 0$, showing that $$ \int_a^b f d \alpha = f(s), $$ as required.

Is my rendering of Rudin's proof correct and to the point? If not, then at which point have I gone astray?

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  • $\begingroup$ Actually, if $|f(x)-f(s)|<\frac{\epsilon}{4},$ then $M_i-m_i\leq \frac{\epsilon} {4}$ $\endgroup$ – Li Chun Min Jun 20 '17 at 16:14
  • $\begingroup$ Write $I$ be that subinterval. Pick any $s \in I$. For all $x\in I$, we have $f(x)\leq f(s)+\epsilon$. Therefore, $f(s)+\epsilon$ is an upper bound. And we have $M_i \leq f(s)+\epsilon$. So you might continue to do that get rid of those ugly coefficients in your proof next time. $\endgroup$ – Li Chun Min Jun 20 '17 at 16:26
  • $\begingroup$ @LiChunMin thanks for your comments. Can you please also comment on whether my proof has any issues? $\endgroup$ – Saaqib Mahmood Mar 21 '18 at 15:28

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