Prob. 1, Chap. 6, in Baby Rudin: If $f(x_0)=1$ and $f(x)=0$ for all $x \neq x_0$, then $\int f\ \mathrm{d}\alpha=0$

Question

Here is Prob. 1, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $\alpha$ increases on $[a, b]$, $a \leq x_0 \leq b$, $\alpha$ is continuous at $x_0$,$f\left( x_0 \right) = 1$, and $f(x) = 0$ if $x \neq x_0$. Prove that $f \in \mathscr{R}(\alpha)$ and that $\int f \ \mathrm{d} \alpha = 0$.

My Attempt:

As $\alpha$ is continuous at $x_0$, so, given $\varepsilon > 0$, we can find a $\delta > 0$ such that $$ \left\lvert \alpha(t) - \alpha \left( x_0 \right) \right\rvert < \frac{\varepsilon}{4} \tag{*} $$ for all $t \in [a, b]$ for which $\left\lvert t-x_0 \right\rvert < \delta$.

Let $n$ be a natural number such that $n > 2$, and let $P = \left\{ t_0, t_1, \ldots, t_{n-1}, t_n \right\}$ be a partition of $[a, b]$ such that $x_0$ is one of the points of $P$, and such that $\Delta t_j < \frac{\delta}{2}$ for each $j = 1, \ldots, n$. [We have included $x_0$ in $P$ in order to account for the case when $x_0$ is either of the endpoints of $[a, b]$.]

Then, for any point $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we have $$ \begin{align} & \left\lvert \sum_{j=1}^n f \left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \right\rvert \\ &\leq \begin{cases} \alpha\left( t_1 \right) - \alpha \left( t_0 \right) \qquad \mbox{ if $x_0 = a$} \\ \left[ \alpha \left( t_{i+1} \right) - \alpha \left( t_i \right) \right] + \left[ \alpha \left( t_i \right) - \alpha \left( t_{i-1} \right) \right] = \alpha \left( t_{i+1} \right) - \alpha \left( t_{i-1} \right) \qquad \mbox{ if $x_0 \in (a, b)$ and $x_0 = t_i$ for some $i \in \{ 1, \ldots, n-1 \}$} \\ \alpha \left( t_n \right) - \alpha \left( t_{n-1} \right) \qquad \mbox{ if $x_0 = b$} \end{cases} \\ &< \frac{\varepsilon}{4}. \qquad \mbox{ [ by (*) above ] } \end{align} $$ Thus we can conclude that, for any choice of points $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we have $$ - \frac{\varepsilon}{4} \leq \sum_{j=1}^n f \left( u_j \right) \Delta \alpha_j \leq \frac{\varepsilon}{4}, \tag{0} $$

But $$ L(P, f, \alpha) = \inf \left\{ \ \sum_{j=1}^n f\left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] \ \mbox{ for } \ j = 1, \ldots, n \ \right\}, \tag{A}$$ and $$ U(P, f, \alpha) = \sup \left\{ \ \sum_{j=1}^n f\left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] \ \mbox{ for } \ j = 1, \ldots, n \ \right\}. \tag{B}$$

Here is the link to my post here on Math SE on how we can obtain (A) and (B)

Riemann-Stieltjes Upper and Lower Sums as Suprema and Infima

So from (0) we can conclude that $$ -\frac{\varepsilon}{4} \leq L(P, f, \alpha) \leq U(P, f, \alpha) \leq \frac{\varepsilon}{4}. \tag{1}$$

Now from (1) we obtain $$ U(P, f, \alpha) - L(P, f, \alpha) \leq \frac{\varepsilon}{2} < \varepsilon; $$ but as $\varepsilon$ is an arbitrary positive real number, so the last set of inequalities, by virtue of Theorem 6.6 in Baby Rudin, implies that $f \in \mathscr{R}(\alpha)$ on $[a, b]$.

Moreover, as $$ \int_a^b f \ \mathrm{d} \alpha = \sup \left\{ \ L(Q, f, \alpha) \ \colon \ \mbox{ Q is a partition of $[a, b]$ } \ \right\} = \inf \left\{ \ U(Q, f, \alpha) \ \colon \ \mbox{ Q is a partition of $[a, b]$ } \ \right\}, $$ so we must also have $$ L(P, f, \alpha ) \leq \int_a^b f \ \mathrm{d} \alpha \leq U(P, f, \alpha). \tag{2} $$

So from (1) and (2), we obtain $$ - \frac{\varepsilon}{4} \leq \int_a^b f \ \mathrm{d} \alpha \leq \frac{\varepsilon}{4}, $$ which implies that $$ \left\lvert \int_a^b f \ \mathrm{d} \alpha \right\rvert < \varepsilon. \tag{3} $$

But $\varepsilon$ was an arbitrary positive real number. So from (3) we can conclude that $$ \int_a^b f \ \mathrm{d} \alpha = 0, $$ as required.

Is my proof good enough? Or, are there any issues with its logic, rigor, or presentation?

Answer 1

Here's a modification to your proof:

Fix $\varepsilon > 0$. As $\alpha$ is continuous at $x_0$, We can find a $\delta > 0$ such that $$ \left\lvert \alpha(t) - \alpha \left( x_0 \right) \right\rvert < \frac{\varepsilon}{4} \tag{*} $$ for all $t \in [a, b]$ for which $\left\lvert t-x_0 \right\rvert < \delta$.

Let $n$ be a natural number such that $n > 2$, and let $P = \left\{ t_0, t_1, \ldots, t_{n-1}, t_n \right\}$ be a partition of $[a, b]$ such that $x_0 = t_k$ is one of the points in $P$, and such that $\Delta t_j < \frac{\delta}{2}$ for each $j = 1, \ldots, n$. [We have included $x_0$ in $P$ in order to account for the case when $x_0$ is either of the endpoints of $[a, b]$.]

Then, for any point $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we note that $f(u_j) = 0$ unless $j = k$ or $j-1 = k$. Moreover, $\alpha$ is increasing, so $\alpha(t_j) - \alpha(t_{j-1}) \geq 0$, and $f(x) \geq 0$. Thus, we have $$ \begin{align} 0<& \sum_{j=1}^n f \left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right]\\ &\leq \begin{cases} \alpha\left( t_1 \right) - \alpha \left( t_0 \right) \qquad \mbox{ if $k=0$} \\ \alpha \left( t_n \right) - \alpha \left( t_{n-1} \right) \qquad \mbox{ if $k=n$}\\ \left[ \alpha \left( t_{i+1} \right) - \alpha \left( t_i \right) \right] + \left[ \alpha \left( t_i \right) - \alpha \left( t_{i-1} \right) \right] = \alpha \left( t_{i+1} \right) - \alpha \left( t_{i-1} \right) \\ \qquad \text{otherwise} \end{cases} \\ &< \frac{\varepsilon}{4}. \qquad \mbox{ [ by (*) above ] } \end{align} $$ Thus we can conclude that, for any choice of points $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we have $$ 0 \leq \sum_{j=1}^n f \left( u_j \right) \Delta \alpha_j \leq \frac{\varepsilon}{4}, \tag{0} $$

[Note: so far I haven't changed much. I might have handled the $x_0 \in \{a,b\}$ case differently but so far so good].

But $$ L(P, f, \alpha) = \inf \left\{ \ \sum_{j=1}^n f\left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] \ \mbox{ for } \ j = 1, \ldots, n \ \right\}, \tag{A}$$ and $$ U(P, f, \alpha) = \sup \left\{ \ \sum_{j=1}^n f\left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] \ \mbox{ for } \ j = 1, \ldots, n \ \right\}. \tag{B}$$

Here is the link to my post here on Math SE on how we can obtain (A) and (B)

Riemann-Stieltjes Upper and Lower Sums as Suprema and Infima

[Note: what you have above is not strictly necessary for this proof. I think it's safe to assume the reader knows what $L(P,f,\alpha)$ and $U(P,f,\alpha)$ are. This is a matter of taste, though].

Now, I condense the rest of your proof to two lines:

So from (0) we can conclude that $$0 \leq L(P, f, \alpha)\leq \int_a^b f\,d\alpha \leq U(P, f, \alpha) \leq \frac{\varepsilon}{4}. \tag{1}$$

But $\varepsilon$ was an arbitrary positive real number. So from (1) we can conclude that $$ \int_a^b f \ \mathrm{d} \alpha = 0, $$ as required.