Rudin 6.1 exercise solution verification Is this an ok proof of this?
Suppose $\alpha$ increases on $[a,b]$, $a \leq x_0\leq b$, $\alpha$ is continuous at $x_0$, $f(x_0)=1$, and $f(x)=0$ if $x \neq x_0$.Prove that $f \in \mathcal{R}(\alpha)$ and that $\int f d\alpha=0$.
Attempt:
Let $\epsilon>0$. Since $\alpha$ is continuous at $x_0$, there is a $\delta>0$ such that $|x-x_0|<\delta \implies |\alpha(x)-\alpha(x_0)|<\frac{\epsilon}{2}$. Choose a partition $P=\{x_0,...,x_n\}$ of $[a,b]$ such that $x_0 \in (x_{k-1},x_k)$ for some $k$ such that $\Delta x_k<\delta$. Then $$|\alpha(x_k)-\alpha(x_{k-1})|\leq|\alpha(x_k)-\alpha(x_0)|+|\alpha(x_{k-1})-\alpha(x_0)|<\epsilon$$
So that $$\begin{equation}\begin{split}U(f,P,\alpha)-L(f,P,\alpha)=&\sum\limits_{i=1}^{n}(M_i-m_i)\alpha(x_i)\\=&0+(\alpha(x_k)-\alpha(x_{k-1}))(1)\\<&\epsilon\end{split}\end{equation}\\
$$
So that $f \in \mathcal{R}(\alpha)$.Since $f \in \mathcal{R}(\alpha)$ and $L(f)=\text{Sup}\{L(f,P)|P \ \text{is  a  partition  of} \ [a,b]\}=0$, we have $\int_{a}^{b}fd\alpha=L(f)=0$.

Strictly speaking, if $x_0 = a$ or $x_0 = b$ you can't have $x_0\in (x_{k-1}, x_k) \subseteq [a,b]$. It's safer (and a bit more rigorous) to explicitly state your partitions. You can say "Choose the partition $P=\{a, \max (x_0-\frac{\delta}{3}, a), \min(x_0+\frac{\delta}{3}, b), b\}$". Other than that, looks good!
