Theorem 6.9 about Riemann integral in PMA Rudin 
How he uses continuity of $\alpha$ on $[a,b]$? I thought for a moment but any results.
By the way theorem 4.23 is Intermediate value property of continuous function. 

Can anyone help me to understand this moment of the proof?
 A: Basically, he's saying that for each $i = 1,\dots, n$, we can find $x_i\in[a,b]$ such that
$$ \alpha(x_i) = \alpha(a) + \frac{\alpha(b)-\alpha(a)}{n}i $$
i.e. we can find $\{x_i\}_{i=1}^{n}$ such that $\{\alpha(x_i)\}_{i=1}^{n}$ are evenly spaced between $\alpha(a)$ and $\alpha(b)$. We can do this because IVT guarantees that we can indeed take on any intermediate value, including those dividing the interval between $\alpha(a)$ and $\alpha(b)$ into evenly spaced intervals.
A: Geometrically, graph $\alpha$, then subdivide the image $[\alpha(a), \alpha(b)]$ into $n$ subintervals of equal length
$$
\Delta \alpha = \frac{\alpha(b) - \alpha(a)}{n}.
$$
Each "break point" $y_{i} = \alpha(a) + i\, \Delta\alpha$ is a value of $\alpha$ by continuity (the intermediate value property), so there exists an $x_{i}$ in $[a, b]$ such that $y_{i} = \alpha(x_{i})$. These $(x_{i})_{i=0}^{n}$ are the desired partition $P$.
Edit: The right-hand diagram shows what can happen if $\alpha$ is discontinuous: The graph of $\alpha$ may miss one or more lines $y = y_{i}$, and so fail to determine a partition point.

