A detail in theorem 6.9 rudin real analysis here is the theorem(Thm 6.9 page 126): if $f$ is monotonic on $[a,b]$ and if $\alpha$ is continuous then  $f \in \mathscr R$ ($\alpha$ is assumed montonic)
the detail that i don't understand in the proof is the following:
$(1*) \enspace$  $\triangle \alpha_i = \frac{\alpha(b) - \alpha(a) }{n} , (i = 1,...,n)$ . This is possible since $\alpha$ is continuous (by theorem 4.23 which states: let $f$
 be a continuous real function on the interval $[a,b]$. if $f(a) < f(b)$ and if $c$  is a number such that $f(a)< c < f(b)$, then there exists a point $x \in (a, b)$ such that  $f(x) = c$
anyhow how does he use theorem 4.23 to conclude that $(1*)$ is possible? what would happen if $\alpha$ is not continuous
 A: (Prove (1*)) Let's analyse to get the solution: Put $c=(\alpha(b)-\alpha(a))/n$. We find $n-2$ points $a=x_0\le x_1\le...\le x_{n-1}\le x_n=b$ such that 
$$ \alpha(x_i)-\alpha(x_{i-1})=c,\quad i=1,n$$
Adding the first k equations gives
$$\alpha(x_k)-\alpha(a)=kc$$
Hence $$\alpha(x_k)=k/n\alpha(b)+ (n-k)/n\alpha(a)$$ (it is clear then $\alpha(x_k)\le \alpha(x_{k+1})$). 
Since $k/n\alpha(b)+ (n-k)/n\alpha(a)\in [\alpha(a),\alpha(b)]$, we can apply the mentioned theorem to get the existence of $x_k$. Since $\alpha(x_k)\le \alpha(x_{k+1})$ and $\alpha$ is increasing monotone we have $x_k\le x_{k+1}$.
A: 
anyhow how does he use theorem 4.23 to conclude that (1∗) is possible?

Define the partition $a=x_0 \leq x_1 \leq \dots \leq x_n=b$ by solving the following equations $$\alpha(x_1)=\alpha(x_0)+\Delta \alpha_1 \\ \alpha(x_2)=\alpha(x_1)+\Delta\alpha_2 \\ \vdots \\ \alpha(x_n)=\alpha(x_{n-1})+\Delta \alpha_n $$
Which is solvable because $\alpha$ has the intermediate value property (theorem 4.23).

what would happen if α is not continuous

Consider, for instance $$\alpha(x)=\begin{cases} 0 & 0\leq x \leq 1/2 \\ 1 & 1/2<x \leq 1  \end{cases}, $$
For any partition $P:0=x_0 \leq x_1 \leq \dots \leq x_n=1$ of the interval $[0,1]$ there must be a stage where $x_k \leq 1/2$ and $x_{k+1}>1/2$. In that stage $\alpha$ "jumps" by $1$, regardless of $n$.
