Need some help on baby Rudin theorem 6.15

Following is theorem 6.15 of baby Rudin:

If $a<s<b$, $f$ is bounded on $[a,b]$. $f$ is continuous at $s$, then $\alpha(x) = I(x-s)$, then $\int_a^b f d \alpha = f(s)$. $\alpha(x)= I(x-s)$ is the unit step function, $\alpha= 0$ if $x \le s ,\alpha= 1$ if $x >s.$

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

Need some help on shedding some light on the bold sentence, especially on how it is used to show that $\int_a^b f d \alpha = f(s)$. I calculated that the difference between upper sum and lower sum is $M_2 - m_2$, but then I am stuck and I do not get the last sentence of the proof.