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.
Best Answer
Because $f$ is continuous at $s$, given $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|s - t| < \delta$ then $|f(s) - f(t)| < \varepsilon/2$. When Rudin says "as $x_2 \rightarrow s$," he means "chosing a partition $P$ such that $x_2 - x_1 < \delta$," because then for all $t \in [x_1, x_2]$, $|f(s) - f(t)| < \varepsilon/2$, so certainly $M_2 - f(s) < \varepsilon$ and $f(s) - m_2 < \varepsilon$. Then $M_2 - m_2 < 2\varepsilon$, so $f$ is integrable.
Furthermore, $m_2 = L(P, f, \alpha) \le \int_a^b f \hspace{0.03 in} d \alpha \le U(P,f, \alpha) = M_2$, so the above implies that $$f(s) - \varepsilon < m_2 \le \int_a^b f \hspace{0.03 in} d \alpha \le M_2 < f(s) + \varepsilon$$ $$|f(s) -\int_a^b f \hspace{0.03 in} d \alpha| < \varepsilon,$$
Which he have shown for arbitrary $\varepsilon$, so $\int_a^b f \hspace{0.03 in} d \alpha = f(s)$.
Edit: now require $|f(s) - f(t)| < \varepsilon/2$ because $f$ does not necessarily take on its supremum or infimum on $[x_1, x_2]$, as proximal noted in his answer.