I have a question about the following proof from Rudin's Principles of Mathematical Analysis.
6.20 Theorem Let $f \in \Re$ on $[a,b]$. For $ a \leq x \leq b$, put
$$F(x) = \int_a^x f(t)dt$$ Then $F$ is continuous on $[a,b]$; furthermore, if
$f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is
differentiable at $x_o$ and $$F'(x_0) = f(x_0)$$(I have omitted the proof of continuity of $F$ on $[a,b]$)
Suppose $f$ is continuous at $x_0$. Given $\epsilon > 0 $ choose
$\delta > 0$ such that$$\vert f(t)- f(x_o) \vert < \epsilon $$
if $\vert t- x_0 \vert < \delta$, and $a \leq t \leq b $. Hence, if
$x_0 – \delta < s \leq x_0 \leq t < x_0 + \delta$ $\enspace$ with:
$a\ \leq s < t \leq b$we have by theorem 6.12(d)
$$\left| \frac{F(t) – F(s)}{t-s} – f(x_0) \right| = \left| \frac{1}{t-s} \int_s^t [f(u) – f(x_0)]du \right| < \epsilon$$
it follows that $F'(x_0) = f(x_0)$
Why does Rudin use $s$ rather than $x_0$ in the epsilon portion of proving the derivative exists? If we are proving the derivative exists at $x_0$, I would expect that we would prove that
$$\left| \frac{F(t) – F(x_0)}{t-x_0} – f(x_0) \right| = \left| \frac{1}{t-x_0} \int_{x_0}^t [f(u) – f(x_0)]du \right| < \epsilon$$
Best Answer
Inequality $$\left| \frac{F(t) - F(s)}{t-s} - f(x_0) \right| < \varepsilon$$ we have for any $s,t$ such, that $x_0-\delta \lt s \leqslant x_0 \leqslant t \lt x_0+\delta $. So nothing prevent to take $s=x_0$ for right limit and $t=x_0$ for left limit and obtain definition of derivative.