# Continuous function integrable on interval (Rudin theorem 6.8)

if $f$ is continuous on $[a,b]$ then $f \in \Re(\alpha)$ on [a,b]

proof:

let $\epsilon > 0$ be given. Choose $\eta > 0$ so that: $[\alpha(b) - \alpha(a)]\eta < \epsilon$.

since $f$ is uniformly continuous on [a,b], there exist a $\delta > 0$ such that:

$\vert f(x) - f(t) \vert < \eta$ $\space$ $\space$ (16)

if $x\in [a,b]$, $t \in [a,b]$, and $[x-t] < \delta$.

If $P$ is any partition of $[a,b]$ such that ${\mathit{\Delta}} x_i < \delta$ for all $i$, then (16) implies:

$M_i - m_i \leq \eta$ $\: \: \:$ $(i-1,.....,n)$

I don't see how (16) implies: $M_i - m_i \leq \eta$ , can someone clarify this to me

• The actual line in Rudin is $M_i - m_i \leq \eta$... – universalset Nov 30 '13 at 22:32
• sorry my mistake – Danny Nov 30 '13 at 22:34
• but why $M_i - m_i \leq \eta$ from what i understand $M_i - m_i < \eta$ , i see that John and ncmathsadist ,gave their answers with $M_i - m_i < \eta$, not as it is stated in the book $M_i - m_i \leq \eta$ – Danny Nov 30 '13 at 23:11
• @Danny: $M_i - m_i\leq \eta$ is weaker then the assertion $M_i - m_i <\eta$. Rudin is correct anyway. Actually we do not need the strict inequality in the argument. – user99914 Nov 30 '13 at 23:32
• @Danny, in fact, if $|f(x)-f(y)|<\epsilon$ for all $x,y\in[a,b]$, then $M-m\le\epsilon$, hence we do not need to invoke continuity argument. I think that's what Rudin does. – Silent Dec 3 '18 at 11:32

You should say "if $x\in[a,b], t\in[1,b]$ and $|x - t| < \delta$". Because of the way you chose the partition, any two points in a given subinterval will be closer that $\delta$ so if you evaluate $f$ at them, the absolute value of the difference is smaller than $\eta$. Hence $M_k - m_k < \eta$, $1\le k \le n$.

As If $P$ is chosen such that $\Delta x_i < \delta$, then

$$M_i - m_i = f(y_1) - f(y_2)$$

where $f$ attains maximum and minimum ($f$ is continuous) at $y_1$ and $y_2$ in the interval $[x_i, x_{i+1}]$ (or $[x_{i-1}, x_i]$?). By (16), $M_i - m_i <\eta$.

Then

$$U(f, P) - L(f, P) = \sum (M_i - m_i) (\alpha(x_{i+1}) - \alpha(x_i)) < \eta \sum (\alpha(x_{i+1}) - \alpha(x_i))$$ $$= \big(\alpha(b) - \alpha(a) \big) \eta < \epsilon$$

I made the same question to myself when I read the proof. I think its just a typo, and the correct statement is $$M_i-m_i<\eta$$ you can see (as other answerers have pointed out) that the proof works also in this case.