2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ The actual line in Rudin is $M_i - m_i \leq \eta$... $\endgroup$ – universalset Nov 30 '13 at 22:32
  • $\begingroup$ sorry my mistake $\endgroup$ – Danny Nov 30 '13 at 22:34
  • 1
    $\begingroup$ 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 $M_i - m_i \leq \eta$ $\endgroup$ – Danny Nov 30 '13 at 23:11
  • 1
    $\begingroup$ @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. $\endgroup$ – user99914 Nov 30 '13 at 23:32
  • $\begingroup$ @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. $\endgroup$ – Silent Dec 3 '18 at 11:32
2
$\begingroup$

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 $M_k - m_k < \eta$, $1\le k \le n$.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.