11
$\begingroup$

Can someone tell me whether this is correct thank you!

We know that if a function f is continuous on $[a,b]$, a closed finite interval, then f is uniformly continuous on that interval. This means that if we're given any $\varepsilon > 0$, there exists $\delta > 0$ such that if $x$ and $y$ are any two points in $[a,b]$ with $|x-y| < \delta$, then $|f(x) - f(y)| < \varepsilon$.

So let's say we're given an epsilon. To show that a continuous function f is integrable, we must find a delta such that: For all partitions $\Gamma= \{x_0< \ldots < x_n\}$ of $[a,b]$ with $|\Gamma|:=\max \{x_{i+1} - x_i\} < \delta$ we have $$\mathrm{S}_{\delta} - \mathrm{s}_{\delta} < \varepsilon,$$ where, $$\mathrm{S}_{\delta}:=\inf \Sigma\ M_i(x_{i+1} - x_i)\ \mathrm{and}\ \mathrm{s}_{\delta}:=\inf \Sigma\ m_i(x_{i+1} - x_i),$$

over all partitions $\Gamma$ that satisfies $|\Gamma|<\delta$, with $M_i:=\max f|_{[x_i, x_i+1]}$ and $m_i:=\min f|_{[x_i, x_i+1]}$.

So, to recap, for a given epsilon we must find a delta such that $\mathrm{S}_{\delta} - \mathrm{s}_{\delta} < \varepsilon$. Since $f$ is uniformly continuous on $[a,b]$, we can choose $\delta$ such that $$|x-y| < \delta \ \Rightarrow \ |f(x) - f(y)| < \varepsilon/(b-a).$$

Then, for this $\delta$, we have, for any partition $\Gamma$ with $|\Gamma| <\delta$, that $M_i - m_i < \epsilon/(b-a)$. Therefore,

$$\mathrm{S}_{\delta}-\mathrm{s}_{\delta} < \frac\varepsilon{(b-a)} (b-a) = \varepsilon.$$

$\endgroup$
5
  • 5
    $\begingroup$ MathJax basic tutorial and quick reference $\endgroup$
    – user31280
    Dec 7, 2012 at 23:56
  • 2
    $\begingroup$ If you're asking whether what you did is correct, without using LaTeX and thus making that very difficult to read, then the answer I can give is: yes, it seems to be correct. $\endgroup$
    – DonAntonio
    Dec 8, 2012 at 0:23
  • 2
    $\begingroup$ This is a classical result in Riemann integration theory. Much better if you look some standard textbooks in real analysis. You may refer Theorem 7.2.6 of the book of Bartle, entitled "Introduction to Real Analysis". $\endgroup$ Dec 8, 2012 at 0:25
  • 1
    $\begingroup$ I wrote your text in tex. The proof is correct! $\endgroup$ Dec 8, 2012 at 0:37
  • 1
    $\begingroup$ @Kelson, you changed a lot of the text as well just adding TeX formatting. I think that is undesirable. $\endgroup$
    – user856
    Dec 8, 2012 at 0:57

1 Answer 1

2
$\begingroup$

Yes, your proof seems correct, I'm going to add a proof that relies only on the definition and on Modulus of continuity which might help :

As far as concern $f$ continuos we know from Extreme value Theorem that $f$ is bounded between it's maximum and minimum, taken since we are on a compact.

Additionally as you said, we know from Heine Cantor that $f$ is uniformely continuos.

Let $\omega(t)$ be a modulus of continuity for $f$, which means that $|f(x)-f(x')| \leq \omega(|x-x'|)$ con $\omega$ continuos and infinitesimal in $0$.

$\forall J \subseteq I = [a,b]$,$\forall x,x' \in J,$

$|f(x)-f(x')| \leq \omega(|x-x'|)\leq\omega(|J|),$

Thanks to $\omega$ we were able to estimate $osc(f,J): = \sup\limits_{x,x' \in J} |f(x)-f(x')|$ $f$ su $J$, $osc(f,J) \leq \omega(|J|)$.

This is because $osc(f,J) = \sup\limits_{x \in J} f(x) - \inf\limits_{x \in J} f(x)= \sup\limits_{x,x' \in J}(f(x)-f(x')) \leq \omega(|x'-x|) = \omega(|J|)$

So, for every $P$ partition of $[a,b], \forall P \in \mathbb{P}([a,b])$ (The set of all partion of $[a,b]$)

Remembering that $|P|:=\max\limits_{1 \leq k \leq n}|I_{k}|$, and defining $\rho(f,P):= S(f,P)-s(f,P)$

(Where $S(f,P)-s(f,P) = \sum\limits_{k=1}^{n}|I_{k}|(\sup\limits_{x \in I_{k}}f(x)-\inf\limits_{x' \in I_{k}}f(x'))$

$$\rho(f,P) = \sum\limits_{k=1}^{n}|I_{k}| \cdot osc(f,I_{k}) \leq \sum\limits_{k=1}^{n} \cdot |I_{k}| \omega(|I_{k}|) \leq \sum\limits_{k=1}^{n} |I_{k}| \cdot \omega(|P|) = (b-a)\cdot \omega(|P|)$$

Because $\omega$ is continuos in $0$,we have $\omega(|P|) = o(1)$ when $|P| \to 0$.

Infact, $\forall |t| \leq \delta,\hspace{0.2cm} \omega(t) < \varepsilon$.

It is enought to choose a partition such that $0 < |P| < t$ and automatically we will have $\omega(|P|) \leq \omega(t) < \varepsilon$.

In other words $\inf\limits_{P \in \mathbb{P}(I)}\rho(f,P) = 0,$ which means that $f$ is Riemann integrable.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .