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