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.
If you already proved the particular case when the curve is a graph, then you can use the implicit function theorem!
By the theorem, every $C^1$ curve is (locally) a graph, either of the form $(x, f(x))$ or of the form $(f(y),y)$.
So, your proof can be expanded as follows:
- Take a curve $(x(t),y(t))$, where $t\in[a,b]$.
- For every point $t_0\in [a,b]$, there exists some $\delta_{t_0}$ such that the $(x(t),y(t))$, $t\in (t_0-\delta_{t_0}, t_0+\delta_{t_0})$, can be parametrized as $(x,f(x))$ or as $(f(y),y)$.
- Since $$[a,b] = \bigcup_{t\in[a,b]} (t-\delta_t, t+\delta_t),$$
and since $[a,b]$ is comact, there exists a finite covering of $[a,b]$, i.e. there exist $t_1,t_2,\dots,t_n$ such that $$[a,b]=\bigcup_{i=1}^n (t_i-\delta_{t_i}, t_i+\delta_{t_i}).$$
- On each interval $(t_i-\delta_{t_i}, t_i+\delta_{t_i})$, you can cover the curve with pieces of a total area of less than $\frac{\epsilon}n$.
- Therefore, you can cover the whole curve with an area of less than $n\cdot\frac\epsilon n = \epsilon$
Best Answer
This isn't really relevant to this specific problem, but I just want to add that this result is still true for any Borel-measurable function. This is realized as a trivial consequence of Fubini's Theorem.
Let $m_1$ and $m_2$ denote 1- and 2-dimensional Lebesgue measure, respectively, and let $f: \mathbb{R} \to \mathbb{R}$ be measurable. Then $$ m_2(G_f) = \int_{\mathbb{R}^2} 1_{G_f} dm_2 = \int_{\mathbb{R}} \int_{\mathbb{R}} 1_{G_f}(x,y) dy dx = \int_{\mathbb{R}} m_1(\{ f(x) \}) dx = \int_{\mathbb{R}} 0 dx = 0$$