I'm starting to learn some basic definitions that are presented on Spivak's Calculus on Manifolds, mainly Measure ant content zero. The definition for measure zero is that, for every $\varepsilon >0$, we have a set of closed rectangles such that $E \subset\bigcup_{k \in \Bbb{N}} R_k$ and $\sum_{k=1}^\infty \mu (R_k) < \varepsilon$. Content zero is the same but with finitely many rectangles.
Now, my question is, does the graph of a continuous function has content zero? I'm aware that the graph of real continuous function has measure zero. I completely understand the OP solution with the correction gave there. But the solution implies that it has content zero as well? Indeed, in a part of the proof, we get that $$G(f) \subset \bigcup_{i=1}^n [x_{i-1},x_i]\times[m_i,M_i],$$ i.e., we can cover $G(f)$ by a finite amount of closed rectangles, mainly, $\{[x_{i-1},x_i]\times[m_i,M_i]:1\le i\le n\}$. Which, by definition, it implies that it has content zero as well. Am I correct or am I making a dumb mistake?