I'm a little lost in here with content zero. It is a question of a test o mine and the professor said it was wrong (I said it had content zero because it was a graph of a function) and I don't know where do I begin correctly. It is the following statement:
Consider constants $a,b>0$. Does the set $$C=\left\{(x,y)\in \mathbb{R}^2 \middle|
\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1\right\}$$ have content zero? Justify your answer!!!
Definition 1. A rectangle is any set of the form $R = [a, b] × [c, d]$. Its area is $(b − a)(d − c)$. Note that rectangles are, by definition, closed (they include their boundary points.)
Definition 2. A bounded set $S \subset \mathbb{R}^2$ has zero content if for any $\varepsilon > 0$ one can find a finite number of rectangles $R_1,\dotsc, R_m \subset \mathbb{R}^2$ such that
- $S \subset R_1 \cup \dotsb \cup R_m$, and
- $\mathrm{area}(R_1) + \dotsb + \mathrm{area}(R_m) < \varepsilon$.
Best Answer
This is an extension of my comments. So assume that you have proved the theorem about union of sets of content $0$ (proof is easy).
Now consider the part of $C$ which lies in first quadrant. This can be covered by $n$ rectangles. The diagonal points of $i$-th such rectangle are $(x_i, f(x_i)), (x_{i+1},f(x_{i+1}))$ where $x_{i} =ia/n$ and $f(x) =b\sqrt{1-(x/a)^{2}}$ (note that $i$ takes values $0,1,2,\dots,n-1$). Prove that the total area of such rectangles is $b/n$ which can be made smaller than any given positive number by choosing a large $n$. You will find that the fact that $f$ is monotone in interval $[0,a]$ will come handy here. Thus you will show that the content of the part in first quadrant is $0$. Similarly parts in other quadrants are also of content $0$.
For those who are well versed, the above procedure is the way we prove that a monotone function is Riemann integrable.