# Prove that the set C has content zero

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

1. $S \subset R_1 \cup \dotsb \cup R_m$, and
2. $\mathrm{area}(R_1) + \dotsb + \mathrm{area}(R_m) < \varepsilon$.
• Can you find a point in $C$ ? Aug 16 '17 at 16:05
• An ellipse is not the graph of a function. It is, however, the union of the graphs of two functions... Aug 16 '17 at 16:19
• What is "content"? Hausdorff content? Minkowski content? Aug 16 '17 at 16:23
• It is the condition for the existence of double integrals. Aug 16 '17 at 16:25
• you can prove that your set has measure zero and its compact because measure zero and compactness implies content zero Aug 16 '17 at 17:15

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.

• Very good point at the end, about Riemann integration. Aug 16 '17 at 17:50

$C$ is the union of the graphs of two functions $f,g$ on $[-a,a]$, with $f=\{(x,y)\in C: y\geq 0\}$ and $g=\{(x,y)\in C:y\leq 0\}.$ Since $g(x)=-f(x)$ for all $x\in [-a,a],$ it suffices to show that the content of the graph of $f$ is $0.$

(i). For $a>r>0$ let $g_r$ be the part of the graph of $f$ restricted to the domain $[-a,-a+r]\cup [a-r,a].$ Now $g_r$ is covered by two rectangles: $[-a, -a+r]\times [0,f(-a+r)],$ and $[a-r,a]\times [0,f(a-r)]$.

So the content of $g_r$ is at most $D(r)=r(f(a-r)+f(-a+r)).$ Since $f$ is continuous at $\pm a$ with $f(\pm a)=0,$ the value $D(r)$ can be as small as desired, by taking sufficiently small $r$.

We can conclude that the content of the graph of $f$ is $0$ if we can show that the content of $f$ resticted to $[-a+r,a-r]$ is $0,$ ....

.... which we do by putting $p=-a+r$ and $q=a-r$ in the following:

(ii). Lemma. For $p<q$ and for differentiable $f:[p,q]\to \mathbb R$ such that $\sup \{|f'(x)|:x\in [p,q]\}=K<\infty ,$ the content of the graph of $f$ on $[p,q]$ is $0.$

Proof: For $n\in \mathbb N$ let $d_n=\frac {q-p}{n}$ and let $a_j=a+jd_n$ for $0\leq j\leq n.$ The graph of $f$ on the interval $[a_j,a_{j+1}]$ is covered by the rectangle $[a_j,a_{j+1}]\times [f(a_j)-Kd_n,\;f(a_j)+Kd_n].$ This rectangle has area $2K(d_n)^2.$ There are $n$ such rectangles, so the content of the graph of $f$ on $[p,q]$ cannot exceed $$n\cdot 2K(d_n)^2 =2K(q-p)^2/n.$$ Since $n$ can be any natural number, the content of the graph of $f$ on $[p,q]$ is zero.

This is not as general as the comments and answer of Paramanand Singh but suffices for the Q.

Remarks: I hope I got "the graph of..." in all the right places. To a set-theorist a function $is$ its graph.... In the Lemma it would also suffice that $f$ is Lipschitz-continuous on $[p,q]$ with Lipschitz constant $K$.

The circumference of an ellipse is approximatelly $B=\pi[3(a+b)-\sqrt{(3a+b)(a+3b)}]$(Ramanujan)

We can say that the circumference is $O(B)$

Now we can find rectangles with sides $1/k$ and area $1/k^2$ which can cover $C$

We need $O(kB)$ of them thus the whole $Vol(C)=O(\frac{B}{k})$