# proving that the graph of a function is of Jordan measure zero

Let $f$ be an integrable function from $B$ to $[0,\inf]$ where $B$ is a sphere in $\mathbb{R^n}$.

Exercise: For $f$ and $B$, the graph $$\Gamma=\{(x,f(x)):x\in B\} \subset \mathbb{R}^{n+1}$$ is of volume zero. Prove it.

I am having a hard time proving and understanding the whole notion of when a group is of Jordan measure zero and when it isn't; on the same note, what is the intuition behind proving that a group is Jordan Measurable? For instance, what about a sphere in $\mathbb{R^n}$? Thanks a lot!

• At the very least, tell us what "as in 6h1" means. Otherwise, this problem is incomplete. – Thomas Andrews Jan 21 '15 at 21:23
• My bad, let f be an integrable function from B->[0,inf], where B is a Sphere in R^n. – Tom Turner Jan 21 '15 at 21:25
• Don't put it in a comment, edit the question. Answerers should not have to read comment threads to get the full question. – Thomas Andrews Jan 21 '15 at 21:26

By definition, if $R\,$ is a rectangle containing $B$ and $\hat f$ is the zero-extension of $f$ to $R$, then, for every $\varepsilon>0$, there exists a partition of $R$ in subrectangles $S$ with $$\sum_S\, [M_S(\hat f)-m_S(\hat f)] \cdot v(S)<\varepsilon$$ Note that the first member of the inequality is the sum of the volumes of the cartesian products $$[m_S(\hat f),M_S(\hat f)] \times S$$ which are rectangles in $\mathbb{R^{n+1}}$.

So the graph of $f$ has content zero since it is included in the union of those rectangles.

As to your question about proving the Jordan measurability of a (euclidean) n-ball, I will use the theorem:

If $B \subset \mathbb R^n$ is a compact set whose boundary has content zero and $f$ is a continuous function on $B$, then $f$ is integrable on $B$ (and so the graph of $f$ has content zero).

I proceed by induction on n to prove that the boundary of the unit n-ball has content zero.

The boundary of the unit 1-ball $[-1,1]$ has obviously content zero.

Now assume that the boundary of the unit n-ball has content zero. The function $f$ defined by $$f(\mathbf x)=\sqrt {1-\|\mathbf x\|^2}$$ is continuous on the unit n-ball. Its graph is the boundary of an half of the unit (n+1)-ball and has content zero (see the above theorem). The boundary of the other half (consider the function $-f$) has also content zero. Hence the boundary of the unit (n+1)-ball has content zero being the union of two sets having content zero.

Any n-ball is obtained from the unit n-ball by a scaling or/and a translation that preserve the zero content.

• Why is the first member of the inequality the sum of the volumes of the cartesian products $$[m_S(\hat f),M_S(\hat f)] \times S?$$ – user42912 Jan 23 '17 at 5:55
• @user42912 We have the union of pairwise disjoint generalized rectangles. – Tony Piccolo Sep 28 '18 at 11:15
• Why does the boundary having content zero and f being integratable on the set $B$ mean that the total content of $B$ is zero? Like i'm just imagining the unit disc, $D^2$,,, wouldn't it's content equal it's volume which is like $\pi r^2$? I think I understand now. It's because we arn't measuring the content of $D^2$ in the plane, we would be measuring it in $\mathbb{R}$^3, so now the elementary sets are cubes and so ofc the content will be zero. – Mathematical Mushroom May 6 at 13:54