7
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ At the very least, tell us what "as in 6h1" means. Otherwise, this problem is incomplete. $\endgroup$ – Thomas Andrews Jan 21 '15 at 21:23
  • $\begingroup$ My bad, let f be an integrable function from B->[0,inf], where B is a Sphere in R^n. $\endgroup$ – Tom Turner Jan 21 '15 at 21:25
  • 2
    $\begingroup$ Don't put it in a comment, edit the question. Answerers should not have to read comment threads to get the full question. $\endgroup$ – Thomas Andrews Jan 21 '15 at 21:26
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Why is the first member of the inequality the sum of the volumes of the cartesian products $$[m_S(\hat f),M_S(\hat f)] \times S?$$ $\endgroup$ – user42912 Jan 23 '17 at 5:55
  • $\begingroup$ @user42912 We have the union of pairwise disjoint generalized rectangles. $\endgroup$ – Tony Piccolo Sep 28 '18 at 11:15
  • $\begingroup$ 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. $\endgroup$ – Mathematical Mushroom May 6 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.