2
$\begingroup$

3-21) "Let $A \subseteq \mathbb{R}^n$ be a closed rectangle, and $C \subseteq A$. Show that $C$ is Jordan measurable if and only if for every $\varepsilon > 0$, there is a partition $P$ of $A$ such that

$$\sum_{S \in \sigma_1}v(S) - \sum_{S \in \sigma_2}v(S) < \varepsilon$$

where $\sigma_1$ is the collection of subrectangles $S$ determined by the partition $P$ such that they intersect $C$, and $\sigma_2$ consists of those which are contained in $C$."

So I think I have an outline of a solution but there are a few steps which I can't justify even though they seem to be true intuitively.

proof of $\implies$ direction:

Suppose $C$ is Jordan measurable and let $\varepsilon > 0$ be given.

Then we can choose a collection of closed rectangles $\{U_i\}_{i \in \mathbb{N}}$ to cover the boundary of $C$ such that $\sum_{i=1}^{\infty} v(U_i) < \varepsilon$. Since bd($C$) is closed and bounded, it's compact thus it has a finite subcover $\{U_{\omega}\}_{\omega \in \Omega}$.

Next, we construct a partition $P$ of $A$ using the endpoints of the closed intervals which make up the closed rectangles $U_{\omega}$; so that each $U_{\omega}$ is a union of subrectangles determined by $P$.

Now, because $\sigma_2 $ is contained in $\sigma_1$, we have that $$\sum_{S \in \sigma_1}v(S) - \sum_{S \in \sigma_2}v(S) = \sum_{S \in \sigma_1 - \sigma_2}v(S) $$

Now here is where I can't properly justify my next step: I want to claim that the RHS is $\leq \sum_{\omega \in \Omega} v(U_{\omega})$, because from the pictures I drew that's what it seems like...

Assuming that step is true, we have that \begin{align} \sum_{\omega \in \Omega} v(U_{\omega}) &\leq \sum_{i=1}^{\infty} v(U_i) \\ &< \varepsilon \end{align}

Since $\varepsilon > 0$ was arbitrary, this completes the "proof" of this direction.

For the other direction, for any given $\varepsilon > 0$, we can choose a partition $P$ such that

$$ \sum_{S \in \sigma_1}v(S) - \sum_{S \in \sigma_2}v(S) < \varepsilon$$ Once again since $\sigma_2 \subseteq \sigma_1$, we have that
$$\sum_{S \in \sigma_1 - \sigma_2}v(S) < \varepsilon$$

Here, I would like to claim that $\sigma_1 - \sigma_2$ covers bd($C$), which once again seems like it should be true from the pictures I've drawn, but I can't prove it. If this is indeed true then we've just shown that bd($C$) has content $0$, which implies it has measure $0$; thus completing the proof.

I'd appreciate any help in justifying those two steps of my proof, and also any comments on any other mistakes which are present. Thanks!

$\endgroup$
3
$\begingroup$

If $x \in \partial C$ ($x$ is a boundary point) then every open rectangle containing $x$ contains points both in $\text{int} \, C$ and $\text{ext} \, C$.

For the second part, the rectangles in $\sigma_1 - \sigma_2$ must cover $\partial C$.

Suppose $x \in \partial C \cap C$ but is not contained in a rectangle in $\sigma_1 - \sigma_2$. It must then belong to a rectangle $S_x$ in $\sigma_2$ which is contained in $C$. This requires $x$ to be on the boundary of that rectangle. Otherwise $x$ is contained in the interior of $S_x$, which is an open rectangle that contains no points in $\text{ext} \, C$, a contradiction. Since $x$ is on the boundary of $S_x$, it belongs to at least one other rectangle which is in $\sigma_1-\sigma_2$. This follows because if $x$ did not belong to a rectangle in $\sigma_1-\sigma_2$, it lies in the interior of a union of rectangles in $\sigma_2$ and there is a subrectangle containing $x$ that contains no points in $\text{ext} \, C$.

On the other hand suppose $x \in \partial C$ and $x \notin C$, but is not contained in a rectangle in $\sigma_1 - \sigma_2$. It must then belong to a rectangle $S_x \notin \sigma_1$ of the partition $P$. Again, this requires $x$ to be on the boundary of that rectangle. Otherwise $x$ is contained in the interior of $S_x$, which is an open rectangle that contains no points in $\text{int} \, C$, a contradiction. By similar reasoning above, $x$ belongs to at least one other rectangle in $\sigma_1- \sigma_2$.

For the first part, the boundary $\partial C$ is covered by a countable collection of open rectangles $U_j$ that contain points both in $\text{int} \, C$ and $\text{ext} \, C$ (since every open rectangle containing a boundary point must have this property) such that $\sum_j v(U_j) < \epsilon$. Since $\partial C$ is compact, there is a finite subcover by closed rectangles $\{\bar{U}_\omega\}_{\omega \in \Omega}$, where each closed rectangle contains both points in $\text{int} \, C$ and $\text{ext} \, C$. Proceed from here.

$\endgroup$
  • $\begingroup$ I don't understand this sentence. "If not x belongs to the interior of a union of rectangles in σ2 and there is a subrectangle containing x that contains no points in extC." didn't you just establish that $x$ lies on the boundary of a rectangle of the partition? could you please elaborate/clarify this point $\endgroup$ – peek-a-boo Jul 10 '18 at 0:45
  • 1
    $\begingroup$ @Avyay: I rewrote a couple of sentences. Suppose $x$ is on the boundary of $S_x$ which is a $\sigma_2$ rectangle contained in $C$. Since it is on the boundary of the rectangle it must belong to one or more adjacent rectangles, and if none of these are in $\sigma_1 - \sigma_2$ then they must all be in $\sigma_2$ which leads to a contradiction. $\endgroup$ – RRL Jul 10 '18 at 3:09

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.