I want to show that if $E_1 \supset E_2 \supset....$ is a monotonic sequence of Jordan measurable sets, so $$Z = \bigcap_{k=1}^\infty E_k$$ is of Lebesgue measure null, then Z is also Jordan measurable and it's volume is 0.

Clearly, if $Z$ is indeed Jordan measurable, then it has no volume. So it's enough to show that $Z$ is Jordan measurable.
I started by showing that $\overline{Z} \subset \bigcap_{k=1}^\infty \overline{E_k}$ . I did it by taking some $x\in\overline{Z}$, since the closure $Z$ is the set of all limits of sequences in $Z$, we get that there is a sequence $\{x_i \} \subset Z$ so $x_i \rightarrow x$ as $i \rightarrow \infty$.
Since for all natural $k$ : $\{ x_i \} \subset E_k$ and by the assumption $x_i \rightarrow x$ we get that for all natural $k$: $x \in \overline{E_k}$. Hence $\overline{Z} \subset \bigcap_{k=1}^\infty \overline{E_k}$.
From the given information that $Z$ is of Lebesgue measure null, we infer that $Z$ has no interior, so $\partial Z = \overline{Z} \smallsetminus int(Z) = \overline{Z}$
By the previews claim - $\partial Z \subset \bigcap_{k=1}^\infty \overline{E_k}$.
Now it's enough to show that $\partial \bigcap_{k=1}^\infty \overline{E_k}$ is of Lebesgue measure null.

I don't know how to proceed.


First of all, I think you're mixing the terms "Jordan measurable" and "measure zero (=null set)". The first usually means that the indicator function is Riemann integrable, and the second means that the set can be covered by a countable union of cubes that with total volume less that $\varepsilon$ for every $\varepsilon > 0$.

The confusion might arise from the fact that a set is Jordan measurable iff it's boundary is of measure zero. That means that $\operatorname{Vol}(E)=\operatorname{Vol}(\operatorname{int}(E))$, not immediately $0$ (but $\operatorname{Vol}(\partial E) = 0$).

EDIT: perhaps it is immediate that the volume is $0$ (once we know it's measurable) since it's a null set... So you were right there :)

EDIT 2: OOPS. I didn't actually proved anything here, since I mistakenly claimed that the discontinuity set of $1_Z$ is $Z$, instead of $\partial Z$, so that proof was absolutely wrong! I'm truly sorry for misleading.

Here's an actual proof:

Defining $A_k = E_1 \setminus E_{k+1}$ for $k\geq 1$ gives us an increasing sequence, so $B_k = A_{k+1} \setminus A_k$ is a sequence of disjoint sets, and $\bigcup _{k=1} ^{\infty} B_k = \bigcup _{k=1} ^{\infty}A_k = E_1 \setminus Z$.

Note that Jordan measure is finitely additive, so $\operatorname{Vol}(A_k) = \operatorname{Vol}(E_1) - \operatorname{Vol}(E_k+1)$, thus:

$\operatorname{Vol}(B_k) = \operatorname{Vol}(A_k+1) - \operatorname{Vol}(A_k) = \operatorname{Vol}(E_k) - \operatorname{Vol}(E_k+1)$

Now: $\operatorname{Vol}(E_1 \setminus Z) = \sum _{k=1} ^{\infty} \operatorname{Vol}(B_k) = \sum _{k=1} ^{\infty} [\operatorname{Vol}(E_k) - \operatorname{Vol}(E_{k+1})] = \operatorname{Vol}(E_1) - \lim_{k \to \infty} \operatorname{Vol}(E_k)$

So far everything was legal because a union of Jordan measurable sets is a Jordan measurable set + Jordan volume is finitely additive (you can also exchange every $\operatorname{Vol}(X)$ by $\int_{X} 1$).

Exchanging sides, we get that $V(Z) = \lim_{k\to\infty} \operatorname{Vol}(E_k)$ if it exists. The limit exists since the sequence of volumes is bounded and monotonicly decreasing: $\operatorname{Vol}(E_k) \geq \operatorname{Vol}(E_{k+1})$, so $\operatorname{Vol}(Z)$ also exists, meaning $Z$ is Jordan measurable.

If this looks like cheating, you can take any other measure (for example, Lebesgue measure, and do the same \ use sigma-additivity) and follow the same steps (replacing $\operatorname{Vol}$ by $\mu$) - a set is Jordan measurable if it's inner measure equals it's outer measure, and here the inner measure of $Z$ is 0, so it's enough to show that the outer measure approches 0, but I believe that this is beyond the material of the calculus 3 course we're taking.

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  • 1
    $\begingroup$ BTW, I'm taking Hedva 3 as well and believe that this solution is enough (if you did in fact took the question from an old exam), since they said you don't have to be 100% formal (but it never hurts). Good luck! $\endgroup$ – Trouble Jan 16 '16 at 22:56
  • $\begingroup$ You say that $Z$ is the discontinuity set of $1_Z : E_1 \rightarrow \mathbb R$. Isn't it $\,\partial Z$ ? $\endgroup$ – Tony Piccolo Jan 17 '16 at 17:44
  • $\begingroup$ It is! That means that my proof wasn't actually a proof but just a circular argument and a lie. I changed it into a different proof which is (hopefully) better. Thank's for noticing! $\endgroup$ – Trouble Jan 18 '16 at 19:13
  • $\begingroup$ $\operatorname{Vol}(E_1 \setminus Z) = \sum _{k=1} ^{\infty} \operatorname{Vol}(B_k)$: aren't you using countable additivity ? $\endgroup$ – Tony Piccolo Jan 18 '16 at 20:02
  • $\begingroup$ I'm using the Monotone convergence theorem (interchanging integral and sum)... anyhow, sigma-additivity can be used if we switch to Lebesgue outer measure. $\endgroup$ – Trouble Jan 18 '16 at 21:30

I know this is an old question, but I have an interest in it as well and I think I have a better solution, or at least a simpler more readable solution.

I will show that $\partial Z \subset Z \cup(\bigcup_{k=1}^{\infty}\partial E_k)$. It will then follow that $\partial Z$ is zero measure since it is a countable union of zero measure sets.

Let $x \in \partial Z$. If $x \in Z$ then we are done, so assume that $x\notin Z$ to handle the non trivial case.

We will show that $x \in \bigcup_{k=1}^{\infty}\partial E_k$. Suppose on the contrary that $x \notin \bigcup_{k=1}^{\infty}\partial E_k$.

This means that for every $k$, $x$ is not on the boundary of $E_k$. So there's either a punctured open ball $C_k = B(x,\epsilon_k)\setminus\{x\}$ that is completely inside $E_k$ or completely outside $E_k$, for all $k$.

This leaves us with three options:

(1) - There's a punctured open ball $C_k$ completely inside $E_k$ for all $k$. Because they are all nested, we have $\lim_{k \to \infty}C_k \subset \bigcap_{i=1}^{\infty}E_i = Z$, so $C_k$ is a punctured open ball completely inside $Z$, so $x \notin \partial Z$ which is a contradiction.

(2) There's a punctured open ball inside some sets, and outside other sets. Because they are nested, this means $C_k$ is completely inside $E_1,E_2,\dots,E_k$ and completely outside $E_{k+1},E_{k+2}, \dots$, then $C_k$ is completely outside $Z$ because it's completely outside $E_{k+1}$ so it's outside of the intersection. Then again $x \notin \partial Z$ as $C_k$ is a punctured open ball centered at $x$ completely outside $Z$.

(3) There's a punctured open ball centered around $x$ that's completely outside all $E_k$. Again, very easy to see that $x \notin \partial Z$.

To sum up (assuming $x\notin Z$) - if $x \notin \bigcup_{k=1}^{\infty}E_k$ it follows that $x \notin \partial Z$. Thus if $x \in \partial Z$ it follows that $x \in \bigcup_{k=1}^{\infty}E_k$.

This concludes the proof.

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