# Infinite intersection of monotonic Jordan measurable sets is also measurable

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.

• 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! – Trouble Jan 16 '16 at 22:56
• You say that $Z$ is the discontinuity set of $1_Z : E_1 \rightarrow \mathbb R$. Isn't it $\,\partial Z$ ? – Tony Piccolo Jan 17 '16 at 17:44
• 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! – Trouble Jan 18 '16 at 19:13
• $\operatorname{Vol}(E_1 \setminus Z) = \sum _{k=1} ^{\infty} \operatorname{Vol}(B_k)$: aren't you using countable additivity ? – Tony Piccolo Jan 18 '16 at 20:02
• I'm using the Monotone convergence theorem (interchanging integral and sum)... anyhow, sigma-additivity can be used if we switch to Lebesgue outer measure. – 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.