1
$\begingroup$

Let $\dots E_N \subset E_2 \subset E_1$, $E = \cap_{i = 1}^{\infty} E_i$ and $E_i$ decreases to $E$. $E$ is a measurable set, and $m(E_k) < \infty$ for some $k$.

Show $$E_1 = E \cup \cup_{j = 1}^{\infty} G_j$$

where $G_j = E_j - E_{j + 1}$ and $G_j$ are disjoint.

We were asked to verify this statement within a proof in my measure theory class, but I am unfortunately quite weak when it comes to set theory. I never took a course in it, and I heard my Discrete Math professor didn't really provide me with sufficient tools. So I'm trying to fill in what I can as problems come up. Can anyone help me with this problem? Thanks!

$\endgroup$
1
$\begingroup$

We have that if $x\in E\cup \bigcup_{i=1}^{\infty} G_i$ then since $E\subseteq E_1$ and $G_i=E_i\setminus E_{i+1}\subseteq E_i\subseteq E_1$ we have $x\in E_1$. So $E\cup \bigcup_{i=1}^{\infty} G_i\subseteq E_1$$

if instead $x\in E_1$ then there are two cases. If $x\in E$ then $x\in E\cup \bigcup_{i=1}^{\infty} G_i$. If $x\notin E$ then $x\notin \bigcap_{i=1}^\infty E_i$ but $x\in E_1$. Since the $E_i$ are a downwards chain under inclusion or rather $E_1\supseteq E_2\supseteq\cdots$ there must be a smallest $n$ such that $x\notin E_n$ if there wasn't a smallest $n$ then $x$ would be in all of the $E_i$ and therefore would be in the intersection. Being the smallest $x\in E_{n-1}$ so $x\in E_{n-1}\setminus E_n=G_{n-1}$ therefore in either case if $x\in E_1 $ then $x\in E\cup \bigcup_{i=1}^{\infty} G_i$. So $E_1 \subseteq E\cup \bigcup_{i=1}^{\infty} G_i$.

This means that the two sets are subsets of each other and thus must be equal

$\endgroup$
1
  • $\begingroup$ Thank you! This is clear $\endgroup$
    – Nolan P
    Mar 25 '21 at 18:00
1
$\begingroup$

Every $x \in E \cup \cup_{j=1}^{\infty} G_j$ is clearly in $E_1$. Now suppose that $x \in E_1$. If $x$ is in every $E_i$ then $x$ is in $E$ and thus in $E \cup \cup_{j=1}^{\infty} G_j$. If $x$ is not in every $E_i$ then let $N$ be the largest $N$ such that $x \in E_N$. Note then that $x$ is not in $E_{N+1}$. Such an $N$ exists since $x$ is not in all $E_{i}$ and the sequence is decreasing with respect to set containment. Hence $x$ is in $G_{N}$ and thus in $E \cup \cup_{j=1}^{\infty} G_j$.

$\endgroup$
1
  • $\begingroup$ Honestly your answer is great, so I just decided to accept the one who answered first. I wish I could accept two because your reply is also good! $\endgroup$
    – Nolan P
    Mar 25 '21 at 18:01

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.