1
$\begingroup$

Let $X$ be a non empty set, and $(X, \mathcal{M})$ a measurable space. Then I define

$$ [x]= \cap \lbrace A \in \mathcal{M} \:|\: x \in A \rbrace$$.

At my class my teacher told us to prove that:

If $x,y \in X$, then $[x] \cap [y] =\emptyset$ or $[x]=[y]$.

I have just made a simple attempt to this as if $[x] \cap [y] =\emptyset$ we are done if $[x] \cap [y] \neq \emptyset$ but Im having issues to continue here as Im not sure how to express $u \in [x] \cap [y]$ in order to conclude $[x]=[y]$.

$\endgroup$
  • $\begingroup$ This definition does not make sense to me. Over what does the intersection runs, i. e., what is the index of the intersection? $\{ A \in \mathcal{M} : x \in A \}$ is one set, so $\cap \{ A \in \mathcal{M} : x \in A \} = \{ A \in \mathcal{M} : x \in A \}$. $\endgroup$ – Jan Sep 11 at 19:10
  • $\begingroup$ @Jan A measurable space is, essentially, a set of sets (with some properties). So we are looking for the intersection of all of the sets $A$ (which belong to that set of sets $\cal M$) containing a particular element $x$. $\endgroup$ – Stinking Bishop Sep 11 at 19:21
  • $\begingroup$ @StinkingBishop The set $\{ A \in \mathcal{M} : x \in A \}$ is the set of all sets which contain $x$. If you take the intersection on this set, you simply get this again, since it is one set. What you have described, has to be written as $[x] = \cap_i A_i$, where $A_i$ are the sets containing $x$. $\endgroup$ – Jan Sep 11 at 19:26
  • $\begingroup$ @Jan I didn't mind the OP's notation, it was clear to me. Possibly $\bigcap_{A\in\cal{M},x\in A}A$ could be what you are after. ($\cap_i A_i$ is unclear, because it is not clear what is $i$ - there is no notion that the set $\cal M$ is indexed.) $\endgroup$ – Stinking Bishop Sep 11 at 19:31
  • $\begingroup$ @StinkingBishop Yes, $\bigcap_{A \in \mathcal{M}: x \in A} A$ is what I understand of the intersection of all sets containing $x$. What OP has written is definitely not this, since this is the intersection about the ONE AND ONLY set which contains all sets containing $x$. $\endgroup$ – Jan Sep 11 at 19:34
2
$\begingroup$

Suppose $u\in[x]$. Then we have:

  • If a measurable set $A$ contains $x$, it contains $u$.
  • If a measurable set $A$ does not contain $x$, it cannot contain $u$, because, otherwise, the measurable set $A'$ would contain $x$ but not $u$.

This means that: $(\forall A\in{\cal M})(x\in A\Leftrightarrow u\in A)$, so the intersections defining $[x]$ and $[u]$ are the intersections of the same set of sets. Thus $[u]=[x]$.

An obvious consequence: if $u\in[x]\cap[y]$, then $[u]=[x]$ and $[u]=[y]$, i.e. $[x]=[y]$.

$\endgroup$
  • $\begingroup$ Thanks a lot! But still I dont see clear how these two points implies $[u]=[x]$. I cannot prove neither contention. Aprecciate if you can develope the details there. @Stinking Bishop $\endgroup$ – Cos Sep 12 at 23:24
  • $\begingroup$ @Cos So if we have proven this: $(\forall A\in{\cal M})(x\in A\Leftrightarrow u\in A)$ then it means that the intersections defining $[x]$ and $[u]$ will be the intersections of the same set of sets - so the result of intersection (i.e. $[x]$/$[u]$) will be the same. I have edited the answer to state that. $\endgroup$ – Stinking Bishop Sep 13 at 9:27

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.