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 $[x]=[y]$.

  • $\begingroup$ 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

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. $[x]=[y]$.

  • $\begingroup$ 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

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