# If $x,y \in X$, then $[x] \cap [y] =\emptyset$ or $[x]=[y]$ for $[x]= \cap \lbrace A \in \mathcal{M} \:|\: x \in A \rbrace$.

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 $$[x]=[y]$$.

• 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 $\cap \{ A \in \mathcal{M} : x \in A \} = \{ A \in \mathcal{M} : x \in A \}$. – Jan Sep 11 at 19:10
• @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 $x$. – Stinking Bishop Sep 11 at 19:21
• @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 $x$. – Jan Sep 11 at 19:26
• @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.) – Stinking Bishop Sep 11 at 19:31
• @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 $x$. – 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. $$[x]=[y]$$.
• 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 – Cos Sep 12 at 23:24
• @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. – Stinking Bishop Sep 13 at 9:27