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$.

Question

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 $[x]=[y]$.

Answer 1

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. $[x]=[y]$.