# How can an element not be a member of its own equivalence class?

I'm working my way through these notes on stochastic calculus:

The following is taken from section 2.20:

In discrete probability, equivalence classes are measurable. (Proof: for any $\omega^\prime$ not equivalent to $\omega$ in $\mathcal{F}$, there is at least one $B_{\omega^\prime} \in \mathcal{F}$ with $\omega \in B_{\omega^\prime}$ but $\omega^\prime \notin B_{\omega^\prime}$. Since there are (at most) countably many $\omega^\prime$, and $\mathcal{F}$ is a $\sigma$-algebra, $A_\omega = \bigcap_{\omega^\prime} B_{\omega^\prime} \in \mathcal{F}$. This $A_\omega$ contains every $\omega_1$ that is equivalent to $\omega$ (why?) and only those.)

Can anyone help me understand this proof?

[ed. note: further context on this question can be found in the notes linked above, or in the answers below.]

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The important point to keep in mind is that $B_{\omega^\prime}$ is not the equivalence class of $\omega^\prime$, but is rather an element of the $\sigma$-filter $\mathcal{F}$ that is used to define the equivalence relation $\sim$. Given two outcomes $\omega , \omega^\prime$ we define $\omega \sim \omega^\prime$ iff for each $A \in \mathcal{F}$ we have $\omega \in A$ implies $\omega^\prime \in A$. (Actually, since $\mathcal{F}$ is a $\sigma$-algebra, it follows that $\omega \sim \omega^\prime$ iff $\omega \in A \Leftrightarrow \omega^\prime \in A$ for all $A \in \mathcal{F}$.)

The proof then goes as follows: Let $\omega$ be any outcome. Note that for any outcome $\omega^\prime$ such that $\omega \not\sim \omega^\prime$ there must be a $B_{\omega^\prime} \in \mathcal{F}$ such that $\omega \in B_{\omega^\prime}$ and $\omega^\prime \notin B_{\omega^\prime}$. Consider $A = \bigcap \{ B_{\omega^\prime} : \omega^\prime \not\sim \omega \}$. Since there are countably many such $\omega^\prime$, and $\mathcal{F}$ is a $\sigma$-algebra, it follows that $A \in \mathcal{F}$. Furthermore, it can be shown that $A$ is the equivalence class of $\omega$, i.e., $A = \{ \omega_0 : \omega \sim \omega_0 \}$. (If $\omega \sim \omega_0$, then by definition of $\sim$ it follows, in particular, that $\omega_0 \in B_{\omega^\prime}$ for all $\omega^\prime$ not equivalent to $\omega$, and therefore $\omega_0 \in A$. If $\omega \not\sim \omega_0$, then $\omega_0 \notin B_{\omega_0} \supseteq A$.)

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Thanks for your explanation of the proof. I get it now! – Tyler Mar 22 '11 at 2:44

The definition of equivalence is that $w\sim w'$ if and only if for every $A\in\mathcal{F}$, $w\in A\Rightarrow w'\in A$. The equivalence class of $x$ is defined to be $$A_x = \{\omega\mid \omega\sim x\}.$$ That is, it is $A_x$ which is the equivalence class of $x$, not $B_x$ (your title suggests you are confused about this point).

Rather, it's saying: if $\omega\not\sim\omega'$, then it is not true that for every $S\in\mathcal{F}$, $\omega\in S\Rightarrow \omega'\in S$. Therefore, there exists some element of $B\in\mathcal{F}$, which depends on $\omega'$, with the property that $\omega\in B$ and $\omega'\notin B$. To keep track of the different $B$s for different $\omega'$, we put a subindex: $B_{\omega'}$ is the witness to the fact that $\omega\not\sim\omega'$. But $B_{\omega'}$ is not the equivalence class of $\omega'$.

(This notation is not compatible with that in section 2.17 (where $B_x = \{\omega\mid X(\omega)=x\}$) which may be why you are confused.)

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Ohhhhhh. Thanks so much. The notation was confusing me. – Tyler Mar 22 '11 at 2:09