Is the event $\{\max\{X_1,X_2\}=X_2\}$ measurable with respect to $\sigma(\max\{X_1,X_2\})$? Let $X_1$ and $X_2$ be arbitrary random variables defined one the same probability space $(\Omega,\mathcal{F},P)$. I am trying to determine if the event $\{\max\{X_1,X_2\}=X_2\}$ is measurable with respect to $\sigma(\max\{X_1,X_2\})$? Normally I would try to decompose the event into a countable union or intersection of events which are clearly elements of the $\sigma(\max\{X_1,X_2\})$. However, the only way that I can think to rewrite the original set is as $\{\max\{X_1,X_2\}=X_2\}=\{X_1\leq X_2\}$. I'm not sure where to go from here.
 A: It is not true. Let $X_1$ be a Bernoulli(1/2) random variable and set $X_2=1-X_1$. Then $\max(X_1,X_2)$ is the constant random variable 1. The generated $\sigma$-algebra  $\sigma(\max\{X_1,X_2\})$ is trivial, but the event $\{\max\{X_1,X_2\}=X_2\}$ is not. 
A: $X_1$ a Bernoulli(p) $p>0$ r.v. taking the two possible values 0 or 1
A: Convince yourself that $\exists A \in \mathscr F$ s.t. we can write:
$$X: = \max(X_1, X_2) = X_1 1_A + X_2 1_{A^C}$$
Note that:


*

*$\sigma(X) \subseteq \sigma(\sigma(X_1) \cup \sigma(X_2) \cup \sigma(A))$

*Almost surely, $1_{A^C} = 1_{\{\max\{X_1, X_2\} = X_2\}}$

*$A^C \in \sigma(\sigma(X_1) \cup \sigma(X_2) \cup \sigma(A))$
Q: $A^C \in \sigma(X)$ ? (Alternatively, $A \in \sigma(X)$ ?)
We want to know if we can determine the maximum of $X_1$ and $X_2$ given that we know the value of $X$. We cannot. Just pick $X_1$ and $X_2$ s.t. the value of $X$ is constant.
Let $B \in \mathscr F$ and $a \in \mathbb R$ s.t.
$$X_1 = a \times 1_B$$
$$X_2 = a \times 1_{B^C}$$
Then $X = a \times 1_{A \cap B} + a \times 1_{(A \cap B)^C} = a \times 1_{\Omega} \to$ (almost?) surely, $X \equiv a$.
$\to \sigma(X) = \{\{X=a\}, \{X \ne a\}, \emptyset, \Omega\} = \{\Omega, \emptyset, \emptyset, \Omega\} = \{\emptyset, \Omega\}$
I don't see $A$ or $A^C$ in $\sigma(X)$.
