A function of random variables $X_1, ..., X_k$ that goes from $\mathcal{R}^k$ to the reals is measurable with respect to $\sigma(X_1, ..., X_k)$ I'm reading Resnick's "A probability Path" and doing exercise 3 on page 85.
The statement is:
Suppose
$f : \mathcal{R}^k \rightarrow \mathcal{R}$ and $f \in \mathcal{B}(\mathcal{R}^k) / \mathcal{B}(\mathcal{R})$
Let $ X_1, ..., X_k$ be random variables on $(\Omega, \mathcal{B})$. Then
$$ f(X_1, ..., X_k) \in \sigma(X_1, ..., X_k) $$
If I'm not mistaken, $\mathcal{B}(\mathcal{R}^k)$ is the (smallest) sigma-algebra generated by the open sets of $\mathcal{R}^k$.

My first doubt is how to characterize $\sigma(X_1, ..., X_k) $ in this context. I think it means this:
$$ \forall i \; \{[x_i \in A ], A \in \mathcal{B}(\mathcal{R}) \}$$
in words: for any combination of $k$ sets $A$ in $\mathcal{B}(\mathcal{R})$, $\sigma(X_1, ..., X_k) $ contains all the sets in $\Omega$ that send take you to this combination of $k$ sets.

What I (think I) need to show is that
$$ \forall B \; \in \mathcal{B}(\mathcal{R}),  f^{-1}(B) \in \sigma(X_1, ..., X_k) $$
I tried constructing such set $B$ as the following:
$$f^{-1}(B) = \{(x_1 \subset M_1) \cap (x_2 \subset M_2) \cap ... \cap (x_k \subset M_k) : f(x_1, x_2, ..., x_k) \subset B  \} $$
in words: the intersection of areas on $\mathcal{R}$ for $(x_1, ..., x_k)$ such that $f()$ takes values inside $B$. $f$ takes values on $B$ for a particular combination of sets on $\mathcal{R}$ for each $x_i$ (I'm not sure intersection is the right concept here).
Now, I know that $\cap_{i=1}^k M_i \in \mathcal{B}(\mathcal{R}^k)$ (otherwise, it couldn't be an argument for $f$.
I think it is enough to show that $ \forall i \; M_i \in \mathcal{B}(\mathcal{R})$, since $\sigma(X_1, ..., X_k)$ contains all the sets in $\Omega$ that go to $\mathcal{B}(\mathcal{R})$.
So now I have two questions:

*

*Is the argument sound?

*How can I show that $\cap_{i=1}^k M_i  \in \mathcal{B}(\mathcal{R})^k$ implies  $\forall i  \; M_i \in \mathcal{B}(\mathcal{R})$?

For (2), my first thought was to use the fact that $\mathcal{B}(\mathcal{R})^k = \mathcal{B}(RECTS)$, where $RECTS$ is the class of open rectangles. For any $B \in RECTS$, $B = I_1 \times ... \times I_k$ and it is clear that $I_i \in  \mathcal{B}(\mathcal{R})$, but I can't quite make the connection with (2).
Thanks for your time!
 A: I'm having trouble following some parts of your argument.  For example, you write $f^{-1}(B) \in \sigma(X_1, ..., X_k)$, but $f^{-1}(B)$ is a subset of $\mathcal{R}^k$ whereas $\sigma(X_1, ..., X_k)$ contains subsets of $\Omega$.  In addition I would characterize $\sigma(X_1, ..., X_k)$ like this: it is the smallest $\sigma$-algebra that makes each of $X_1, \ldots, X_k$ measurable. I.e. it is the smallest $\sigma$-algebra that contains $[X_j \in A]$ for all $j = 1, \ldots, k$ and $A \in \mathcal{B}(\mathcal{R})$.
I think the problem can be made simple by recognizing that the object in question is a composition of maps.  We can then use what we know about the measurability of compositions.  We are given that $f:(\mathcal R^k, \mathcal{B}(\mathcal R^k)) \to (\mathcal R, \mathcal{B}(\mathcal R))$ is measurable.  Define the mapping $g:(\Omega, \sigma(X_1, \ldots, X_k)) \to (\mathcal R^k, \mathcal{B}(\mathcal R^k))$ by $g(\omega) := (X_1(\omega), \ldots, X_k(\omega))$.  The problem asks us to show that the composition $f \circ g = f(X_1, \ldots, X_k)$ is measurable with respect to $\sigma(X_1, \ldots, X_k)$.  The composition is measurable if each of $f$ and $g$ are measurable (see Proposition 3.2.2 of Resnick), so it suffices to show that $g$ is measurable.  Since $\mathcal B(\mathcal R^k)$ is generated by rectangles, we need only show that for any rectangle $A = I_1 \times \cdots \times I_k$ in $\mathcal B(\mathcal R^k)$ (where each $I_j$ is a rectangle in $\mathcal R$) we have $[g \in A] \in \sigma(X_1, \ldots, X_k)$. We have
\begin{align}
[g \in A] & = \{\omega: (X_1(\omega), \ldots, X_k(\omega)) \in A) \\
& = \{\omega: X_1(\omega) \in I_1, \ldots, X_k(\omega) \in I_k) \\
& = [X_1 \in I_1] \cap \ldots \cap [X_k \in I_k].
\end{align}
Given the characterization of $\sigma(X_1, \ldots, X_k)$ above, we know $\sigma(X_1, \ldots, X_k)$ must contain $[X_j \in I_j]$ for each $j$ and therefore it contains their intersection $[g \in A]$.
