# 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:

1. Is the argument sound?
2. 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).

• (2) is false. As a counterexample, let $M_1$ be any non-measurable subset of $\mathcal R$ and take $M_2=\mathcal{R}\setminus M_1.$ Their intersection, the empty set, is axiomatically measurable. Part of the problem here is that you have yet to explain even what the $M_i$ are. Another problem is the notation: what does $\mathcal{B}(\mathcal{R})^k$ even mean? The notation of (2) implies this is some subset of $\mathcal{B}(\mathcal R),$ which in the context is strange. – whuber Sep 27 '20 at 19:52
• @whuber I should put the $k$ inside to make it: $\mathcal{B}(\mathcal{R}^k)$. This is Resnick's notation and I believe it's the sigma algebra that is generated by the open sets in $\mathcal{R}^k$. That's fixed. I've also explained a bit more about the $M_i$, but I see that "intersection" might not make much sense here. – cd98 Sep 27 '20 at 22:28
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 $$[g \in A]$$.
• thanks, very clear answer! I understand my confusion now. If I'm not mistaken $g$ is the map that takes random variables and converts them into a random vector. Might be useful to point that out for other readers. – cd98 Sep 28 '20 at 3:18