I'm reading Resnick's "A probability Path" and doing exercise 3 on page 85.

The statement is:


$f : \mathcal{R}^k \rightarrow \mathcal{R}$ 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).

Thanks for your time!

  • $\begingroup$ (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. $\endgroup$
    – whuber
    Commented Sep 27, 2020 at 19:52
  • $\begingroup$ @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. $\endgroup$
    – cd98
    Commented Sep 27, 2020 at 22:28

1 Answer 1


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 $[g \in A]$.

  • $\begingroup$ 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. $\endgroup$
    – cd98
    Commented Sep 28, 2020 at 3:18

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