Suppose that $X_{1}, X_{2}, \ldots, X_{n}$ are random variables with $$X_{i}(\omega) : (\Omega, \mathcal{F})\to (\mathbb{R},\mathcal{R})$$ where $\Omega$ is the sample space, $\mathcal{F}$ is the $\sigma$-field, and $\mathcal{R}$ is the Borel $\sigma$-field.

Now suppose that $f(X_{1}, X_{2}, \ldots, X_{n})$ is a measurable function, $f:(\mathbb{R}^{n}, \mathcal{R}^{n})\to (\mathbb{R},\mathcal{R})$, and consider the $\sigma$-fields generated by the random variables $X_{1}, X_{2}, \ldots, X_{n}$, which I will denote $\sigma(X_{1}), \sigma(X_{2}), \ldots, \sigma(X_{n})$. Finally, let $Y = f(X_{1}(\omega), X_{2}(\omega), \ldots, X_{n}(\omega))$ be a random variable with $Y: (\Omega, \mathcal{F}) \to (\mathbb{R},\mathcal{R})$.

I want to show that $$\sigma(Y) \subset \sigma\left(\bigcup_{i=1}^{n} \sigma(X_{i})\right)$$

I am having trouble with the exact formulation of the proof.

Can anyone provide a rigorous proof of this?


In Durrett's Probability: Theory and Examples (fourth edition), the proof of theorem 2.1.6. basically says that if $\mathcal{G} = \sigma\left(\bigcup_{j} \sigma(X_{j})\right)$, then $f(X_{1}, X_{2}, \ldots, X_{n}) \in \mathcal{G}$ (I believe Durrett meant to put $f^{-1}(X_{1}, X_{2}, \ldots, X_{n}) \in \mathcal{G}$). I want to make sense of why this is true.

  • $\begingroup$ In the current form, the question doesn't make sense. $\sigma(f)$ is a $\sigma$-algebra in $\mathbb{R}^n$ and $\sigma(X_i)$ is a $\sigma$-algebra in $\Omega$. $\endgroup$
    – Dominik
    Commented Dec 5, 2016 at 15:55
  • $\begingroup$ @Dominik Perhaps what I wrote did not reflect what I had in mind. I have revised the question $\endgroup$
    – möbius
    Commented Dec 5, 2016 at 16:00

1 Answer 1


edit: This first answer was for the first version of the quesion, see below for the second version.

Your claim is false in general:

Take $X_i \equiv 0$ then $\sigma(X_i)$ is trivial for each $i$ hence $$\sigma\left(\bigcup_{i=1}^{n} \sigma(X_{i})\right) = \{\emptyset,\Omega\}$$.

Do you really want to claim that $\sigma(f) = \{\emptyset,\Omega\}$ for each measurable function $f$? :-)

Let $X = (X_1,\ldots,X_n)$ then $Y = f\circ X$ so $Y^{-1} = X^{-1} \circ f^{-1}$

Hence $$\sigma(Y) = \sigma(\{ (X^{-1} \circ f^{-1})(A) | A \in \mathcal{R}\}) = \sigma(\{ (X^{-1}\left(f^{-1}(A)\right) | A \in \mathcal{R}\})$$

And $$\sigma\left(\bigcup_{i=1}^{n} \sigma(X_{i})\right) = \sigma(X_i^{-1}(A) | A \in \mathcal{R}, i\in\{1,\ldots,n\})$$

Because $f$ is measurable it holds $f^{-1}(A) \in \mathcal{R}$ so $$\sigma(Y) \subseteq \sigma\left(\bigcup_{i=1}^{n} \sigma(X_{i})\right)$$

  • $\begingroup$ Question about the last step: I agree $f^{-1}(A) \in \mathcal{R}$. For simplicity let $B=f^{-1}(A)$. Then basically we want to show that $\{X^{-1}(B) | B\in\mathcal{R}\} \subset \{X_{i}^{-1}(B) | B\in\mathcal{R}, i\in \{1,\ldots,n\}\}$. Can you elaborate on how to do this final step? $\endgroup$
    – möbius
    Commented Dec 5, 2016 at 16:18
  • $\begingroup$ T.b.h you have to do some more paperwork in there due to the fact that $X^{-1}(B) \subseteq \Bbb R^n$ but $X_i^{-1}(B) \subseteq \Bbb R$ or more detailed $X^{-1}(B) \in \mathcal{R}^n$ but $X_i^{-1}(B) \in \mathcal{R}$. So to be correct you should state $\sigma(Y) \subseteq \sigma\left((\sigma(X_1),\sigma(X_2),\ldots,\sigma(X_n)\right)$ $\endgroup$
    – Gono
    Commented Dec 5, 2016 at 16:29

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