Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Sigma$ be a $\sigma$-algebra over $\mathbb R$ and $\mathcal A \subset \mathcal P(\mathbb R)$. Let also $f: \mathbb R \to \mathbb R$ be any function.

If $\mathcal A$ generates $\Sigma$, is it true that $\widetilde{f^{-1}}(\mathcal A)$ generates $\widetilde{f^{-1}}(\Sigma)$? That is, do these symbols commute:

$$\sigma(\widetilde{f^{-1}}(\mathcal A)) = \widetilde{f^{-1}}(\sigma(\mathcal A))\quad?$$

share|improve this question
Still not had the time to go to math.stackexchange.com/q/47332/6179 and accept an answer? –  Did Sep 10 '11 at 18:48
Is having your answer accepted that important? I find this rather abrasive, though - admitedly - I'm new to the discussion... –  The Chaz 2.0 Sep 10 '11 at 18:57
@TheChaz, Indeed you are. And the answer to your question is: No, in general. –  Did Sep 10 '11 at 19:04
What does the tilde represent here? –  Arturo Magidin Sep 10 '11 at 19:53
@Luke: I know this is tangential to the question here, but the idea that you must accept an answer to every question is far from universally agreed upon. In particular, if you still do not understand the answers, I believe it is reasonable to wait, either until you are able to work out how the existing answers help you, or possibly for other answers that you find more helpful. (If you never accepted answers, it might be another story, but you have an extraordinarily high accept rate.) –  Jonas Meyer Sep 12 '11 at 4:02

1 Answer 1

up vote 5 down vote accepted

For any set-theoretic function $f\colon A\to B$, the inverse image function $f^{-1}\colon\mathcal{P}(B)\to\mathcal{P}(A)$ given by $f^{-1}(Y) = \{a\in A\mid f(a)\in Y\}$ is extremely well-behaved relative to set operations. In particular, for all $X,Y\subseteq B$ and all families $\{X_i\}\subseteq \mathcal{P}(B)$, $$\begin{align*} f^{-1}(X\cup Y) &= f^{-1}(X)\cup f^{-1}(Y),\\ f^{-1}(X\cap Y) &= f^{-1}(X)\cap f^{-1}(Y),\\ f^{-1}(\cup X_i) &= \cup f^{-1}(X_i),\\ f^{-1}(\cap X_i) &= \cap f^{-1}(X_i),\\ f^{-1}(X-Y) &= f^{-1}(X) - f^{-1}(Y),\\ f^{-1}(X^c) &= (f^{-1}(X))^c,\\ f^{-1}(X\triangle Y) &= f^{-1}(X)\triangle f^{-1}(Y) \end{align*}$$ where $\triangle$ is the symmetric difference. As such, the inverse image of a family that generates a $\sigma$-algebra will generate the inverse image of the $\sigma$-algebra generated: you can justify the details by looking at the "bottoms-up" description of the $\sigma$-algebra generated by a family that appears in Asaf's answer to this question.

share|improve this answer
A bit quicker, and not dependent on having a transfinite hierarchy construction, is to use the "good sets principle" (google it). This will give a proof from the better known top-down formulation of $\sigma$-algebras. –  Dave L. Renfro Sep 12 '11 at 14:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.