Sign up ×
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|cite|improve this question
What does the tilde represent here? – Arturo Magidin Sep 10 '11 at 19:53
@Arturo Magidin $f^{−1}$ only exists if $f$ is bijective; $\widetilde{f^{−1}}$, inverse image, always exists. But indeed, I have a strong impression that's not such an important distinguishment, since even my teacher doesn't care — and you don't either, as I see below. – Luke Sep 10 '11 at 23:15
@Luke: $f^{-1}$ is often used for the induced function on the power sets, but precisely to avoid confusion I explicitly said what I meant by it. It's not that "I don't care", it's that from context it is clear what is meant. For example, Halmos (in Naive Set Theory) first introduces the notation $\underline{f}$ and $\overline{f}$ for the direct and inverse image functions. But because the inverse image function takes subsets as arguments whereas the inverse function (if it exists) takes elements as arguments, there is almost never any ambiguity. – Arturo Magidin Sep 10 '11 at 23:36
@Arturo Magidin Exactly, that's why it's easy to distinguish, and the tilde is not crucial. – Luke Sep 11 '11 at 0:10
Possible duplicate:… (although I am late :) ) – Listing Nov 23 '11 at 10:11

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|cite|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.