Let $f$ be a continuous function on $\mathbb{R}$. Define

$\mathcal{A}=\left \{ E\subseteq \mathbb{R} : f^{-1}(E)\in \mathcal{B}(\mathbb{R})\right \}$.

I want to show that $\mathcal{A}$ is a $\sigma$-algebra.

I think I'm just being silly, but this exercise almost seems trivial to me. I mean since $f$ is continuous, then for every open $E\subseteq \mathbb{R}$ we know that $f^{-1}(E)$ is open in $\mathbb{R}$. Then wouldn't $\emptyset, E^c \in \mathcal{A}$ if $E\in \mathcal{A}$, and that $\mathcal{A}$ is closed under countable unions automatically follow? What am I missing here?

  • $\begingroup$ Just a small observation: if f is continuous, then for every open $E\subseteq \mathbb{R}$ we know that $f^{-1}(E)$ is open in $\mathbb{R}$. $\endgroup$ Sep 30, 2013 at 0:54
  • $\begingroup$ @AldoGuzmánSáenz Yes, I stated that above. That's why the exercise seems trivial to me. Am I right? $\endgroup$
    – Hassan
    Sep 30, 2013 at 0:55
  • $\begingroup$ You should prove that it's trivial if it is :P $\endgroup$ Sep 30, 2013 at 0:56
  • $\begingroup$ @MartinaK. I'm sorry, I fail to see the openness condition stated anywhere, where is it? Because not every set in the Borel algebra is open. $\endgroup$ Sep 30, 2013 at 0:57
  • $\begingroup$ @EnjoysMath Well I can basically prove this in one line, right? I just want to make sure I'm not making a mistake by overlooking a detail. $\endgroup$
    – Hassan
    Sep 30, 2013 at 0:58

1 Answer 1


Let $f$ be any map, not necessarily continuous.

Let $B \in \mathcal{A}$, then $f^{-1}(\mathbb{R} - B) = f^{-1}(\mathbb{R}) - f^{-1}(B) = \mathbb{R} - f^{-1}(B)$, so $\mathcal{A}$ is closed under complement. Notice that it also contains $\mathbb{R}$, and so $\emptyset$ as well.

If $\{A_i\}_{i\geq 1} \subset \mathcal{A}$ is a countable sequence of sets, then $f^{-1}(A_1 \cup \dots) = f^{-1}(A_1) \cup \dots \ $ So $\mathcal{A}$ is closed under countable union as well. Thus it is a $\sigma$-algebra.

You want to know if it's obvious though. I'm not sure about that.


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