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The question asks to show that a sigma-algebra $\mathcal A$ consisting of $A$ s.t. $A=f^{-1}(B)$, where $B$ is in $\mathcal B$ are Borel subsets of $R$ and $f$ is continuous, is contained in $\mathcal B$.
I proceeded to simply argue that $f^{-1}(B)$ will be open for open $B$ since $f$ is continuous and that would put $\mathcal A$ in $\mathcal B$.

The publisher of the text has also made available a solution for this problem. In there, after arguing that $f^{-1}(B)$ is open just like I had done, a sigma-algebra $\mathcal L$ consisting of $B$ s.t $A=f^{-1}(B)$ is constructed and it is shown that $\mathcal L$ contains $\mathcal B$. It is claimed that this proves that $f^{-1}(B)$ is in $\mathcal B$ and only after that it is argued that $\mathcal A$ is in $\mathcal B$. I am not sure why $\mathcal L$ needed to be constructed in the first place and why a direct argument based on the openness of $f^{-1}(B)$ is not sufficient to answer the problem.

drhab
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befuddled
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  • "and that would put $\mathcal A$ into $\mathcal B$" The continuity of $f$ implies that $\left{ f^{-1}\left(B\right)\mid B\in\tau\right} \subset\tau\subset\sigma\left(\tau\right)=\mathcal{B}$ where $\tau$ denotes the topology on $\mathbb{R}$. That is not enough yet. Unfortunately I have no time anymore. Later I will come back. – drhab Jan 26 '15 at 09:00
  • "a direct argument based on the openness of f−1(B) is not sufficient to answer the problem" Hard to say since you do not provide such a direct argument. – Did Jan 26 '15 at 09:02
  • See here for a general discussion. – Alp Uzman Jan 26 '15 at 09:16
  • Also I must add, if you don't have much experience with measure theory you might find it useful to study an elementary book on measure theory alongside, like Bartle's The Elements of Measure Theory. I had the unfortunate incident of not knowing any measure theory (analysis actually, me at the time being in my third undergraduate term) when I took a course with Jacod & Protter as the primary textbook, which needless to say, did not end well. You might find these lecture notes helpful as well. – Alp Uzman Jan 26 '15 at 09:23

2 Answers2

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Let $\mathcal{C}:=\{C\subseteq\mathbb{R}|f^{-1}(C)\in\mathcal{B}\}$. $\emptyset=f^{-1}(\emptyset)\in\mathcal{B}\implies \emptyset\in\mathcal{C}$. If $C\in\mathcal{C}$, then $f^{-1}(C)\in\mathcal{B}$, and as $\mathcal{B}$ is a $\sigma$-algebra, $f^{-1}(C)^c= \mathbb{R}\setminus\{x\in\mathbb{R}|f(x)\in C\}= \{x\in\mathbb{R}|f(x)\not\in C\}= \{x\in\mathbb{R}|f(x)\in C^c\}= f^{-1}(C^c)\in \mathcal{B}$. Hence $C^c\in\mathcal{C}$. Finally if $\{C_n\}_n\subseteq\mathcal{C}$, then $\forall n: f^{-1}(C_n)\in\mathcal{B}$. Again, since $\mathcal{B}$ is a $\sigma$-algebra, $\bigcup_n f^{-1}(C_n)= \bigcup_n \{x\in\mathbb{R}|f(x)\in C_n\}= \{x\in\mathbb{R}| f(x)\in\bigcup_n C_n\}=f^{-1}(\bigcup_n C_n)\in\mathcal{B}$, so that $\bigcup_n C_n\in\mathcal{C}$. Hence $\mathcal{C}$ is a $\sigma$-algebra (Observe that we make extensive use of how nice $f^{-1}$ is regarding set operations).

$\mathcal{B}$ is by definition the $\sigma$-algebra generated by the open sets, i.e., it is the smallest $\sigma$-algebra containing all open sets. If $O\subseteq\mathbb{R}$ is open, then by the continuity of $f$, $f^{-1}(O)$ too is open. As $\mathcal{B}$ contains all open sets, we have that all open sets are in $\mathcal{C}$. Then $\mathcal{C}$ turns out to be a $\sigma$-algebra containing all open sets, so that $\mathcal{B}\subseteq\mathcal{C}$ (since $\mathcal{B}$ is the smallest such $\sigma$-algebra; the ordering here is inclusion).

Let $B\in\mathcal{B}$. Then $B\in\mathcal{C}$, and hence $f^{-1}(B)\in\mathcal{B}$, from which the result follows.

Note: What you show (state) is that the continuous preimages of open sets are open, which is true by the definition of continuity. But you are asked for something stronger, namely that the continuous preimages of Borel sets are Borel.

Alp Uzman
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Let $f:X\rightarrow Y$ be a function and let $\mathcal V\subseteq\wp(Y)$.

A nice lemma that can be applied in cases as mentioned in your question is:$$\sigma(f^{-1}(\mathcal V))=f^{-1}(\sigma(\mathcal V))$$

Here $\sigma(\mathcal U)$ is a notation for the "smallest" $\sigma$-algebra that contains $\mathcal U$, and it is also referred to as the $\sigma$-algebra generated by $\mathcal U$.

Two facts are important here and form the underground for the proof of this lemma:

  • If $\mathcal B\subseteq\wp(Y)$ is a $\sigma$-algebra then $f^{-1}(\mathcal B)\subseteq\wp(X)$ is a $\sigma$-algebra.
  • If $\mathcal A\subseteq\wp(X)$ is a $\sigma$-algebra then $\{B\mid f^{-1}(B)\in\mathcal A\}\subseteq\wp(Y)$ is a $\sigma$-algebra.

Proving these facts I leave up to you. The first tells us that $f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)$ is a $\sigma$-algebra and this of course with $f^{-1}\left(\mathcal{V}\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)$ so it immedeately follows that $$\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)$$ Conversely the second tells us that $\mathcal{A}:=\left\{ B\mid f^{-1}\left(B\right)\in\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\right\} $ is a $\sigma$-algebra and this of course with $\mathcal{V}\subseteq A$ so it immedeately follows that $\sigma\left(\mathcal{V}\right)\subseteq\mathcal{A}$, or equivalently $$f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)\subseteq\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)$$

In the case you mention we are dealing with a continuous function $f:\mathbb R\rightarrow\mathbb R$ and for $\mathcal V$ we take the topology. The continuity of $f$ tells us that $f^{-1}(\mathcal V)\subseteq\mathcal V$ and applying the lemma we find $$f^{-1}\left(\mathcal{B}\right)=f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\subseteq\sigma\left(\mathcal{V}\right)=\mathcal{B}$$

Warning: the $\mathcal A$ in your question is in fact $f^{-1}(\mathcal B)$ and has no connection with the $\mathcal A$ used in my answer.

drhab
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