# Showing if $f$ is Borel measurable and $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set.

The following problem is from Royden & Fitzpatrick (4 ed.). I am stuck on showing (ii), can someone please help me prove it? Thank you.

$$\def\R{{\mathbb R}}$$ Page 59, problem 8. (Borel measurability) A function $$f$$ is said to be $$\textbf{Borel measurable}$$ provided its domain $$E$$ is a Borel set and for each $$c,$$ the set $$\{x\in E | f(x) > c\}$$ is a Borel set. Verify that Proposition 1 and Theorem 6 remain valid if we replace "(Lebesgue) measurable set" by "Borel set." Show that: (i) every Borel measurable function is Lebesgue measurable; (ii) if $$f$$ is Borel measurable and $$B$$ is a Borel set, then $$f^{-1}(B)$$ is a Borel set; (iii) if $$f$$ and $$g$$ are Borel measurable, so is $$f\circ g;$$ and (iv) if $$f$$ is Borel measurable and $$g$$ is Lebesgue measurable, then $$f\circ g$$ is Lebesgue measurable.

$$\textit{Proof.}$$ Every Borel measurable set is Lebesgue measurable since $$B\in B(\R),$$ then $$B$$ is a Lebesgue measurable set except perhaps on a set of measure $$0.$$ For (iii), assume $$g: \mathbb{R} \to \mathbb{R}$$ and $$f: \mathbb{R} \to \mathbb{R}.$$ Then, $$(f\circ g)^{-1}((c,\infty)) = g^{-1}\circ f^{-1} ((c,\infty)).$$ By the hypothesis, $$f^{-1}((c,\infty)) = B\in B(\R).$$ By definition of Borel set, any member of $$B(\R)$$ is the result of countable set operations or a member of the topology on $$\R.$$ Any member of the topology on $$\R$$ may be written as the countable result of set operations on $$(a,\infty)$$ for some $$a\in \R,$$ so $$g^{-1}(B) \in B(\R).$$ Thus, $$f\circ g$$ is Borel measurable. Now to prove (iv), assume $$f: (X,T) \to (\R,U)$$ with $$(X,T)$$ a general topological space, and $$U$$ the standard topology on $$\R.$$ By definition, any Borel set $$B\in B(\R)$$ is a result of countable set operations as an open set. Now given that $$f^{-1}((c,\infty)) \in B(x),$$ any open set may be written in terms of open rays and any Borel set in $$\R$$ can be written in terms of these open sets. Hence, the inverse image of a Borel set in $$\R$$ is the countable set theoretic result of operations on $$f^{-1}((c,\infty))$$ which is a Borel set as $$B(x)$$ is a $$\sigma$$-algebra.

Let $$\mathcal A$$ be the set of all Borel subset $$B$$ of $$\Bbb R$$ such that $$f^{-1}(B)$$ is also a Borel subset of $$\Bbb R$$. Since $$f$$ is Borel-measurable we have $$(c,\infty)\in \mathcal A$$ for all $$c\in\Bbb R$$.

Let $$\sigma(\mathcal A)$$ be the smallest $$\sigma$$-algebra containing the set $$\mathcal A$$. Since, the operation $$f^{-1}$$, i.e. operation of taking inverse commutes with the countable union operation and taking complement operation, so we have $$\sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)=\big\{f^{-1}(X): X\in\sigma(\mathcal A)\big\}.$$

Now, since $$\sigma(\mathcal A)$$ is a $$\sigma$$-algebra we have $$(a,\infty)\cap (b,\infty)=(a,b)\in \sigma(\mathcal A)$$ for all $$a,b\in\Bbb R$$.

Similarly, $$(-\infty,a']=\Bbb R\backslash (a',\infty)$$ is also in $$\sigma(\mathcal A)$$ for all $$a'\in\Bbb R$$ as $$\sigma$$-algebra is closed under complement.

Hence, $$(-\infty,a)=\bigcup_{n=1}^\infty\big(-\infty,a-\frac{1}{n}\big]$$ is also an element of $$\sigma(\mathcal A)$$ for all $$a\in\Bbb R$$ as $$\sigma$$-algebra is closed under countable union.

Also, every open subset of $$\Bbb R$$ can be written as a countable union of open intervals of $$\Bbb R$$ and every $$\sigma$$-algebra is closed under countable union. Therefore, every open subset of $$\Bbb R$$ is an element of $$\sigma(\mathcal A)$$. In other words, the set $$\tau(\Bbb R)$$ of all open subsets of $$\Bbb R$$ is a subset of $$\mathcal A$$.

But, the Borel-$$\sigma$$ algebra $$\mathcal B(\Bbb R)$$ of $$\Bbb R$$ is the smallest $$\sigma$$-algebra containing all open subsets of $$\Bbb R$$, i.e. $$\sigma\big(\tau(\Bbb R)\big)=\mathcal B(\Bbb R)$$. Hence, $$\sigma(\mathcal A)\supseteq \mathcal B(\Bbb R)$$ as $$\mathcal A\supseteq \tau(\Bbb R)$$.

Finally, For any $$Y\in\mathcal B(\Bbb R)\implies Y\in \sigma(\mathcal A)\implies f^{-1}(Y)\in \sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)\subseteq \mathcal B(\Bbb R)$$. The last inclusion is due the fact that each set $$f^{-1}(B)\in \mathcal B(\Bbb R)$$ for all $$B\in \mathcal A$$ from definition of $$\mathcal A$$. Hence, $$\sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)\subseteq \sigma\big(\mathcal B(\Bbb R)\big)=\mathcal B(\Bbb R)$$.

• Thank you, do you think my explanation for the rest is enough? Sep 22, 2020 at 19:04
• You are welcome. Yes, whatever you have written in your question I mean proofs of (i),(iii), and (iv) are correct. Sep 22, 2020 at 19:06