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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.

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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, $\sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)\subseteq \sigma\big(\mathcal B(\Bbb R)\big)=\mathcal B(\Bbb R)$.

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

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