Exercise 0.17 (Continuous functions are measurable)

As always, we use the measure-theoretic induction from the Remark 1.4.15 of An Introduction to Measure Theory. To do so, we first prove that the theorem assertion, i.e.,

holds for the generator of the Borel $\sigma$-algebra, namely the collection of all open sets $O(X)$ of $X$. But this follows directly from the definition of a continuous function (see Theorem 2.1.5 from Analysis II). Thus, it is only left to verify that this property $P(S):=f^{-1}(S)\in F$ is indeed a $\sigma$-algebra property.

• $P(\varnothing )=f^{-1}(\varnothing )\in F$ is true, since $f^{-1}(\varnothing )=\varnothing$ and the empty set is always contained in any $\sigma$-algebra $F$.

• Suppose that $P(S)=f^{-1}(S)\in F$ is true for some $S\subseteq Y$. Using the set-theoretic property from the Exercise 3.4.4 of Analysis I we see that

  $$P(Y∕S)=f^{-1}(Y∕S)=f^{-1}(Y)∕f^{-1}(S)=X∕f^{-1}(S).$$

  Since $F$ is a $\sigma$-algebra that contains $f^{-1}(S)$, it must also contain its complement.

• Let $S_1,S_2,\dots \subseteq X$ be such that $P(S_i)=f^{-1}(S_i)\in F$ is true for all $i\in \mathbb{N}$. Then, we also have by Exercise 3.4.3 of Analysis I that

  $$P\left(\bigcup _{i=1}^{\infty }{S}_i\right)=f^{-1}\left(\bigcup _{i=1}^{\infty }{S}_i\right)\in F=\bigcup _{i=1}^{\infty }{f}^{-1}\left(S_i\right)\in F$$

  is true as well since $F$ contains the countable unions of any of its elements.

Thus, the property $f^{-1}(O)\in F$ is a $\sigma$-algebra property that is true for any $O\in O(X)$, it must also be true for all $S\in ⟨O(X)⟩$ as well.