Exercise 1.1.1 (Continuous functions are Borel measurable)

Proof. In Remark 1.4.15 of An Introduction to Measure Theory, Terence Tao introduces one of the most practical proof methods for working with properties pertaining $\sigma$-algebras (heuristically calling it measurable induction). It states that it is sufficient to take a family $F$ which generates the Borel $\sigma$-algebra $B[X]$ in question and to demonstrate that

(i)

$f^{-1}(\varnothing )$ is Borel measurable in $Y$.

(ii)

If $f^{-1}(E)$ is Borel measurable in $Y$ for some $E\subseteq X$, then $f^{-1}(X\setminus E)$ is Borel measurable in $Y$ also.

(iii)

If $E_1,E_2,\dots \subseteq X$ are such that $f^{-1}(E_n)$ is Borel measurable in $Y$ for all $n\in N$, then $f^{-1}\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\right)$ is Borel measurable in $Y$ also.

(iv)

$f^{-1}(E)$ is Borel measurable in $Y$ for all open sets $E\in F$.

By doing so, we will have demonstrated that $f^{-1}(E)$ is Borel measurable for any Borel measurable $E\in ⟨F⟩=B$. By Exercise 1.4.14 of An Introduction to Measure Theory, we can set $F$ to be the family of open sets on $X$.

The first property follows trivially since $f^{-1}(\varnothing )=\varnothing$ is Borel measurable. The second property follows since $f^{-1}\left(X\setminus E\right)=Y\setminus f^{-1}(E)$ and since the complements of Borel measurable sets are Borel measurable. Similarly, the third property follows since $f^{-1}\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\right)={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{f}^{-1}\left(E_n\right)$ and since countable unions of Borel measurable sets are Borel measurable.

Finally, the fourth property follows by the fact that the inverse image of any open set is open under a continuous function; thus, it must be Borel measurable as well. □