Borel measurable vs continuous functions

Question

I have the following statement which I think (given the definitions below) should hold. I would appreciate your expert opinions on whether the proof is valid.

Definition. 1 (Continuoity). Given topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$, a function $f:X\to Y$ is continuous if $f^{-1}(O)\in\tau_X$ for every open set $O\in\tau_Y$.

Definition. 2 (Measurablity). Given measurable spaces $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$, a function $f:X\to Y$ is measurable if $f^{-1}(A)\in\Sigma_X$ for every subset $A\in\Sigma_Y$.

Theorem. Let $\Sigma_i$ be the Borel $\sigma$-algebra induced by the topology $\tau_i$ on the space $i$ for $i=X,Y$. A function $f:X\to Y$ is continuous if and only if it is Borel measurable.

Proof. Assume that $f$ is Borel measurable. Since the Borel $\sigma$-algebras contain all open subsets, $f^{-1}(A)\in\Sigma_X$ for every subset $A\in\Sigma_Y$ implies that $f^{-1}(O)\in\tau_X$ for every open set $O\in\tau_Y$.

To show the converse, we usethe following proposition (see e.g. Proposition 2.4 in here).

Proposition. Suppose that $f^{-1}(O)\in\Sigma_X$ for every open set $O\in\tau_Y$. Then, $f^{-1}(A)\in\Sigma_X$ for every subset $A\in\Sigma_Y$.

Now assume that $f$ is continuous. Then, $\forall O\in\tau_Y, \, f^{-1}(O)\in\tau_X \Rightarrow f^{-1}(O)\in\Sigma_X \, \overset{Prop.}{\Rightarrow} \, \forall A\in\Sigma_Y, \, f^{-1}(A)\in\Sigma_X$. QED.

Answer 1

If you consider the Euclidean topology on $\Bbb R$, then the function $f:\Bbb R\to \Bbb R$, $f(x)=\begin{cases}1&\text{if }x>0\\ 0&\text{if }x\le 0\end{cases}$ is Borel and discontinuous.

Answer 2

Definitely false. Continuity is much stronger than measurability. For example, consider the indicator function $f= \chi_{[0,1]}$ on $\mathbb{R}$ which is measurable but non-continuous.