Why Luzin's theorem can not be changed from $|\mathbb{R} \backslash B| < \epsilon$ to $|\mathbb{R} \backslash B| = 0$

Question

I am looking to prove the following however I am struggling. Any help would be appreciated. Thanks in advance.
We let $G$ be an open set so that $\mathbb{Q} \subseteq G$ and $|G| < 1$, and let $f : \mathbb{R} \rightarrow \mathbb{R}$ be given by $f = \chi_G$. I am wanting to show that there is no Borel set $B$ with $|\mathbb{R} \backslash B| = 0$ so that $f|B$ is continuous.

Answer 1

My attempt bellow. It was more delicate than I originally thought (do not look my edit, lol). Probably there is a simpler solution. I would appreciate any comments criticizing it as I might have made some mistake along the way, I'm a bit new to measure theory and the subtleties often escape me.

By abuse of notation, I denote the restriction of $f$ to $B$ as $f$, since I wont use the original function very much. I hope its clear.

The point is that as $B$ carries all of the mass of $\mathbb{R}$, there is no way to make $f$ "constant enough" on $B$ (and the only way for a characteristic function to be continuous is if its constant in each connected segment), because every subset of finite positive measure will intersect $B$ non trivially (otherwise $|\mathbb{R}\setminus B|>0$), and it cannot contain $B$ (because the measure of B is large). So, if $E=B\cap G$, $E$ is always not empty, $f(E) = \{1\}$, and there is always a $x \in B\setminus G$ such that $f(x) = 0$. (This proofs the result in case B is a connected borel subset of $\mathbb{R}$). Another observation is that $G=\cup I_n$ for disjoint open non empty intervals $I_n$, each containing a rational number..

I'll proof the following two cases:

First Case: $B$ is totally disconnected.

Now take any $x \in B\setminus G$ and a sequence $(r_k)$ of rationals converging to $x$. From this we can construct a sequence $(x_k)$ in $E= B\cap G$ (so $f(x_k)=1$ for every $k$) converging to $x$. Thus $f$ is not continuous at $x$, and as $x$ is arbitrary, it cannot be continuous on $B$. (i. e,$ \lim f(x_k) = 1 \neq 0= f(x) $ )

This construction is not hard, basically for every interval $J_\varepsilon$ centered in $x$ and lenght $\varepsilon$, there is a rational point $r_k \in G=\cup I_n$ (in fact infinitely many), so $J_\varepsilon \cap I_n$ is not empty for some $n$. This intersection is a interval of finite measure, so it has some point $x_1$ of $B$ sufficiently close to $x$, by the observations above. Repeating the argument for smaller and smaller $\varepsilon$, we get the desired sequence.

Second Case: $B$ has connected components.

If some connected component $A$ of $B$ contains some $I_n$ properly. In this case, we clearly have that $f$ is not continuous in this component, and we are done.

Suppose it doesn't happen. That is, every connected component of $B$ is contained or equal to some $I_n$. Them removing all of those from $B$, we are left with a totally disconnect set $B^\prime$ with most of the mass of $B$ because the mass of $G$ is small compared to $B$ (in other words, $B$ is "mostly disconnected"). But them, by the first case, $f$ cannot be continuous on $B^\prime$, so it also cannot be continuous on $B$. So we have no problems here.

I think that covers all the cases, but again, I might be missing something. I hope its helpful.