Lusin Theorem Conclusion (Confusion on different versions) 
According to Royden, the conclusion of Lusin Theorem is that:
(1) $f=g$ on $F$, where $g$ is a continuous function on $\mathbb{R}$.
However, according to Wikipedia, the conclusion of Lusin Theorem is:
(2) $f$ restricted to $F$ is continuous.
I have seen somewhere that (1) and (2) are different, e.g. $\chi_\mathbb{Q}$ restricted to $\mathbb{Q}$ is continuous, but $\chi_\mathbb{Q}$ is not equal to a continuous function $g$ on $\mathbb{R}$.
How do we resolve this paradox? Thanks.
 A: Lusin's theorem contains the condition that $F$ be a closed set. That makes a difference.
In sufficiently nice spaces $X$ ($T_4$ or normal spaces, whichever is the weaker in the nomenclature in use), every continuous function $f \colon F \to \mathbb{R}$, where $F\subset X$ is closed, has a continuous extension $F \colon X \to \mathbb{R}$. This is Tietze's extension theorem.
All metric spaces are normal, so for a function $f \colon X \to \mathbb{R}$, where $X$ is a metric space, and a closed subset $F \subset X$, the two assertions


*

*$f\lvert_F$ is continuous, and

*there is a continuous $g \colon X \to \mathbb{R}$ with $g\lvert_F = f\lvert_F$


are equivalent.
Royden, it seems, considers only functions defined on (subsets of) $\mathbb{R}$ at that point.
The wikipedia formulation that the restriction of $f$ to $E$ is continuous holds for Radon measures on arbitrary Hausdorff spaces. Since arbitrary Hausdorff spaces can be quite complicated, it is not surprising that only a weaker assertion can be made in the most general setting. The wikipedia article mentions the - at least formally - stronger conclusion in case of local compactness, so the author is aware of its availability in favourable circumstances.
