The following is an exercise from Bruckner's Real Analysis:
For (a) if we take $ε<1$ in the hypothesis of Lusin's Theorem then $μ(E \backslash F)=0$ and this doesn't cause a problem I think? The reason for fail of Lusin's Theorem perhaps is there is no closed $F \subset Q_1$ such that $μ(Q_1 \backslash F)=0$ since $Q_1$ is dense in $X$.
For (b) the same wrong hypothesis mentioned in (a) holds so I don't know why the author thinks that there is a $g(x)$ fulfilling Lusin's Theorem hypotheses and thus its conclusion?
Lusin's Theorem that is mentioned in the book :
A detailed explanation for both parts of the question would be much appreciated.
(a) "The reason for fail of Lusin's Theorem perhaps is there is no closed $F \subset Q_1$ such that $μ(Q_1 \backslash F)=0$ since $Q_1$ is dense in $X$." Yes, that is why Lusin fails here. But the problem is asking what hypotheses in Lusin do not hold here. I would say the hypothesis that is missing is the assumption that $\mu(X)<\infty.$
(b) This time the hypothesis $\mu(X)<\infty$ is satisfied. Let $\epsilon>0.$ Choose $N$ such that
$$\tag 1 \sum_{i>N} 2^{-i} <\epsilon.$$
Let $F_1 = Q_1\cap \{r_i:i\le N\},$ $F_2 = Q_2\cap \{r_i:i\le N\}.$ Note that $F_1,F_2$ are disjoint closed sets. Define $g= 1$ on $F_1,$ $0$ on $F_2.$ Then $g$ is continuous on $F_1\cup F_2.$ By Tietze, $g$ extends to be continuous on $X.$ Notice that $\chi_{Q_1} =g$ on $\{r_i:i\le N\}.$ Thus the set where $\chi_{Q_1} \ne g$ is contained in $\{r_i:i> N\},$ a set of measure $<\epsilon$ by $(1).$