Proof of Lusin's Theorem
Solution 1:
It may be useful to note that the construction employed here has striking similarities with the construction of a fat Cantor set.
Regarding your first question: I assume that by complements you mean complements w.r.t. $A$. The sets $K_n, K'_n$ are explicitly chosen so that $$ \mu \Bigg( \bigcup_{n=1}^{\infty} (K_n \cup K'_n)^{c} \Bigg) \le \sum_{n=1}^\infty \mu((K_n \cup K'_n)^c) < \sum_{n=1}^\infty \frac{\epsilon}{2^n} = \epsilon $$ so the complements don't really cover much of $A$ if $\epsilon$ is small.
For the second question: I think the reason given for continuity is rather cryptic and confusing, so I found a different way of seeing it. The sets $V_n$ form a base of the topology of $\mathbb R$, so to prove continuity, it is enough to show that the (restricted) inverse image of each $V_n$ is open in $K$. It is an easy exercise in set theory to prove that $$ K \cap f^{-1}(V_n) = K \cap K_n = K \setminus K'_n $$ and $K'_n$, as a compact subset of $\mathbb R$, is closed, so its complement is open.
To get a better perspective of how continuity is achieved, consider that for every $n$ we have
- $K_n \cap K'_n = \emptyset$
- $K \subset K_n \cup K'_n$
- $K_n$ and $K'_n$ are both closed.
It follows that whenever $K \cap K_n$ and $K \cap K'_n$ are nonempty, they form a separation of $K$. This suggests that $K$ tends to have very little connectivity and indeed if $f$ is injective, $K$ will be totally disconnected.