Lusin's theorem states that for every $\varepsilon$, for every borel measure $\mu$, for every function $f:\mathbb{R}^n\to\mathbb{R}^m$, for every open set $A$ of finite measure, there exists a compact set $K$ such that $\mu(A-K)<\varepsilon$ and $f$ restricted to $K$ is continue.
Now, I'm wondering about what's the form of the compact $K$ in the case of the Dirichlet's function. Can you help me?
[Math] Lusin’s theorem
real-analysis
Best Answer
I found an answer by myself. First, let's order the rationals, say $\{q_n\}$. Then consider the covering $O_n=(q_n-\frac{\varepsilon}{2^n},q_n+\frac{\varepsilon}{2^n})$ for $n=1,2,\ldots$. It's straightforward to check that the measure of $\cup O_n$ is less than $\epsilon$. Now it suffices to choose the complementary $K$ of $\cup O_n$ in $A$ which is obviously compact. The Dirichlet's function on $K$ is $0$.