Let $f\in L^1(R^n)$. Let $\epsilon>0$ be given. Show that there exists a Lebesgue measurable set A with $\mu(A)< \epsilon$ such that the restriction of f to $R^n$ \ $A$ is continuous. (Suggestion : Consider Tchebyshev's inequality: $\mu({x:\vert g(x)\vert>\alpha})\le \frac{\Vert g \Vert}{\alpha}$)
I want to prove this problem by using Lusin's Theorem. But, I can't use this suggestion ....
Suppose $n=1$ just to make life simpler. By Lusin, in each interval $(m,m+1)$ there is a compact subset $K_m$ such that $\mu ((n,n+1)\setminus K_m)<\epsilon/2^m,$ and such that $f$ is continuous on $K_m.$ Then $f$ is continuous on $\cup_{m\in \mathbb {Z}}K_m,$ and $\mu (\mathbb {R}\setminus [\cup_{m\in \mathbb {Z}}K_m])< 3\epsilon.$ (We don't need $f\in L^1$ for this, just measurability.)