We have measure theory in this semester.I found the statement of Lusin's theorem on the internet to be:
Let $f:\mathbb{R\to R}$ be a Lebesgue measurable function.Then for each $\epsilon>0$ there exists a closed set $F
_\epsilon\subset \mathbb R$ such that $f|_{F_{\epsilon}}$ is continuous and $|\mathbb R\setminus F_{\epsilon}|<\epsilon$.
But in another book I saw the following version:
Let $f:\mathbb{R\to R}$ be a Lebesgue measurable function .Then for each $\epsilon>0$ there exists a compact set $K_{\epsilon}\subset \mathbb R$ such that $f|_{K_{\epsilon}}$ is continuous and $|\mathbb R-K_\epsilon|<\epsilon$.
Things turned out getting worse when our instructor told us the following version of Lusin's theorem:
Any continuous function on $\mathbb R$ with compact support is Lebesgue integrable.
Now I am really confused.I cannot understand why these all are equivalent.I also tried to prove these results but couldn't.In the book by Sheldon Axler I found a proof but that proof is given for Borel measurable functions not Lebesgue measurable functions.How can I prove these results and how to show they are indeed same?
Best Answer
The second statement is wrong. There is no compact subset $K$ in ${\bf R}$ such that $|{\bf R} \setminus K| < \varepsilon$. A compact subset of the real line is bounded and his complement is of infinite Lebesgue measure.
The first statement is correct. The last statement is correct but is unrelated to Lusin's theorem. These three statements are definitely not the same and they can't be deduced from each others.