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?


2 Answers 2


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.

  • $\begingroup$ So,can you tell me the proof of the first statement? $\endgroup$ May 6, 2022 at 14:40
  • $\begingroup$ This is done for example in the book of Rudin, real analysis. It would be a bit long to fit here. $\endgroup$
    – coudy
    May 6, 2022 at 14:55
  • $\begingroup$ I did not find it in Rudin,please write it here. $\endgroup$ May 6, 2022 at 15:08
  • $\begingroup$ This is theorem 2,23 in Rudin. Start working on a finite interval $[-N, N]$. $\endgroup$
    – coudy
    May 6, 2022 at 15:15
  • $\begingroup$ @Kishalay Sarkar: Proofs of the first statement should be easy to find by googling, and there are probably several such in MSE answers. However, I have a proof in some notes I wrote about 21 years ago, notes that contain other things of possible interest -- send me an email if you are interested. Also, my answer to Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions? might be of interest. $\endgroup$ May 6, 2022 at 15:17

Dave L. Renfro's comments and hints made me answer this question.First we prove the following:

Let $f:\mathbb{R\to R}$ be measurable and $\epsilon>0$,then there exists $E\subset \mathbb R$ measurable such that $\lambda(X\setminus E)<\epsilon$ and $f|_E$ is continuous.

If we could prove this,then we have a measurable set $E$ such that $\lambda(X\setminus E)<\epsilon/2$.Now $E$ can be expressed as a disjoint union $E=K\cup L$ where $\lambda(L)<\epsilon/2$ and since $f|_E$ is continuous,so is $f|_K$ and $\mathbb R\setminus K=\mathbb R\setminus (E^c\cup L)$ so that $\lambda(\mathbb R\setminus K)<\epsilon$.

Having said that,let us prove the statement given above:

Let $\epsilon>0$ and let $(I_n)$ be an enumeration of the open intervals in $\mathbb R$ with rational endpoints.

$f^{-1}(I_n)$ is measurable as $I_n$ is Borel and $f$ measurable.Then $\exists G_n$ open such that $f^{-1}(I_n)\subset G_n$ and $\lambda(G_n\setminus f^{-1}(I_n))<\epsilon/2^n$.

Let $E$ be the complement of $\bigcup\limits_{n=1}^\infty (G_n\setminus f^{-1}(I_n))$,then $E$ is measurable and $\lambda(\mathbb R\setminus E)<\epsilon$.Now it can be shown that $(f|_E)^{-1}(I_n)=f^{-1}(I_n)\cap E=E\cap G_n$ which is open in $E$ and since $\{I_n\}$ is a basis of $\mathbb R$with usual topology,so $f|_E$ is continuous.


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