# Is my proof really circular-reasoning?

I proved for any measurable function $\mathbb{R}\rightarrow \mathbb{R}$, it is the limit of a sequence of continuous functions outside a set of measure zero.

Suppose $f$ is a measurable function on $\mathbb{R}$, $\mu(A)<\infty$, $f=0$ if $x\not\in A$, and $\epsilon>0$. Then there exist a $g\in C_{c}(\mathbb{R})$ such that $$\mu(\{x:f(x)\not=g(x)\})<\epsilon$$

By Lusin's theorem(see page 77, Tao, page 53, Rudin, page 140, Macdonald), we may select a sequence of sets $A_{i}=[-i,i]$ and $g_{i}$ such that $g_{i}\in C(A_{i})$ while $\mu(\{x:f(x)\not=g_{i}(x)\})<\frac{1}{i^{2}}$. Then since $f,g_{i}$ are both measurable, $f\pm g_{i}$ is measurable. In particular the set $K_{i+}=x:f(x)-g_{i}(x)>0$ and $K_{i-}=x:f(x)-g_{i}(x)<0$ are both measurable. Then $K_{i}=K_{i+}\cup K_{i-}$ is measurable and has measure less than $\frac{1}{i^{2}}$. By Tieze extension theorem from Munkres, because $\mathbb{R}$ is a normal topological space, we may extend $g_{i}$ to a continuous function $h_{i}$ from $\mathbb{R}$ to $\mathbb{R}$ such that for $x\in [-i,i]$, $h_{i}(x)=g_{i}(x)$.

Let $L_{n}=\bigcup_{i=n}^{\infty} K_{i}$ and $K_{\infty}=\bigcap_{n=1}^{\infty}L_{n}$. Then $K_{\infty}$ has measure 0 by last problem in last test, and we have $\sum \mu(K_{i})<\pi<\infty$. Let me cite the lemma: Suppose $E_{i}$ are Lesbegue measurable sets with $\sum \mu(E_{i})<\infty$, then $$\mu(\bigcap^{\infty}_{n=1}\bigcup^{\infty}_{k=n}E_{k})=0$$

Then I claim $h_{i}\rightarrow f$ on $R-K_{\infty}$. This is clear because for any $x\in \mathbb{R}$, it is in some interval $[-n,n]$ and thus $g_{n}(x)=h_{n}(x)=f(x)$ if $x\not\in K_{n}$. If $x\in K_{n}$ but not in $K_{\infty}$, then there is some $N>n$ such that for any $i>N$ we have $h_{i}(x)=g_{i}(x)=f(x)$. So the only chance $h_{n}(x)\not=f(x)$ as $n\rightarrow \infty$ is $x\in K_{\infty}$. However this is the case we excluded. So we proved $h_{i}\rightarrow f$ on $R-K_{\infty}$.

The professor and I got into a huge dispute because he think my proof is circular reasoning. He claimed that this statement is used in proving Lusin's theorem. I felt frustrated. Can someone take a look?

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Can you provide a proof of Lusin's theorem that does not use the statement? –  Andres Caicedo Dec 7 '12 at 2:19
Tao's proof in his book does not use this statement. And I think Rudin's proof also does not. –  Bombyx mori Dec 7 '12 at 2:22
Well, there you go. –  Andres Caicedo Dec 7 '12 at 2:32
He is not satisfied. He said Lusin's theorem is not proved during class or on exercises. So we cannot use it. I pointed out Tao's book is his own choice of reference book. But he still say this, and insist I have to show $f$ differ from a Borel function almost everywhere (to finish his version of proof). I felt very frustrated. –  Bombyx mori Dec 7 '12 at 2:35
I tried to prove every Lesbegue measurable function is the limit of a Borel function. It seems I cannot avoid the bootstrap setup of step function - simple function - measurable function, etc. I see his point now. –  Bombyx mori Dec 7 '12 at 4:10