# What does Luzin's theorem imply?

Luzin's theorem states that: let $f:[a,b]\rightarrow R$ be an a.e. finite function, $f$ is measurable iff $\forall \epsilon \geq 0: \exists \phi_\epsilon$ continuous on $[a,b]$ and $\mu\{x: f(x)\neq \phi_\epsilon (x)\} \leq \epsilon$

Why doesn't it imply that a measurable function $f$ equals a.e. to a continuous function $\phi_\epsilon$?

-
$\phi_\epsilon$ might be different for every $\epsilon$. How would you use them to pick a single function $\phi$? I think you'll find that your technique doesn't preserve continuity. – Nate Eldredge Oct 8 '12 at 3:44
@NateEldredge I don't necessarily mean to pick a single $\phi$. – Anita Oct 8 '12 at 3:51
Well, each $\phi_\epsilon$ could be different from $f$ on a set of positive measure (but at most $\epsilon$). I guess I don't understand why you think Luzin's theorem could possibly imply that $f$ is a.e. equal to some continuous function. And of course, you must know of counterexamples. – Nate Eldredge Oct 8 '12 at 3:57
Anna: In addition to the answer Seirios gave, you might find some things of interest in my essay on Luzin's Theorem. – Dave L. Renfro Oct 8 '12 at 20:17

Let $f : [0,1] \to [0,1]$ such that $f(x)= 0$ if $x \leq 1/2$ and $f(x)=1$ otherwise. Then, let $f_{\epsilon} : [0,1] \to [0,1]$ be such that $f(x)=0$ if $0 \leq x \leq \frac{1}{2}- \epsilon$, $f(x)= 1$ if $\frac{1}{2}+ \epsilon \leq x \leq 1$ and $f_{\epsilon}$ affine on $[1/2- \epsilon, 1/2+ \epsilon]$.
So $f_{\epsilon}$ is continuous and $\mu \{x : f(x) \neq f_{\epsilon}(x) \}=2 \epsilon$.
However, there is no continuous function $\phi$ such that $f=\phi$ almost everywhere. Indeed, if it was the case, there would be sequences $x_n < 1/2$ and $y_n > 1/2$ converging to $1/2$ and such that $f(x_n)=\phi(x_n)$, $\phi(y_n)=f(y_n)$ (otherwise $\{ x : f(x) \neq \phi(x) \}$ would contain an interval); for $n \to + \infty$, you find $0= \phi(1/2)=1$.