0
$\begingroup$

Lusin's Theorem. Let $f$ be a real-valued measurable function on $E$. Then for each $\varepsilon>0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E\sim F)<\varepsilon$.

can anybody guide me the proof of the theorem or can advice me some book from where i can go through the proof.

$\endgroup$
2
  • $\begingroup$ It's in Rudin's Real and Complex Analysis, theorem 2.24 $\endgroup$
    – Aweygan
    Mar 4, 2017 at 16:36
  • $\begingroup$ It's in pretty much any measure-theoretic real analysis reference. I suggest Royden/Fitzpatrick. $\endgroup$
    – Ian
    Mar 4, 2017 at 16:43

1 Answer 1

0
$\begingroup$

Here is a proof of the Lusin's Theorem where $f:[a,b]\to \mathbb{C}$.

Lusin's Theorem: If $f:[a,b]\to \mathbb{C}$ is Lebesgye measurable and $\epsilon > 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) < \epsilon$ and $f|E$ is continuous. (Note I will use Egoroffs theorem, Theorem 2.26, theorem 1.18, and Corollary 2.32 in Folland's Real Analysis, I am sure there are the same theorems in your book)

Proof: Let $f:[a,b]\to \mathbb{C}$ be Lebesgue measurable and $\epsilon >0$. By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\to f \in L^1$$ Then by corollary 2.32 there is a subsequence $\{g_{n_j}\}$ of $\{g_n\}$ such that $g_{n_{j}}\to f$ a.e. Now by Egoroff's theorem, for any $\epsilon > 0$ there exists a set $F\subset [a,b]$ with $\mu(F) < \epsilon/2$ such that $g_{n_j}\to f$ uniformly on $F^{c}$.

Now by theorem 1.18, since $\mu([a,b]) < \infty$, there is $E$ compact subset of $[a,b]$ such that $E\subset F^{c}$ and $$\mu(F^c) - \epsilon/2 < \mu(E) \leq \mu(F^c)$$ So $F\subset E^c$ and we have \begin{align*} \mu(E^c) &= \mu(F) + \mu(E^c\setminus F)\\ &= \mu(F) + \mu(E^c\cap F^c)\\ &= \mu(F) + \mu(F^c\setminus E)\\ &= \mu(F) + (\mu(F^c) - \mu(E))\\ &\leq \epsilon/2 + \epsilon/2\\ &= \epsilon \end{align*} Note since $E\subset F^c$ and $g_{n_j}\to f$ uniformly on $F^c$, we have that $g_{n_j}\to f$ uniformly on $E$. Since, for all $j$, $g_{n_j}$ is continuous, we have that $f$ is continuous on $E$, that is $f|E$ is continuous.

Hope this helps.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .