# Lusin's theorem and its consequences.

Suppose we have already proved the following theorem:

Lusin Theorem. Let $$(X,\mathcal{G})$$ be a topological space., $$(X,\mathcal{L},\overline{\mu})$$ the associated Lebesgue measure space. We suppose that $$\overline{\mu}(X)<\infty$$ and $$\overline{\mu}$$ regular. Let $$f\colon X\to \mathbb{R}$$ Lebesgue measurable function. Then for all $$\delta>0$$ exists a closed set $$C\subseteq X$$ such that $$\overline{\mu}(X\setminus C)<\delta$$ and $$f_{|C}$$ is a continuous function.

I'm looking for texts, or (if it's simple) a proof by you of the following results:

Theorem 1. Let $$K\subseteq \mathbb{R}^n$$ be a compact set; $$f\colon K\to \overline{\mathbb{R}}$$ Lebesgue measurable. Then for all $$\delta>0$$ exists a continuous function $$\hat{f}\colon K\to \mathbb{R}$$ such that:

$$(i)\;$$ $$\lambda_n(\{f\ne \hat{f}\})<\delta$$;

$$(ii)\;$$ $$\sup_{x\in K} |\hat{f}(x)|\le \sup_{x\in K} |f(x)|$$.

Question How can I prove this theorem 1 using Lusin's theorem? What other result is needed?

Theorem 2. Let $$f\colon \mathbb{R}^n\to\mathbb{R}$$ Lebesgue measurable, such that $$\lambda_n({f\ne\hat{f}})<\infty$$. Then for all $$\delta > 0$$ exists $$\hat{f}\colon \mathbb{R}^n\to \mathbb{R}$$, continuous and compact support, such that:

$$(i)\;$$ $$\lambda_n(\{f\ne\hat{f}\})<\delta$$;

$$(ii)\;$$ $$\sup_{x\in\mathbb{R}^n}|f(x)|\le \sup_{x\in\mathbb{R}^n} |f(x)|$$.

$$\lambda_n$$ denotes the Lebesgue measure on $$\mathbb{R}^n$$.

Question How can theorem 2 be proved by using theorem 1? Do you know any text that goes exactly like this? Do you know any text that goes exactly like this?

Lusin's theorem gives you a closed subset $$C\subset K$$ (since $$K$$ is compact, $$C$$ will also be compact) s.t. $$\left. f\right|_{ C}$$ is continuous with $$\lambda\!\left( X \,\setminus\, C\right) < \delta$$. Now extend this to a continuous function $$\hat{ f} \colon K \to \overline{ \mathbf{R}}$$ using e.g. the Tietze extension theorem (this is very general, since we are dealing with $$\mathbf{R}$$ you could also prove it more directly) s.t. \begin{aligned} \sup_{ x \in C}\big|\left. f\right|_{ C} ( x)\big| = \sup_{ x \in K} |\hat{ f}( x) | .\end{aligned} Moreover, we have $$\hat{ f} = f$$ on $$C$$ meaning \begin{aligned} \lambda ( \{ f \neq \hat{ f} \}) \leqslant \lambda\!\left( X \setminus C\right) < \delta .\end{aligned}
The second theorem probably has some typo since $$\hat{ f}$$ seems to be fixed in the first sentence but what we try to find in the second sentence.