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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?

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1 Answer 1

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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.

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