Lusin's theorem. The following steps furnish a proof of the following result. Theorem (Lusin). Let (mu) be
Question:
Lusin's theorem. The following steps furnish a proof of the following result.
Theorem (Lusin). Let \(\mu\) be an outer regular measure on the space \((X, d, \mathscr{B}(X))\). For every \(f \in \mathcal{L}^{p}(\mu), 1 \leqslant p0\) there is some \(\phi_{\epsilon} \in \mathcal{L}^{p}(\mu) \cap C_{b}(X)\), such that
\[\left\|\phi_{\epsilon}ight\|_{\infty} \leqslant\|f\|_{\infty} \leqslant \infty, \quad \mu\left\{f eq \phi_{\epsilon}ight\} \leqslant \epsilon \quad \text { and } \quad\left\|f-\phi_{\epsilon}ight\|_{p} \leqslant \epsilon\]
(i) Let \(A \in \mathscr{B}(X)\) such that \(\mathbb{1}_{A} \in \mathcal{L}^{p}(\mu)\). Then there is an open set \(U \supset A\) such that \(\mu(U)
(ii) Let \(f \in \mathcal{L}^{p}(\mu)\) such that \(0 \leqslant f \leqslant 1\). By the sombrero lemma (Corollary 8.9) there is a uniformly convergent sequence of simple functions which can be 'smoothed' using part (i).
(iii) Let \(f \in \mathcal{L}^{p}(\mu)\) such that \(c=\|f\|_{\infty}
(iv) Let \(f \in \mathcal{L}^{p}(\mu)\). Apply part (iii) to \(f_{R}:=(-R) \vee f \wedge R\).
Data from corollary 8.9
Data from lemma 17.3
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