Let $\Omega \subseteq \mathbb R^N$ be open and $\varphi$ be a simple function defined on $\Omega$ which vanishes outside a set of finite Lebesgue measure. Then given $\varepsilon \gt 0$ there exists a continuous function $g \in C_c (\Omega)$ with compact support contained in $\Omega$ such that $g = \varphi$ except possibly on a set whose Lebesgue measure is less than $\varepsilon$ and $\|g\|_{\infty} \leq \|\varphi\|_{\infty}.$
This is clearly Lusin's theorem except the last part. Can we somehow manage to find a continuous function which has the same property but it is bounded above by the measurable function we have started with? Any suggestion in this regard would be warmly appreciated.
Thanks for your time.
Best Answer
Suppose that $g : \mathbb \Omega \to \mathbb C$ is a continuously compactly-supported function such that $g(x) = \varphi(x)$ for all $x \in \mathbb \Omega \setminus A$, where $\mu(A) < \epsilon$.
Let $R = \sup_{x \in \Omega} |\varphi(x)|$ (which is finite, since $\varphi$, being simple, only attains finitely many values), and define another function $\widetilde g : \Omega \to \mathbb C$ by $$ \widetilde g(x) = \begin{cases} g(x) & \text{if } |g(x)|< R \\ \frac{Rg(x)}{|g(x)|} & \text{if } |g(x)| \geq R\end{cases}. $$ Then $\widetilde g$ is continuous. Furthermore, $\widetilde g$ has the same support as $g$, hence this support is a compact subset of $\Omega$. We also have $\widetilde g(x) = g(x) = \varphi(x)$ for all $x \in \Omega \setminus A$. Finally, it's clear that $\| \widetilde g \|_\infty \leq \| \varphi\|_\infty$ from the way that $\widetilde g$ is constructed. Thus $\widetilde g$ satisfies all your requirements.