# An application of Lusin's theorem.

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.

• Lusin's theorem is quite different, though. It asserts that $f$ is $\mu$-measurable if and only there exists a compact set $K$ such that $f\mid_K$ is continuous and $K^\complement$ is $\mu$-small. The issue is that $f\mid_K$ is regarded as a function $K \to \mathbf{R},$ and in your exercise $g$ has the same domain as $f.$... Commented May 17, 2023 at 16:32
• ...To put things more solid, consider $f = \mathbf{1}_\mathbf{Q\cap [0,1]}.$ Then $f$ is Lebesgue-measurable and therefore for some compact $K \subset [0,1],$ such that $\lambda(K)\geq 1 - \varepsilon,$ $f$ is continuous when regarded as a function $K \to \mathbf{R}.$ However, on no subset with an interior point will $f$ be continuous. Commented May 17, 2023 at 16:33
• In regards to your problem, the usual way would be to reduce $\varphi$ to the indicator function of a measurable set and then use something like monotone class theorem. To deal with $\mathbf{1}_E,$ you can use the regularity of Lebesgue measure to approximate $E$ with open and compact set, and then define something along the lines of the Tietze-Urysohn lemma math.stackexchange.com/questions/3428804/…. Commented May 17, 2023 at 16:47
• @WilliamM.: Have you seen the version of Lusin's theorem mentioned in Rudin's RCA? Commented May 17, 2023 at 18:20
• @WilliamM.$:$Please have a look at math.stackexchange.com/q/4444504/512080 Commented May 17, 2023 at 18:38

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.
• Showing $g$ exists is pretty much the exercise. Commented May 17, 2023 at 17:21
• @WilliamM. I thought the OP already has a proof that $g$ exists, and is asking how to construct $\widetilde g$ from $g$? That's how I interpreted the question. For example, the title of the question is "An application of Lusin's theorem". And the OP writes, "This is clearly Lusin's theorem except the last part." - suggesting that the OP knows how to construct $g$ but not $\widetilde g$. Commented May 17, 2023 at 17:24
• How do you know that $\widetilde {g} (x) = g (x)$ for all $x \in \Omega \setminus A\$? I think you have to change the definition of $\widetilde g$ slightly. Just define it to be $g(x)$ when $|g(x)| \leq R.$ Commented May 17, 2023 at 18:37
• Sorry for all $x \in \Omega$ with $|g(x)| = R$ we have $\widetilde g (x) = g(x)$ as well. I missed that point. Commented May 17, 2023 at 18:43