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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.

Anacardium
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  • 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 $f.$... – William M. May 17 '23 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. – William M. May 17 '23 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 https://math.stackexchange.com/questions/3428804/easier-proof-of-tietze-theorem-in-metric-case. – William M. May 17 '23 at 16:47
  • @WilliamM.: Have you seen the version of Lusin's theorem mentioned in Rudin's RCA? – Anacardium May 17 '23 at 18:20
  • @WilliamM.$:$Please have a look at https://math.stackexchange.com/q/4444504/512080 – Anacardium May 17 '23 at 18:38
  • I see this version: https://math.stackexchange.com/questions/1847916/lusins-theorem-from-rudin-rca

    I wouldn't call that Lusin's theorem (and I don't think that is the common way of stating Lusin's theorem) but rather than $C_c$ is dense in $L^1.$ (In fact, $C_c^\infty$ is dense in $L^p$ for all $p.$) The proof of that post is a Tietze-Urysohn style proof as I was suggesting. If you already have denseness of $C_c$ in $L^1,$ then bounding is easy as the answer below shows.

     – William M. May 17 '23 at 21:04
  • @WillianM.$:$ The result I posted is actually used to prove that $C_c$ is dense in $L^p$ for any $1 \leq p \lt \infty.$ I don't know where did you get your version. The version of Lusin's theorem I am familiar with can be found in any standard text in measure theory. For instance, Rudin's RCA, Royden's Real Analysis or Measure Theory by Elias M. Stein and Rami Shakarchi etc. – Anacardium May 18 '23 at 06:58
  • Your post is actually Lusin's theorem from Rudin's RCA as you can check yourself in the link I shared. The answer below is the last paragraph in Rudin's RCA's proof. The version of Lusin's theorem that I know I took it from Ash's Real Analysis and Probability; and I just checked Stein and Shakarchi's book, and their version and Ash's version is the same (and different from Rudin's); Dieudonne's book Treatise on Analysis, vol 2, also has the same version of Lusin's theorem as Ash's. I never really worked/used either of Rudin's or Royden's books (because of diverse reasons). – William M. May 18 '23 at 14:57

1 Answers1

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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.

Kenny Wong
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  • Showing $g$ exists is pretty much the exercise. – William M. May 17 '23 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 $\widetilde g$. – Kenny Wong May 17 '23 at 17:24
  • Anyway, I'm happy to edit if the OP responds with clarifications. – Kenny Wong May 17 '23 at 17:27
  • 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 $|g(x)| \leq R.$ – Anacardium May 17 '23 at 18:37
  • For all $x \in \Omega \setminus A$, $g(x) = \varphi(x)$. Hence $|g(x)| = |\varphi(x)|\leq R$. So $\widetilde g(x) = g(x)$. – Kenny Wong May 17 '23 at 18:42
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    Sorry for all $x \in \Omega$ with $|g(x)| = R$ we have $\widetilde g (x) = g(x)$ as well. I missed that point. – Anacardium May 17 '23 at 18:43
  • Actually, as written, it is true that $\widetilde g(x) = g(x)$ when $|g(x)|\leq R$. Think about it... – Kenny Wong May 17 '23 at 18:44
  • Right. I agree. – Anacardium May 17 '23 at 18:44
  • So this is an application of Lusin's theorem. Right? I don't get what William M. tried to say. – Anacardium May 17 '23 at 18:45