fmaths Posted April 8, 2021 Share Posted April 8, 2021 (edited) Hi, I've been working on some Lebesgue measure and Lebesgue integral exercises for a few days and I have some doubts. I need to say if the statements are true (I need to prove it) or false (I need to give a counterexample). Let $f:E\subset\mathbb{R}^n\rightarrow\mathbb{R}^n$ such that $|f(x)-f(y)|\le|x-y|$ for $x,y\in\mathbb{R}^n$, then $f$ transforms null measure sets into null measure sets. If two integrable functions agree in a dense set, the value of the integrals is the same. If two functions agree in a dense set, one of then is measurable if and only if the other one is also measurable. For the first one, \textbf{I have no idea. I only now it's true since Lipschitz function is like that}. For the second one, I know it's false. I've thought of considering two functions $f,g:[0,1]\rightarrow\mathbb{R}$ defined as follow: \begin{equation*} f(x) = \begin{cases} 1 & \text{if $x\in\mathbb{Q}$}\\ 0 & \text{if $x\notin\mathbb{Q}$} \end{cases} \end{equation*} \begin{equation*} g(x) = \begin{cases} 1 & \text{if $x\in\mathbb{Q}$}\\ \displaystyle{\frac{1}{q}} & \text{if $x\notin\mathbb{Q}$, $q\in\mathbb{N}$} \end{cases} \end{equation*} These functions agree in $\mathbb{Q}\cap[0,1]$. \textbf{Here, I don't know how to prove that $\mathbb{Q}\cap[0,1]$ is dense on $[0,1]$}. For the last one, I also know it's false. We could take a not measurable set $E$ et let $f=\chi_{E}\cdot\chi_{\mathbb{I}}$ et $f=g$, where $g$ s'annule sur $\mathbb{Q}$. \textbf{I don't know how to prove that $f$ is not measurable}. I'll be duly grateful for any help. Edited April 8, 2021 by fmaths Link to comment Share on other sites More sharing options...
mathematic Posted April 8, 2021 Share Posted April 8, 2021 1 hour ago, fmaths said: Hi, I've been working on some Lebesgue measure and Lebesgue integral exercises for a few days and I have some doubts. I need to say if the statements are true (I need to prove it) or false (I need to give a counterexample). Let $f:E\subset\mathbb{R}^n\rightarrow\mathbb{R}^n$ such that $|f(x)-f(y)|\le|x-y|$ for $x,y\in\mathbb{R}^n$, then $f$ transforms null measure sets into null measure sets. If two integrable functions agree in a dense set, the value of the integrals is the same. If two functions agree in a dense set, one of then is measurable if and only if the other one is also measurable. For the first one, \textbf{I have no idea. I only now it's true since Lipschitz function is like that}. For the second one, I know it's false. I've thought of considering two functions $f,g:[0,1]\rightarrow\mathbb{R}$ defined as follow: \begin{equation*} f(x) = \begin{cases} 1 & \text{if $x\in\mathbb{Q}$}\\ 0 & \text{if $x\notin\mathbb{Q}$} \end{cases} \end{equation*} \begin{equation*} g(x) = \begin{cases} 1 & \text{if $x\in\mathbb{Q}$}\\ \displaystyle{\frac{1}{q}} & \text{if $x\notin\mathbb{Q}$, $q\in\mathbb{N}$} \end{cases} \end{equation*} These functions agree in $\mathbb{Q}\cap[0,1]$. \textbf{Here, I don't know how to prove that $\mathbb{Q}\cap[0,1]$ is dense on $[0,1]$}. For the last one, I also know it's false. We could take a not measurable set $E$ et let $f=\chi_{E}\cdot\chi_{\mathbb{I}}$ et $f=g$, where $g$ s'annule sur $\mathbb{Q}$. \textbf{I don't know how to prove that $f$ is not measurable}. I'll be duly grateful for any help. The second and third easily false. The two function could agree on the rationals, but not elsewhere. You need to use another deliminator for Latex $ doesn't work here [math] and [math] with / work. Link to comment Share on other sites More sharing options...
joigus Posted April 8, 2021 Share Posted April 8, 2021 Is it Lebesgue day today? Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now