fmaths 0 Posted April 8 Share Posted April 8 (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 by fmaths Link to post Share on other sites

mathematic 104 Posted April 8 Share Posted April 8 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 post Share on other sites

joigus 460 Posted April 8 Share Posted April 8 Is it Lebesgue day today? Link to post Share on other sites

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