# Lebesgue measure and Lebesgue integral

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Posted (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 by fmaths
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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.

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