Return to Answer

The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

The answer is no.

Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.

Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1_A*1_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus \begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*} Yet, $\nu$ is carried $\{(x,z) \in \mathbb{R}^2 : x+z \in \mathbb{Q}\}$, whose $\lambda_2$ measure is null. Furthermore, $\lambda_2$ is carried by the complement of $\{(x,z) \in \mathbb{R}^2 : x+z\in \mathbb{Q}\}$, which has null $\nu$-measure. Hence, the measures $\lambda_2$ and $\nu$ are mutually singular.

The measure $(1_{I^2})\nu$ yields a couterexample.

The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

The answer is no.

Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.

Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1_A*1_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus \begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*} Yet, $\nu$ is carried by $\{(x,z) \in \mathbb{R}^2 : x+z \in \mathbb{Q}\}$, whose $\lambda_2$ measure is null. Furthermore, $\lambda_2$ is carried by the complement of $\{(x,z) \in \mathbb{R}^2 : x+z\in \mathbb{Q}\}$, which has null $\nu$-measure. Hence, the measures $\lambda_2$ and $\nu$ are mutually singular.

The measure $(1_{I^2})\nu$ yields a couterexample.

The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

The answer is no.

Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.

Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1_A*1_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus \begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*} Yet, $\nu$ is carried $\{(x,z) \in \mathbb{R} : x+z \in \mathbb{Q}\}$, whose $\lambda_2$ measure is null. Furthermore, $\lambda_2$ is carried by the complement of $\{x,z) \in \mathbb{R}^2 : x+z\}$, which has null $\nu$-measure. Hence, the measures $\lambda_2$ and $\nu$ are mutually singular.

The measure $(1_{I^2})\nu$ yields a couterexample.

The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

The answer is no.

Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.

Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1_A*1_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus \begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*} Yet, $\nu$ is carried $\{(x,z) \in \mathbb{R}^2 : x+z \in \mathbb{Q}\}$, whose $\lambda_2$ measure is null. Furthermore, $\lambda_2$ is carried by the complement of $\{(x,z) \in \mathbb{R}^2 : x+z\in \mathbb{Q}\}$, which has null $\nu$-measure. Hence, the measures $\lambda_2$ and $\nu$ are mutually singular.

The measure $(1_{I^2})\nu$ yields a couterexample.

The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

The answer is no.

Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.

Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1_A*1_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus \begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*} Yet, $\nu$ is carried $\{(x,z) \in \mathbb{R} : x+z \in \mathbb{Q}\}$, whose $\lambda_2$ measure is null.

The measure $(1_{I^2})\nu$ yields a couterexample.

The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

The answer is no.

Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.

Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1_A*1_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus \begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*} Yet, $\nu$ is carried $\{(x,z) \in \mathbb{R} : x+z \in \mathbb{Q}\}$, whose $\lambda_2$ measure is null. Furthermore, $\lambda_2$ is carried by the complement of $\{x,z) \in \mathbb{R}^2 : x+z\}$, which has null $\nu$-measure. Hence, the measures $\lambda_2$ and $\nu$ are mutually singular.

The measure $(1_{I^2})\nu$ yields a couterexample.