Measuribility of Set

Question

let $X=Y=\mathbb R$ and $\mu=\nu=$Lebesgue measure, and $\mu*\nu$ is a two dimensional Lebesgue measure on $X*Y=R^2$. Then Show that

For each measurable subset $E$ of $\mathbb R$, the set $\sigma(E)=$$\{(x,y):x-y\in E\}$ is a measurable subset of $\mathbb R^2$. If $f$ is measurable function on $\mathbb R$, then function $F$ defined by $F(x,y)=f(x-y)$ is a measurable function on $\mathbb R^2$.

i was trying to approach it as: given $E$ to be measurable in $\mathbb R$ consider $\chi_E$. then $\chi_E$ is a measurable function and consider $g: \mathbb R^2 \rightarrow \mathbb R$ as $g(x,y)=x-y$ then $\chi_E\circ g=\chi_{\sigma_(E)}$ will be a borel measurable function given $E$ to be borel measurable but i got stuck when $E$ will be Lebesgue measurable then how can we talk about measurability of $\sigma(E)$.

Any type of suggestion would be appreciated. Thanks in advance.

Answer 1

Consider $f:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x,y)=x-y$. $f^{-1}(E)=\{(x,y):x-y\in E\}$. Since $f$ is continue, $f^{-1}(E)$ is measurable.

Answer 2

To show that $\sigma (E)$ is measurable it suffices to consider the case where $E$ is bounded. Because if $E^*(n)=\sigma (E\cap [-n,n])$ is measurable for each $n\in \mathbb N$ then $\sigma (E)=\cup_{n\in \mathbb N}E^*(n)$ is measurable.

The following is primitive, but it works.

Assume $E$ is bounded.

(1). Let $F=\{E_n:n\in \mathbb N\}$ be a family of closed subsets of $\mathbb R$ with $\cup F\subset E$ and $\mu (E$ \ $\cup F)=0.$

Each $\sigma (E_n)$ is closed in $\mathbb R^2$, for if $((y_j+e_j, y_j))_j$ is a sequence in $\sigma (E_n)$ converging to $(x,y)$, with $e_j\in E_n$ for every $j$, then there is a sub-sequence $(e_{f(j)})_j$ of $(e_j)_j$ converging to some $e\in E_n$ (because $E_n$ is compact). Then $(x,y)=(y+e,y)\in \sigma (E_n).$

So $\cup_{n\in \mathbb N}\sigma (E_n)$ is a countable union of closed sets of $\mathbb R^2$ and therefore is measurable.

(2). It remains to show that $\sigma(E)$ \ $\cup_{n\in \mathbb N}\sigma(E_n)$ is measurable. We shall show that in fact it has zero outer measure.

(2-i). For brevity let $\cup_{n\in \mathbb N}\sigma (E_n)= G.$ We have $$\sigma (E) \backslash \sigma (G)\subset \{(y+e,y)\in \mathbb R^2: e\in E \backslash G\}=\sigma (E \backslash G).$$

So it will suffice to show that the outer measure $(\mu*\nu)^o(\sigma (E$ \ $G)$ is $0$. (Notation: For $X\subset \mathbb R^2$ the outer measure of $X$ is $(\mu*\nu)^o(X)\;$.)

(2-ii). We have $\mu (E$ \ $G)=0.$ For any $r>0$ let $C_r$ be an open subset of $\mathbb R$ with $\mu(C_r)<r$ and with $(E$ \ $G)\subset C_r.$ Obviously $\sigma (E$ \ $G)\subset \sigma (C_r).$

For $z\in \mathbb Z$ and $r>0$ define $$S(z,r)=\sigma (C_r)\cap (\mathbb R\times (z,z+2))=\{(y+e,y):y\in (z,z+2))\land e\in C_r\}.$$

I will leave it as an exercise that $S(z,r)$ is open in $\mathbb R^2$ and that $(\mu*\nu)(S(z,r))=2\mu (C_r).$

(2-iii). For any $z\in \mathbb Z$ and any $r>0$ we have $S(z,r)\supset \sigma (E$ \ $G)\cap (\mathbb R\times [z,z+1].$ So if $r_z>0$ for each $z\in \mathbb Z$ then $\cup_{z\in \mathbb Z}S(z,r_z)\supset \sigma (E$ \ $G).$

(2-iv). So for each $z\in \mathbb Z$ take $r_z>0$ such that $\sum_{z\in \mathbb Z}r_z=1/2.$ Then for any $s>0$ we have $$(\mu*\nu)^o(\sigma (E)\backslash \sigma(G))\leq (\mu*\nu)^o(\sigma ( E \backslash G)\leq$$ $$\leq (\mu*\nu)(\cup_{z\in \mathbb Z}S(z,sr_z)\leq$$ $$\leq \sum_{z\in \mathbb Z}2sr_z=s.$$ Since $s>0$ is arbitary, we have $(\mu*\nu)^o(\sigma (E)$ \ $\sigma (G))=0.$