3
$\begingroup$

I want to prove the following:

Suppose $E$ is a subset of $\Bbb R$, let $\gamma(E)=\{ (x,y)\in \Bbb R \times \Bbb R :x-y\in E\}$. If $E\in \Bbb B$ (Borel/Lebesgue measurable set), show that $\gamma(E)\in \Bbb B \times \Bbb B$ .

Here’s my idea:

I want to first prove the following: Let $(X,\Bbb X)$ be a measurable space, $f$ be a measurable function on $X$ to $\Bbb R$, and let $\phi$ be a Borel measurable function, want to show that $\phi \circ f$ is measurable. Then I will take $\phi=\chi_E$ and $f(x,y)=x-y$ to come up with the result.

But I’m not sure how to prove $\phi \circ f$ is $X$-measurable. Could someone help to provide a proof please? I’m also not sure if I’m on the right track. Thanks.

$\endgroup$
  • $\begingroup$ $\gamma(E)=f^{-1}(E)$, and $f$ is Borel measurable. The case of a $E$ being Lebesgue measurable is a little harder but can be done (show that $\lambda^2(f^{-1}(A))=0$ whenever $\lambda(A)=0$, where $\lambda$ is the Lebesgue measure and $\lambda^2$ is the product measure). $\endgroup$ – Luiz Cordeiro May 3 '16 at 15:35
  • $\begingroup$ @Luiz Cordeiro So is this a different way of proving the problem? Would be great if you could write it up as an answer. $\endgroup$ – nora012 May 3 '16 at 15:49
  • $\begingroup$ $\,\,(X,X)?\,\,$ $\endgroup$ – zhw. May 3 '16 at 16:49
  • $\begingroup$ @zhw. I’ve edited the question, $X$-measurable is simply measurable. $\endgroup$ – nora012 May 3 '16 at 16:54
1
$\begingroup$

The function: $f:\mathbb{R}^2\to\mathbb{R}$ is Borel-measurable (it is continuous), so if $E$ is Borel, then $\gamma(E)=f^{-1}(E)$ is Borel.

For the Lebesgue-measurable part, let $\lambda$ be the Lebesgue measure. Let's first show that if $A$ is Borel in $\mathbb{R}$ and $\lambda(A)=0$, then $\lambda^2(f^{-1}(A))=0$ (where $\lambda^2$ is the Lebesgue measure on $\mathbb{R}^2$).

Consider the linear function $T:\mathbb{R}^2\to\mathbb{R}^2$, $T(x,y)=(x-y,y)$ (this transforms $f^{-1}(A)$ in a rectangle). Then $$\mu(f^{-1}(A))=\int_{\mathbb{R}^2}\chi_A(x)dx$$ Making the substitution $x=T^{-1}y$, we have $$\mu(f^{-1}(A))=\int_{\mathbb{R}^2}\chi_A(T^{-1}y)|\det T^{-1}|dyx=\int_{\mathbb{R}^2}\chi_{T(A)}(y)dy=\lambda^2(T(A))=\lambda^2(A\times\mathbb{R})=0$$

It follows that if $N$ is null in $\mathbb{R}$ (i.e., $N\subseteq A$, where $A$ is Borel and $\lambda(A)=0$), then $f^{-1}(N)$ is null in $\mathbb{R}^2$, and in particular it is Lebesgue measurable.

Since every Lebesgue set in $\mathbb{R}$ is the union of a Borel set and a null set, we are done.

$\endgroup$
  • $\begingroup$ Also, I don’t get the first part: The function: $f:\mathbb{R}^2\to\mathbb{R}$ is Borel-measurable (it is continuous), so if $E$ is Borel, then $\gamma(E)=f^{-1}(E)$ is Borel. Is this a theorem that’s assumed to be true? $\endgroup$ – nora012 May 3 '16 at 16:49
  • $\begingroup$ @nora012 Yes that's a basic result. The collection of sets $\{f^{-1}(E): E \text { is Borel }\}$ is a $\sigma$-algebra, and it contains the open sets. Therefore it contains all Borel sets. $\endgroup$ – zhw. May 3 '16 at 16:54
0
$\begingroup$

Another proof of: Let $f(x,y) = x-y.$ If $E\subset \mathbb R$ is a Borel set with $\lambda_1(E)=0,$ then $f^{-1}(E)$ is a Borel subset of $\mathbb R^2$ with $\lambda_2(f^{-1}(E))=0.$

Proof: I take as known that the continuity of $f$ implies $f^{-1}(E)$ is a Borel set. So we can apply Fubini:

$$\tag 1 \lambda_2(f^{-1}(E)) = \int_{\mathbb R^2} \chi_{f^{-1}(E)}\, d \lambda_2 = \int_{\mathbb R}\int_{\mathbb R}\chi_{f^{-1}(E)}(x,y)\, dx\, dy.$$

If we fix $y,$ then the inner integrand as a function of $x$ is simply $\chi_{E+y}.$ The $x$ integral is then $\lambda_1(E+y) = \lambda_1(E) = 0.$ Thus $(1)$ equals $\int_{\mathbb R} 0\, dy = 0$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.