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Suppose that $f\in \mathcal{L}(\mathbb{R})$ and $g\in \mathcal{L}(\mathbb{R})$, $\phi(x,y)=f(y-x)g(x)$, prove that $\phi$ is measurable in $\mathbb{R}^2$.

I try to prove this problem by the definition of measurable function:

For any $\alpha\in\mathbb{R}$, $E=\{(x,y)\in\mathbb{R}^2|\phi(x,y)<\alpha\}$ is measurable in $\mathbb{R^2}$.But I failed.

Are there some better ideas to solve this problem about measurable function in $\mathbb{R^2}$ ?

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There is a short discussion here: – Siminore May 18 '12 at 13:44
@Siminore.Thank you for your help.But I still cannot make it clear.I have read Page 1 of this document.I have known that $F(x,y)=f(y-x)$ is measurable on $\mathbb{R^2}$.However,why $f(y-x)g(x)$ is measurable? – Juntao Huang May 18 '12 at 15:20
Hewitt & Stromberg, REAL AND ABSTRACT ANALYSIS, have this result. See (21.31). For the Lebesgue-measurable case it is about half a page. The main thing to be proved: If $\phi(x,y) = x-y$, and $M \subseteq \mathbb R$ has measure zero in the line, then $\phi^{-1}(M) = \{(x,y)\colon x-y \in M\}$ is Lebesgue measurable in the plane. It is not enough to use sigma-algebra Lebesgue x Lebesgue in the plane: you have to use the completion of that sigma-algebra. – GEdgar May 18 '12 at 15:48
up vote 1 down vote accepted

$\newcommand{\R}{\Bbb R}\newcommand{\Z}{\Bbb Z}\newcommand{\N}{\Bbb N}\newcommand{\Q}{\Bbb Q}$ I don't know how many tools you have available, but a solution can go like this.

If $F$ and $G$ are measurable functions defined in $\R^d$ then $FG$ is a measurable function. A proof of this fact can be found here.

Then it's enough to show that the maps $$(x,y)\mapsto f(y-x),\qquad (x,y)\mapsto g(x)$$ are measurable functions.

First, we need the following Lemma.

Lemma. Let $E\subseteq \R$ a measurable set. If $[a,b[$ is a finite interval, then $E\times [a,b[$ and $[a,b[\times E$ are measurables in $\R^{2}$.

Notice that it's enough to prove the Lemma for the interval $[0,1[$. Then the proof of the Lemma is in stages:

  1. First consider $E$ with $m(E)=0$
  2. Then consider $E=(c,d)$ a finite open interval.
  3. $E$ open set with finite measure.
  4. $E$ a $G_\delta$ set with finite measure.
  5. $E$ a measurable set with finite measure.
  6. $E$ a measurable set.

Notice that from the Lemma follows:

Corollary. If $E\subseteq \R$ is a measurable set, then $E\times \R$ and $\R\times E$ are a measurable sets of $\R^2$.

To prove the Corollary, note that $$E\times\R=\bigcup_{n\in\Z} E\times [n,n+1[.$$


Lemma. Let $f:\R\to\R$ a measurable function. Then the functions $F,G:\R^2\to\R$ defined by $$F(x,y)=f(y),\qquad G(x,y)=f(x)$$ are measurable.

Proof. Let $\alpha\in\R$. Notice that $$\lbrace (x,y):F(x,y)\lt\alpha \}=\R\times \lbrace y:f(y)\lt\alpha\},$$ by the corollary above, we are done.

To finish note that $$\lbrace (x,y)\in\Q^2: f(y-x)\lt\alpha\}=\bigcup_{x\in\Q}\lbrace x\}\times(\lbrace y\in\Q : f(y)\lt \alpha\}+x),$$ is measurable, therefore $(x,y)\mapsto f(y-x)$ is measurable.

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Thank you,leo!I just forget that if $f$ and $g$ are measurable in $\mathbb{R^d}$, then $fg$ is measurable in $\mathbb{R^d}$. – Juntao Huang May 18 '12 at 17:20
You're welcome! Glad to help – leo May 18 '12 at 17:23
@leo Can you please explain how the result about $\mathbb Q^2$ implies that $\lbrace (x,y)\in\R^2: f(y-x)\lt\alpha\}$ is measurable? – Dimitrios Nt Dec 1 '13 at 3:25
@DimitrisDallas yes look at this – leo Dec 1 '13 at 19:51
In the link, they are dealing with $a\in\mathbb Q$, not $x\in\mathbb Q$, so they are using $$\{x\in E:f(x)>\alpha\}=\bigcup_{n=1}^\infty\{x\in E:f(x)>q_n\}$$ But in our case, we have proved the result for $x\in \mathbb Q$. I can't see how to generalize for $\mathbb R$. – Dimitrios Nt Dec 2 '13 at 2:06

First of all, the $\sigma$-algebras $\mathcal{B}(\mathbb{R}^2)$ and $\mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$ are the same, since $\mathbb{R}$ is a separable metric space.

So, a function which takes values in $\mathbb{R}^2$ is measurable if and only if its two coordinates functions (which take values in $\mathbb{R}$) are measurable.

The map $(x,y) \in \mathbb{R}^2 \to x - y \in \mathbb{R}$ is continuous, and so is also measurable.

Hence, the map $(x,y) \to f(x-y)$ is measurable (by composition). The map $(x,y) \to g(y)$ is measurable (composition of the second coordinate map with $g$). We can deduce that the map $(x,y) \to (f(x-y),g(y))$ is measurable.

Since the map $(x,y) \to xy$ is measurable (because continuous), by composition we have that $(x,y) \to f(x-y)g(y)$ is measurable.

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And what do you do if $\mathcal{L}(\mathbb{R})$ denotes Lebesgue measurable functions, not Borel measurable functions? – t.b. May 18 '12 at 11:49
The point being that in $(x,y) \mapsto f(x-y)$ you're composing a Lebesgue measurable function with a continuous function "the wrong way around": in general the pre-image of a Lebesgue measurable set under a continuous function need not be Lebesgue measurable. – t.b. May 18 '12 at 11:57
I am sorry that I do not make it clear.$f$ and $g$ are in the $\mathcal{L^1}(\mathbb{R})$,i.e. $f$ and $g$ are Lebesgue measurable functions and absolutely integrable functions in $\mathbb{R}$. – Juntao Huang May 18 '12 at 12:46
@t.b. every equivalence class of Lebesgue measurable functions contains a Borel measurable function, so apply the result to those. Using that function is Lebesgue measurable iff it is equal to a Borel measurable function a.e. finishes it off. – Chris Janjigian May 18 '12 at 15:30
I haven't written out all the details of that argument in a while. It may be relevant that both mappings take sets of measure zero to sets of measure zero. – Chris Janjigian May 18 '12 at 15:41

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