2
$\begingroup$

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a Lebesgue measurable function, and define a function $\phi:\mathbb{R}^2\rightarrow\mathbb{R}$ by $\phi(x,y):=f(x-y)$. I want to prove that $\phi$ is Lebesgue measurable, i.e., given an open set $\mathcal{O}\subseteq\mathbb{R}$, $\phi^{-1}\left[\mathcal{O}\right]\subseteq\mathbb{R}^2$ is Lebesgue measurable. I first attempted to take advantage of the fact that the function $g:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $g(x,y):=x-y$ is continuous, as here $\phi=f\circ g$. However, this approach failed, as the composition is in the wrong order; the continuous function $g$ is on the inside, so $\phi^{-1}\left[\mathcal{O}\right]=\left(f\circ g\right)^{-1}\left[\mathcal{O}\right]=g^{-1}\left[f^{-1}\left[\mathcal{O}\right]\right]$, which is not helpful, as the preimage of a Lebesgue measurable set under a continuous function need not be Lebesgue measurable. My next attempt involved taking $\mathcal{O}$ to be a basic open interval of the form $\left(a,b\right)\subseteq\mathbb{R}$ and considering $\phi^{-1}\left[\mathcal{O}\right]$ directly, but I have as yet been unable to determine the nature of $\phi^{-1}\left[\mathcal{O}\right]$. I’m not sure whether I’m on the right track or not here, so any suggestions would be appreciated.

$\endgroup$
2
$\begingroup$

Let $g(x,y)=x-y$ and A be a Lebesgue measurable set, $g^{-1}(A)$ is a rectangle for the axes $\{(1,1),(-1,1) \}$ since it can be written as $$ g^{-1}(A) = \{ (a/2,-a/2): a \in A \} \times \{ (x,y): x = y \}. $$ Thus, $g^{-1} \circ f^{-1} \mathcal O $ is Lebesgue measurable if $f^{-1} \mathcal O$ is, which is true if $\mathcal O$ is open.

$\endgroup$
0
$\begingroup$

Continuous implies Lebesgue-mesurable and the composition of two mesurable functions is mesurable.

$\endgroup$
  • $\begingroup$ The problem is the definition of Lebesgue-measurable function: a function $f$ is Lebesgue-measurable if the pre-image of every Borel set is Lebesgue-measurable (but not necessarily Borel): en.wikipedia.org/wiki/Measurable_function $\endgroup$ – Pedro M. Mar 7 '13 at 23:11
  • $\begingroup$ I have another idea: show that the preimage of a Lebesgue measurable set under this particular continuous function will be Lebesgue measurable, probably using the $\varepsilon$-$\mathcal{O}$ definition for $\mathbb{R}^2$. $\endgroup$ – anonymous Mar 7 '13 at 23:30
  • $\begingroup$ Ok my bad I thought that every Borel set was Lebesgue mesurable (you can't be sure of anything...). But isn't it $$ (f \circ g)^{-1} \mathcal O = f^{-1} \circ g^{-1} \mathcal O $$ $\endgroup$ – roger Mar 7 '13 at 23:32
  • $\begingroup$ Yeah, every Borel set is Lebesgue measurable, but not every Lebesgue measurable set is Borel. $\endgroup$ – anonymous Mar 7 '13 at 23:33
  • $\begingroup$ No, taking the inverse image of a composite map reverses the order of the composition. $\endgroup$ – anonymous Mar 7 '13 at 23:35

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.