# A question on Lebesgue measurable functions

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.

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.
• 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 – Pedro M. Mar 7 '13 at 23:11
• 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 $\mathbb{R}^2$. – anonymous Mar 7 '13 at 23:30
• Ok my bad I thought that every Borel set was Lebesgue mesurable (you can't be sure of anything...). But isn't it $$(f \circ g)^{-1} \mathcal O = f^{-1} \circ g^{-1} \mathcal O$$ – roger Mar 7 '13 at 23:32