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Prove that if $f$ is Lebesgue measurable on $\mathbb{R^n}$, then $f(x-y)$ is Lebesgue measurable on $\mathbb{R^n}\times \mathbb{R^n}$.

My Work:

Define $S:\mathbb{R^n}\times \mathbb{R^n}\rightarrow \mathbb{R^n}$ by $S(x,y)=x-y$. Since $S$ is continuous it is Borel measurable. Hence, if $f$ is Borel measurable on $\mathbb{R^n}$ then $f\circ S=f(x-y)$ is Borel measurable on $\mathbb{R^n}\times \mathbb{R^n}$. But how can I prove that Lebesgue measurability of $f$ implies Lebesgue measurability of $f(x-y)$? Any help would be appreciated.

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    $\begingroup$ @ajotatxe No. Composition of 2 Lebesgue measurable functions is not always Lebesgue measurable. $\endgroup$ – Extremal Feb 5 '15 at 23:03
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    $\begingroup$ Hint: Consider the linear transformation $T(x,y)=(x-y,y)$. $\endgroup$ – PhoemueX Feb 5 '15 at 23:20
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    $\begingroup$ @Mathi That depends on what you mean by Lebesgue measurable function. If you prove that your $S$ is Lebesgue measurable in the sense that preimages of Lebesgue measurables are Lebesgue measurables then the composition is Lebesgue measurable. The dictum of composition of Lebesgue measurable functions is not necessarily Lebesgue measurable is true when one defines Lebesgue measurable as preimage of Borel is Lebesgue. $\endgroup$ – Pp.. Feb 5 '15 at 23:29
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    $\begingroup$ @Pp..I am using the definition as "preimage of Borel is Lebesgue", I have never seen the other as the definition and two inconsistent definitions cannot exist for a term $\endgroup$ – Extremal Feb 5 '15 at 23:32
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    $\begingroup$ If you already know the change of variables theorem (at least for linear transformations) then you know that $g\circ T$ is Lebesgue measurable if $g$ is. Apply this to $g(x,y)=f(x)$. $\endgroup$ – PhoemueX Feb 6 '15 at 6:32
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Let $h:\mathbb{R}^{2d}\to\mathbb{R}^d$ be such that $h(x,y)=x-y$ and $f:\mathbb{R}^d\to\mathbb{R}$ a Lebesgue measurable function. We want to show that $K:\mathbb{R}^{2d}\to\mathbb{R}$ with $ K(x,y):=f\circ h(x,y)=f(x-y) $ is Lebesgue measurable.

Now let $E$ be a Borel set of $\mathbb{R}$. It suffices to show that $K^{-1}(E)=h^{-1}\circ f^{-1}(E)$ is Lebesgue measurable in $\mathbb{R}^{2d}$. Observe that $f^{-1}(E)$ is Lebesgue measurable. It thus suffices to show that $h^{-1}(F)$ is Lebesgue measurable whenever $F\subset\mathbb{R}^d$ is Lebesgue measurable.

This is true when $F$ is Borel measurable since $h$ is continuous. On the other hand, one can check that the preimage of a set of zero outer measure is again of zero outer measure, so null sets pull back to null sets. Since Lebesgue measurable sets differ by a null set from a Borel measurable set, the claim for Lebesgue measurable sets then follows from the corresponding claim for Borel sets.


[Added for the "one can check" part] Note that $h(x,y)=\psi\circ T(x,y)$ with $$ \psi(x,y)=x,\quad T(x,y)=(x-y,x). $$ It is also known that $m(T(E))=|\det T|m(E)$. To deal with $\psi$, a warming-up question would be to show that if $E\subset\mathbb{R}$ has zero outer measure, then the set $E\times[0,1]\subset \mathbb{R}^2$ has zero outer measure; using the countable subadditivity of outer measure, this implies that $E\times\mathbb{R}$ also has zero outer measure.

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    $\begingroup$ I like to put "one can check" for the tricky part of a proof. $\endgroup$ – GEdgar May 7 '16 at 17:20

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