Vtyjkyuk

Suppose $EsubsetmathbbR$ is measurable and $E=E+frac1n$ for every natural number $ngeq1$. Show that either $m(E)=0$ or $m(mathbbRsetminus E)=0$.

Let $m(E)>0,ain
%TCIMACROU211d %
%BeginExpansion
mathbbR
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$ and $f(x)=m(Ecap lbrack a,x])$ for $aleq x<infty $. If $a<x<y$ then $$
f(y+frac1n)-f(x+frac1n)=m(Ecap (x+frac1n,y+frac1n])$$

$$=m(E-frac1ncap (x,y])=m(Ecap (x,y])$$ and $$f(y-frac1n)-f(x-%
frac1n)=m(Ecap (x-frac1n,y-frac1n])$$

$=m(E+frac1ncap (x,y])=m(Ecap (x,y])$. It follows that $$f(y+frac%
1n)-f(x+frac1n)=f(y-frac1n)-f(x-frac1n)$$ . Note that $$%
leftvert f(y)-f(x)rightvert leq leftvert y-xrightvert $$ so $f$ is
absolutely continuous. Hence it is differentiable almost everywhere and
using above equation we conclude that its derivative is a constant $c$ a.e..
Since $$fracm(x-delta ,x+delta )2delta =fracf(x+delta )-f(x-delta
)2delta $$ and almost all points of $E$ have metric density $1$ we see
that $c=1$ Thus $f(y)-f(x)=int_x^yf^prime (t)dt=y-x$. This
gives $f(y)=f(a)+y-a$ $forall x>a$. Thus $f(y)-f(x)=y-x$ or $m(Ecap
(x,y])=m((x,y])$ for $a<x<y$. This gives $m(E^ccap (x,y])=0$ for $a<x<y$
which clearly implies that $m(E^c)=0$.