# If E is measurable then so it's each of it's translates

1. Dec 2, 2007

### SiddharthM

I'm working with caratheodory's definition of measurability of sets as given in Royden. I'm trying to prove that given any measurable set E we have that E+y={x+y|where x is contained in E} is also measurable. I'm looking for a hint not the entire solution please.

I think I've actually done this problem before in my real analysis class and i remember there being a 'special set' (the silver bullet) that relates E with E+y through operations allowed in sigma algebras...i could be wrong

thanks for any help, again no solutions just hints please.

2. Dec 2, 2007

### SiddharthM

actually I think it's simpler than I thought. We use the fact that lebesgue outer measure is translation invariant, that is m(A)=m(A+y) for any set A and any real number y. We also use two identities

(E+y)^c=E^c+y (that is the compliment of the translate is the translate of the compliment) and
[(A-y) intersect (E)]+y=A intersect (E+y)

I will not prove these things as it is simple to do so.

Let A be any subset of reals, put B=A-y, then since E is measurable
mB=m(B intersect E)+m(B intersect E^c)
since m is translation invariant
mB=m([B intersect E]+y)+m([B intersect E^c]+y)
Now using the first identity above we see that
mB=m(A intersect (E+y))+m(B intersect (E^c+y))
using the 2nd identity we have
mB=m(A intersect (E+y))+m(B intersect (E+y)^c)
and mB=mA because of trans invariance. QED