# Measurable sets

1. Oct 1, 2011

### CornMuffin

1. The problem statement, all variables and given/known data
Prove:
If $A$ is $\lambda ^*$-measurable and $x\in \mathbb{R} ^n$
then $x+A$ is $\lambda ^*$-measurable.

My attempt at the proof is below, but i feel like it is not a correct proof.

2. Relevant equations
Notation:
$\lambda ^*$ is the lebesgue outer measure

3. The attempt at a solution
Proof:
let $A$ be a $\lambda ^*$-measurable set, and let $x\in \mathbb{R} ^n$ and let $S$ be the entire space.
Then $\forall T\subset S$, $\lambda ^* (T) = \lambda ^* (T\cap A)+\lambda ^* (T\cap A^c )$
Lesbesgue outer measure is translation invariant,
so, $\lambda ^* (T-x) = \lambda ^* ((T-x)\cap A) + \lambda ^* ((T-x)\cap A^c)$
$=\lambda ^* (T\cap (A+x)) + \lambda ^* (T\cap (A^c +x))$
$= \lambda ^* (T\cap (A+x)) + \lambda ^* (T\cap (A+x)^c)$
$= \lambda ^* (T)$

so, $x+A$ is $\lambda ^*$-measurable

2. Oct 1, 2011

### Kreizhn

Using Caratheodory's Criterion is the correct way of doing this, but all you've concluded is what you knew to begin with: the outer measure is translation invariant.

You know that A is measurable right? And to show that x+A is measurable you must show that
$\lambda^*(T \cap (x+A)) + \lambda^*(T \cap (x+A)^c ) = \lambda^*(T)$
for every set $T \subseteq \mathbb R^n$. So try starting with $\lambda^*(T \cap (x+A)) + \lambda^*(T \cap (x+A)^c )$ and see if you can get the $\lambda^*(T)$

Edit: Your proof has the right ideas, but not the right set-up.