Proving: If A is \lambda ^* -measurable, Then x+A is \lambda ^* -measurable

[CornMuffin]

Homework Statement

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.

Homework Equations

Notation:
\lambda ^* is the lebesgue outer measure

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)
<br /> =\lambda ^* (T\cap (A+x)) + \lambda ^* (T\cap (A^c +x)) <br />
<br /> = \lambda ^* (T\cap (A+x)) + \lambda ^* (T\cap (A+x)^c)<br />
<br /> = \lambda ^* (T)

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

 

[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.