Question

Show that lebesgue measure is translation invariant.

Answer 1

If ℓ(I)ℓ(I) denotes the length of the interval II, remember that:

m∗E=inf{∑j≥1ℓ(Ij) : (Ij)j≥1 are open intervals and E⊆⋃j≥1Ij}

m∗E=inf{∑j≥1ℓ(Ij) : (Ij)j≥1 are open intervals and E⊆⋃j≥1Ij}

If I call, for simplicity the above set AA, and BB the same with respect to E+yE+y, we can prove that A=BA=B.

Let x∈Ax∈A. So, we have x=∑j≥1ℓ(Ij)x=∑j≥1ℓ(Ij) for some open intervals IjIj such that E⊆⋃j≥1IjE⊆⋃j≥1Ij. Now, that gives:

E+y⊆(⋃j≥1Ij)+y=⋃j≥1(Ij+y)

E+y⊆(⋃j≥1Ij)+y=⋃j≥1(Ij+y)

Notice that the Ij+yIj+y are also open, and ℓ(Ij)=ℓ(Ij+y)ℓ(Ij)=ℓ(Ij+y). This way we have:

x=∑j≥1ℓ(Ij)=∑j≥1ℓ(Ij+y)∈B

x=∑j≥1ℓ(Ij)=∑j≥1ℓ(Ij+y)∈B

and hence A⊆BA⊆B. The same argument, starting with E+yE+y instead of EE and (−y)(−y) instead of yy gives B⊂AB⊂A. So we finally have A=BA=B, that is, infA=infBinfA=infB, which is exactly the equality m∗E=m∗(E+y)m∗E=m∗(E+y) that we wanted.