# How do i prove that completion of two lebesgue measure is another lebesgue measure?

I'm trying to prove this very precisely.

Let $\Sigma_n$ be the sigma-algebra of n-dimensional Lebesgue measurable sets.

Let $F:\mathbb{R}^n\times\mathbb{R}^m \rightarrow \mathbb{R}^{n+m}$ be a function such that $F(x,y)=(x_1,\cdots,x_n,y_1,\cdots,y_m)$.

(This function is meaningful since $\mathbb{R}^n\times\mathbb{R}^m \neq \mathbb{R}^{n+m}$ precisely.)

I'm trying to prove the following statement:

Define $H=\{F(A):A\in \Sigma_n \otimes \Sigma_m\}$

Then, $\mathscr{B}_{\mathbb{R}^{n+m}}\subset H \subset \Sigma_{n+m}$

(Of course, $\mathscr{B}$ denotes Borel algebra)

Is it possible to prove this not invoking Borel hierarchy?

Let $S\subset P(\mathbb{R}^n\times\mathbb{R}^m)$

It's easy to prove $\sigma(\{F(A) : A\in S\})\subset \{F(A) : A\in \sigma(S)\}$ but i think it's impossible to prove the converse not invoking Borel hierarchy.

This process would be clear, but it is not if one does not know Borel hierarchy (just like me). I don't understand why people say this so trivially.

Anyway, is there a easy way to prove this?

You don't need the Borel hierarchy here, all you need is that $F$ is bijective. Thus the induced maps between the power sets are also bijective, and commute with complements, unions, and intersections, and hence for every $\sigma$-algebra $\mathcal{A}$ on $\mathbb{R}^n\times \mathbb{R}^m$, the family

$$F_\ast(\mathcal{A}) = \left\{ F(A) : A\in\mathcal{A}\right\}$$

is a $\sigma$-algebra on $\mathbb{R}^{n+m}$, and for every $\sigma$-algebra $\mathcal{B}$ on $\mathbb{R}^{n+m}$, the family

$$F^\ast(\mathcal{B}) = \left\{ F^{-1}(B) : B\in\mathcal{B}\right\}$$

is a $\sigma$-algebra on $\mathbb{R}^n\times\mathbb{R}^m$, and $F^\ast$ and $F_\ast$ are inverses of each other, $F^\ast(F_\ast(\mathcal{A})) = \mathcal{A}$, and $F_\ast(F^\ast(\mathcal{B})) = \mathcal{B}$.

So $F_\ast(\sigma(S))$ is a $\sigma$-algebra containing $F(A)$ for every $A\in S$, hence $\sigma(F_\ast(S)) \subset F_\ast(\sigma(S))$ ($F_\ast$ and $F^\ast$ are defined as above for all families of subsets). Conversely, $F^\ast(\sigma(F_\ast(S)))$ is a $\sigma$-algebra containing $F^\ast(F_\ast(S)) = S$, whence $\sigma(S) \subset F^\ast(\sigma(F_\ast(S)))$ and consequently $F_\ast(\sigma(S)) \subset F_\ast(F^\ast(\sigma(F_\ast(S)))) = \sigma(F_\ast(S))$, altogether,

$$F_\ast(\sigma(S)) = \sigma(F_\ast(S)).$$

• Wonderful!! Thank you! – John. p Feb 20 '14 at 19:16