# Falling back on the Lebesgue measure from the abstract theory?

1. Sep 26, 2007

### quasar987

I am studying the abstract theory of measure and I was wondering how the Lebesgue case for real functions of a real variable arises. But I did not find it.

In the original theory of Lebesgue, a function f:E-->R was said to be measurable if for every real constant b, the preimage of $]-\infty, b]$ by f was measurable. Let the collection of all measurable sets be denoted $$\mathcal{L}_{\mathbb{R}}$$ (the Lebesgue sigma-algebra). The pair $$(\mathbb{R},\mathcal{L}_{\mathbb{R}})$$ is a measurable space.

In the abstract theory, we consider a function f btw two measurable spaces:

$$f:(X_1,\mathcal{T}_1)\rightarrow (X_2,\mathcal{T}_2)$$

and say that it is measurable if, given a family of subsets of X_2 $G_2$ that generates the sigma-algebra $\mathcal{T}_2$ (i.e. $\mathcal{T}(G_2)=\mathcal{T}_2$), we have

$$f^{-1}(G_2)\subset \mathcal{T}_1$$

If I set X_1 = E a subset of R and X_2 = R, I am trying to find which sigma-algebras $\mathcal{T}_1, \mathcal{T}_2$ will make Lebesgue's definition and the abstract definition coincide. Obviously, we must take $\mathcal{T}_2=\mathcal{T}(\{\{[-\infty,b]\}:b\in\mathbb{R}\})=\mathcal{B}_{\mathbb{R}}$ (the borelian sigma-algebra). Now, if I were allowed to take $\mathcal{T}_1=\mathcal{L}_{\mathbb{R}}$ I would have succeeded, but $\mathcal{L}_{\mathbb{R}}$ is not a sigma-algebra on E. The next best thing is the trace of $\mathcal{L}_{\mathbb{R}}$ on E (aka maybe the induced sigma-algebra on E by $\mathcal{L}_{\mathbb{R}}$) defined by $\mathcal{L}_E=\{EM:M\in \mathcal{L}_{\mathbb{R}}\}$.

But this does not seem to work. I need to check now that we have the equivalence (for all b in R, the preimage of $]-\infty, b]$ by f is in $\mathcal{L}_{\mathbb{R}}$) <==>(for all b in R the preimage of $]-\infty, b]$ by f is in $\mathcal{L}_E$).

The ==> part is trivial but I don'T know how to prive the <== part, and actually, I would think that it is not necessarily true, for instance if E is not a part of $\mathcal{L}_{\mathbb{R}}$.

Any ideas???

2. Sep 26, 2007

### quasar987

I just noticed that the trace of a sigma-algebra on a set E is only defined if E is itself an element of the sigma-algebra, such that the problem I rise concerning part <== does not exists.