Htykuut

So, it would appear I have forgotten the basic requirements of Lebesgue integration!

Let $(Omega,Sigma,mu)$ be an arbitrary $sigma$-finite measure space. I'm particularly interested in the real numbers equipped with the Borel measure, which we could denote $(mathbb{R},mathcal{B},beta)$. Similarly, let $(overline{mathbb{R}},overline{mathcal{B}},beta)$ denote the set of extended reals equipped with the Borel measure.

Question 1. What measurability conditions need to be true of a function $f:Omegato[-infty,infty]$ in order for its Lebesgue integral $int_Omega f;dmu$ to be well-defined?

Definition. Let $(Omega_1,Sigma_1)$ and $(Omega_2,Sigma_2)$ be measure spaces, i.e. nonempty sets equipped with $sigma$-algebras. A function $f:Omega_1toOmega_2$ is called $(Sigma_1,Sigma_2)$-measurable if and only if $f^{-1}(A)inSigma_1$ for all $AinSigma_2$.

I have been laboring for some time under the impression that $f$ needs to be $(Sigma,overline{Lambda})$-measurable, where $(overline{mathbb{R}},overline{Lambda},lambda)$ denotes the extended real numbers equipped with the Lebesgue measure. However, I've just been reading both Bogachev and Royden, and they seem to be requiring only that $f$ be "measurable" in the following sense: that ${x:f(x)<c}inSigma$ for every $cinmathbb{R}$, and also $f^{-1}(infty),f^{-1}(-infty)inSigma$. This turns out to be equivalent to saying that $f$ is $(Sigma,overline{mathcal{B}})$-measurable.

But there are functions $f:mathbb{R}to[-infty,infty]$ which are $(mathcal{B},overline{mathcal{B}})$-measurable but not $(mathcal{B},overline{Lambda})$-measurable. For instance, take the usual example of the homeomorphism $g:[0,2]to[0,1]$ built out of the Cantor "devil's staircase" function, and extend it to all of $mathbb{R}$ by letting $g(x)=g(0)$ whenever $x<0$ and $g(x)=g(2)$ whenever $x>2$. Then $g$ is continuous and hence $(mathcal{B},overline{mathcal{B}})$-measurable, but it is not $(mathcal{B},overline{Lambda})$-measurable. Is $g$ really integrable?

So, I guess Question 1 can be made more specific by breaking it up into the following:

Question 2. Does the definition of the Lebesgue integral of a function $f:Omegato[-infty,infty]$ require that $f$ be $(Sigma,overline{Lambda})$-measurable, or is $(Sigma,overline{mathcal{B}})$-measurability good enough?

I've also noticed that some of the Lemmas and Propositions involved with the Lebesgue integral require that $Sigma$ be complete, i.e it contains all subsets of sets of measure zero. The Borel $sigma$-algebra is the one I'm most interested in, but it is not complete. So:

Question 3. Does the definition of the Lebesgue integral of a function $f:Omegato[-infty,infty]$ require that $(Omega,Sigma)$ be complete? In particular, is $(mathbb{R},mathcal{B})$ allowed to be used for the domain of $f$?

According to my reading of the Royden and Bogachev texts, the answer to question 2 appears to be that we only need $(Sigma,overline{mathcal{B}})$-measurability, not $(Sigma,overline{Lambda})$-measurability. And the answer to question 3 appears to be that no, $(Omega,Sigma)$ need not be complete, and that $(mathbb{R},mathcal{B})$ is allowed to be used. But this seems suspicious, and I am not convinced I have understood the texts correctly.

Any help would be much appreciated. Thanks in advance!

To define the Lebesgue integral you need the domain to be a measure space. You don't need (and never use) a measure on the codomain. What you need of the codomain is that it is a topological space, so that you can take limits (my mantra: an integral is a limit of sums of "value of the function times size of the region").

So, initially what you want for the integral of $f$ to make sense is that the preimage of open sets is measurable. This can be seen if you write for instance
$$
int_Omega f,dmu=lim_{ntoinfty}sum_{k=1}^{n^2}tfrac kn,mu(f^{-1}((tfrac kn,tfrac{k+1}n])
$$
(which is a great way to see why one needs measurability to define the integral).

Because the codomain has a topology, you can consider the Borel $sigma$-algebra on it, and as you mention measurability in the above sense is equivalent to $(Sigma,mathcal B)$ in your notation.

In summary: you need a $sigma$-algebra in the codomain. You don't need a topology in the domain.

On a separate note, you mention integration of non-positive value functions. That's a different beast. All the above applies when $fgeq0$. When $f$ takes both positive and negative values, you need both $f^+$ and $f^-$ to have finite integral, and then you can define
$$
int_Omega f,dmu=int_Omega f^+,dmu-int_Omega f^-,dmu.
$$