# confusion about measurability requirements for Lebesgue integral

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 $$f:\Omega\to[-\infty,\infty]$$ in order for its Lebesgue integral $$\int_\Omega f\;d\mu$$ 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 $$f:\Omega_1\to\Omega_2$$ is called $$(\Sigma_1,\Sigma_2)$$-measurable if and only if $$f^{-1}(A)\in\Sigma_1$$ for all $$A\in\Sigma_2$$.

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 $$\{x:f(x) for every $$c\in\mathbb{R}$$, and also $$f^{-1}(\infty),f^{-1}(-\infty)\in\Sigma$$. This turns out to be equivalent to saying that $$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 $$f:\Omega\to[-\infty,\infty]$$ 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 $$f:\Omega\to[-\infty,\infty]$$ 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!

• take a look at the definition of the Bochner integral, what generalize and simplifies the Lebesgue integral – Masacroso Dec 10 '18 at 14:53

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 $$\int_\Omega f\,d\mu=\lim_{n\to\infty}\sum_{k=1}^{n^2}\tfrac kn\,\mu(f^{-1}((\tfrac kn,\tfrac{k+1}n])$$ (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 $$f\geq0$$. When $$f$$ takes both positive and negative values, you need both $$f^+$$ and $$f^-$$ to have finite integral, and then you can define $$\int_\Omega f\,d\mu=\int_\Omega f^+\,d\mu-\int_\Omega f^-\,d\mu.$$