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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 $\{x:f(x)<c\}\in\Sigma$ 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!

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  • $\begingroup$ take a look at the definition of the Bochner integral, what generalize and simplifies the Lebesgue integral $\endgroup$ – Masacroso Dec 10 '18 at 14:53
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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 $$ \int_\Omega f\,d\mu=\lim_{n\to\infty}\sum_{k=1}^{n^2}\tfrac kn\,\mu(f^{-1}((\tfrac kn,\tfrac{k+1}n]) $$ (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 $$ \int_\Omega f\,d\mu=\int_\Omega f^+\,d\mu-\int_\Omega f^-\,d\mu. $$

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  • $\begingroup$ Thank you! And just one more thing: Does the domain measure space need to be complete? $\endgroup$ – Ben W Dec 10 '18 at 14:35
  • $\begingroup$ Just to define the Lebesgue integral, no. $\endgroup$ – Martin Argerami Dec 10 '18 at 15:05

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