4
$\begingroup$

In some books (like Rudin's or Cohn's) the integral of a function $f\geq 0$ is defined as $$\int f \text{ } d\mu =\sup\{\int \phi \text{ } d\mu;\phi\leq f \wedge\phi \text{ is a simple function} \}$$ moreover in others books (like Halmos' or Aliprantis') they defined the integral by $$\int f \text{ } d\mu=\sup\{\int \phi \text{ } d\mu;\phi\leq f \wedge\phi \text{ is a simple function } \wedge \mu(\{x;\phi(x) \neq 0\}<\infty ) \}$$ A remark in the book Infinite Dimensional Analysis of Aliprantis says that this definitions are equals in the $\sigma$-finite case, but may be different in other cases (page 411). This is quite surprising for me as my intuition says that this cannot happen because the integral has a naive definition as the measure of the set under the function. Therefore every definition of the integral should satisfy $\int f d\mu = \mu \times \lambda (\{(x,y);0\leq f(x) \leq y \})$ were $\mu \times \lambda$ is the measure of the product space (and I think that a function should be integrable if and only if $\{(x,y);0\leq f(x) \leq y \}$ is measurable).

So, what's going on with this definitions?, do we always have these naive properties?, Which definition is better or more usual?

$\endgroup$
  • 3
    $\begingroup$ Given how ubiquitous the $\sigma$-finite case is, I would say it doesn't particularly matter. Those insistent upon non $\sigma$-finite measure spaces will undoubtedly pick whichever definition suits their purposes for whatever result they're trying to prove. If that were me (it isn't) I would gravitate towards the second when it simplified my proofs and the first when the extra property wasn't used in order to make the theorems more general. $\endgroup$ – Adam Hughes Oct 3 '14 at 19:05
  • $\begingroup$ Your second definition doesn't actually match either Halmos's or Aliprantis and Border's definition. Both books consider approximating sequences $(\phi_n)$ of simple functions such that $\mu(\{x \colon \phi_n(x) \neq 0\}) < \infty$ for each $n$, and define $\int f\,d\mu$ as $\lim_{n\to\infty}\int \phi_n\,d\mu$. There are additional stipulations beyond those in your definition in both books: Halmos requires that $(\phi_n)$ is $L^1$-Cauchy and converges in measure to $f$, while Aliprantis and Border stipulate that the $\phi_n$ increase and converge pointwise almost everywhere to $f$. $\endgroup$ – epimorphic Feb 16 '15 at 17:31
  • 3
    $\begingroup$ An example where these definitions differ dramatically: For any nonempty set $X$, consider the measure $\mu\colon\mathcal P(X)\to[0,\infty]$ that assigns every nonempty subset of $X$ infinite measure, and any function $f \colon X \to [0,\infty)$ that is not identically zero everywhere. Then your first definition would give $\int f\,d\mu=\infty$, and your second $\int f\,d\mu=0$. Neither Halmos's nor Aliprantis–Border's regime can integrate $f$. $\endgroup$ – epimorphic Feb 17 '15 at 0:52
  • $\begingroup$ @epimorphic Do you know if the actual definition in Halmos can be reconciled with the usual modern definition (in the $\sigma$-finite case)? $\endgroup$ – Ryan Unger Dec 30 '16 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.