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What does it mean to say a measurable function is integrable with respect to a measure, such as $\mu$? I know the definition of integrability, but I'm still not sure what exactly what "with respect to a measure" means.

For example, with Riemann integration, for a function of one variable, $f(x)$ say, we know we are integrating with respect to $x$ and $x$ is clearly in the integrand. How should I interpret $d\mu$? Or $d\mu(f^{-1}(B))$?

This question is to help me understand Radon Nikodym derivatives.

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  • $\begingroup$ Well, the definition of integrability on a measurable space involves a measure. In a measurable space you can introduce different measures, and they can change the collection of integrable functions. $\endgroup$ – Pedro Tamaroff Mar 28 '16 at 1:36
  • $\begingroup$ Thank you. I edited my question because it wasn't complete. To expand, what exactly does $d\mu$ mean? Is it the same interpretation as $dx$? $\endgroup$ – Live Free or π Hard Mar 28 '16 at 1:41
  • $\begingroup$ in a first time, you should interpret $\int_{[a,b]} f d\mu$ as a generalization of $\int_a^b f(x) \mu'(x) dx$ where $\mu'(x) \ge 0$. $\endgroup$ – reuns Mar 28 '16 at 1:42
  • $\begingroup$ What is $\mu^{'}(x)$? I'm doing a Probability Theory course, so our measure is $\mathbb{P}$. Thanks for your patience. $\endgroup$ – Live Free or π Hard Mar 28 '16 at 1:47
  • $\begingroup$ the derivative of some function. at first, try with $\mu'(x) = 1$ ... the generalization is instead of considering some non-negative function $\mu'(x)$, defining a "measure" $\mu([a,b]) = \int_a^b \mu'(x) dx$, defining the axioms of the LHS : $\mu(A) \ge 0$ for any subset $A \subset \mathbb{R}$, $\mu(A \cup B) = \mu(A) + \mu(B)$ when the sets are disjoints, and after you defined what is a measure, forgetting about the RHS (and hence of the existence of that derivative) and generalizing even more with the Lebesgue integral $\endgroup$ – reuns Mar 28 '16 at 1:50
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A measurable function is integrable with respect to the measure $\mu$ if

$$\int_\Omega |f(x)| d\mu < \infty$$

If you change the measure, you change which functions are integrable.

For exemple if $\mu$ is the dirac in 0, then every function is integrable (assuming a function at value in $\Bbb R$)

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  • $\begingroup$ Thank you. I did edit my question above: is there an intuitive interpretation of $d\mu$? Or $d\mathbb{P}$? $\endgroup$ – Live Free or π Hard Mar 28 '16 at 1:49
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    $\begingroup$ @LiveFreeorπHard $d\mu$ is a merely symbol. It is there to remind you which measure you're integrating against. $\endgroup$ – Pedro Tamaroff Mar 28 '16 at 2:44
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Unlike the $dx$ in Riemann integration, $d\mu$ is simply a notation. It does not mean the infinitesimal change in $\mu$. It is written that way to resemble Riemann integration and avoid the necessity to introduce more symbols.

To define integrability, let's look at simple functions. An example would be $$ \phi = \sum_{k=0}^{n-1} c_k \chi_{S_k} $$ where $\chi$ represents a characteristic function. Notice that $\bigcup_{k=0}^{n-1} S_k = \Omega$ and $S_k$s are disjoint is another criterion for $\phi$ to be a simple function. The integral of $\phi$ is $$ \int_\Omega \phi\,d\mu = \sum_{k=0}^{n-1} c_k \mu(S_k) $$ Now, for a measurable function $f:\Omega \to [0,\infty]$, its integral is $$ \int_\Omega f\,d\mu = \sup\left\{ \int \phi\,d\mu \mid \phi \leq f, \phi \text{ simple} \right\} $$ That is, the supremum of the integrals of all simple functions less than $f$. Finally, if $f:\Omega \to \mathbb{R}$ is measurable, $f$ is integrable if $$ \int_\Omega |f|\,d\mu < \infty $$ and if $f$ is integrable, its integral is given by $$ \int_\Omega f\,d\mu = \int_\Omega \max(0,f)\,d\mu - \int_\Omega \max(0,-f)\,d\mu $$

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    $\begingroup$ In the Riemann integral the dx does certainly not mean any infinitesimal variation of anything, either! $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '16 at 1:49
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    $\begingroup$ @MarianoSuárez-Alvarez Maybe in nonstandard analysis? $\endgroup$ – Henricus V. Mar 28 '16 at 1:50
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    $\begingroup$ @LiveFreeorπHard Yes. There is a long chain of theorems that determine this, but it is all based on simple functions. $\endgroup$ – Henricus V. Mar 28 '16 at 1:55
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    $\begingroup$ @HenryW. : he already read that in his course... I don't see the point of writing the course again. $\endgroup$ – reuns Mar 28 '16 at 1:58
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    $\begingroup$ @LiveFreeorπHard It means you are integrating w.r.t. $\mu$. This will make more sense in Radon-Nikodym derivatives. $\endgroup$ – Henricus V. Mar 28 '16 at 2:05

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