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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$\begingroup$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 $\Bbb R$)
$\endgroup$ 2 $\begingroup$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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