Interpretation of $ \int_A f \ d\mu $ in case $\mu (A) = \infty$

Question

Let $\mu$ be a measure and $f$ measurable and real-valued. If $\mu (A) < \infty$ then $\frac{1}{\mu (A)} \int_A f \ d\mu $ (assuming the integral exists) is the $\mu$-average value of $f$.

Is there any other (except "area\volume") interpretation of $ \int_A f \ d\mu $ in case $\mu (A) = \infty$?

Answer 1

The $\mu$-average $\left(\frac{1}{\mu (A)} \int_A f \ d\mu \right)$ may exists even if $\mu(A)=\infty.$

Consider, as an example, $A=[1,\infty)$, $f(X)=\frac1x,$ and let $\mu$ the Lebesgue measure.

We may say that

$$\frac{1}{[1,\infty)} \int_A f \ d\mu=_{\text{def}}\lim_{b\to\infty}\frac1{\lambda([1,b))}\int_1^b\frac1x \ dx=\lim_{b\to\infty}\frac{\ln(b)}{b-1}=0. $$