When is the integral of a function well defined with respect to a signed measure?

Question

When is the integral of a function well defined with respect to a signed measure?

Hello friends, I have a question. I appreciate who can guide me a little. Is the next:

Let $ (\Omega, \Sigma, \mu) $ be a measure space, where $ \mu $ is a signed measure. If $f $ is measurable in $ (\Omega, \Sigma) $ when does it make sense to talk about the integral of $f $ with respect to $\mu$? The natural thing is to think of defining it in the following way: $$\int_{\Omega} fd\mu: = \int_{\Omega}fd\mu_{+}-\int_{\Omega}fd\mu_ {-},$$ where $\mu = \mu_ {+}-\mu_{-},$ but something like $ \infty- \infty $ can happen, right? Are there other conditions to define the integral in signed measures?

Answer 1

First of all the definition is natural and correct as for the Jordan Decomposition Theorem we have that $\mu = \mu_ {+}-\mu_{-}$ is always possible and unique decomposition. Also keep in mind that $\mu_ {+}$ and $\mu_{-}$ so defined are mutually singular.

Something like $\infty - \infty$ is not defined such as in the case of integral for signed functions in a normal measure space.

Linearity still works but sadly a lot of the basic properties of integration are lost because of the absence of montonicity:

If $A$ is measurable and $\mu (A) < 0$ we can see that $$\int \chi_A \ d\mu= \mu(A) <0$$ Although $\chi_A \ge 0$

There's an excercise here in Royden's "Real Analysis" that asks this:

Provided an $f$ integrable over a set $X$ both for $\mu_+$ and $\mu_-$ then show that if $|f| \le M$ on $X$ then $$\left|\int_X f\ d\mu\right| \le |\mu|(X)$$ where $|\mu|(X)$ is the total variation as defined in the Jordan Decomposition.

Moreover if $|\mu|(X) < \infty$ there is a measurable function $f$ with $|f| \le 1$ for which $$\int_X f\ d\mu = |\mu|(X)$$

I think it may be interesting to resolve.