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

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?

• It makes sense when at least one of the integrals $\int f d\mu_+$ or $\int f d\mu_-$ is finite. Commented May 20, 2020 at 21:10

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.

• Hi friend, thanks for replying, don't you think I should ask that $\int_ {\Omega} fd \mu_+$ or $\int _ {\Omega} fd \mu_-$ be finite? Commented May 20, 2020 at 21:39
• When for example we define the Lebesgue integral for signed functions we require that at least one of $\int f^+ d\nu - \int f^- d\nu$ to be finite. So one can be finite while the other can be infinite and the sign of the overall integral is negative if $\int f^- \ d\nu$ is infinite or positive if $\int f^+ \ d\nu$ is. The same applies to a signed measure, one can be infinite and we will have the overall integral to be infinite with sign accordingly (for ex. we have that $\int_\Omega f \ d\mu_- = -\infty$ and ofc $\int_\Omega f \ d\mu_+$ is finite, then the overall integral will be $+\infty$) Commented May 21, 2020 at 6:42