Theorem. If $\nu$ has a density $f$ with respect to $\mu$, then $$\int gd\nu=\int gfd\mu$$ holds for non-negative $g$. Moreover, $g$ (not necessarily non-negative) is integrable with respect to $\nu$ if and only if $gf$ is integrable with respect to $\mu$, in which case both $$\int gd\nu=\int gfd\mu\qquad\text{and}\qquad \int_A gd\nu=\int_A gfd\mu$$ both hold. For non-negative $g$ $$\int_A gd\nu=\int_A gfd\mu$$ always holds.

Reference. Theorem 16.11 Pg 214 Probability & Measure, Patrick Billingsley

I am trying to understand why the special case of $g$ non-negative was mentioned separately. For any general $g$, both claims are true whenever either side is defined. Is there any relaxed assumption or additional property for the special case when $g$ is non-negative?

Note. I do not want a proof of the theorem. Once it is shown to be true for non-negative $g$, we can simply write any general $g$ as $g=g^+-g^-$.


1 Answer 1


Normally, if $g$ is non-negative and measurable (rather than integrable) then one allows $\int g d\nu = +\infty$, where this means that the set $$ \{\int sd\nu: s \text{ a simple function}\}\subseteq \mathbb R $$ is unbounded. On the other hand, it is built into the definition of integrablity that $$ \int |g|d\nu = \int g^++g^- d\nu < \infty. $$ Thus the equality for non-negative $g$ includes the assertion that $\int g d\nu$ is infinite if and only if $\int gfd\mu$ is infinite.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .