Product of non-negative measurable and integrable functions

Question

Here is the problem that I have

Let $f \in \Sigma^+$ and $h \in \Sigma$ and the measures $\mu$ and $\nu$ are such that $\nu(E)=\mu(\mathbb{1}_E f)$ for any $E \in \Sigma$. Then $h \in \mathcal{L}^1(S,\Sigma, \nu)$ iff $hf \in \mathcal{L}^1(S,\Sigma,\mu)$, in which case one has $\nu(h) = \mu(hf)$.

A few clarifications that I think are very specific to the notation my professor uses: $\nu(h)$ means the integral of $h$ w.r.t. a measure $\nu$. When referring to a function, $\Sigma^+$ is the set of measurable non-negative functions, $\Sigma$ is the set of measurable functions. When referring to a set, the later is simply a sigma-algebra. $\mathbb{1}_E$ is the characteristic (identificator) function w.r.t. set $E$

The way I see it, I have to prove that the product $hf$ is integrable w.r.t. measure $\mu()$ if and only if $\nu(h)=\mu(hf)$ .

I'm trying to go through the definition of integrable but i don't get it. I denote the integral of $h$ w.r.t. measure $\nu()$ and by definition I know that a function is integrable if for

$$h = h^+ - h^-$$

where $h^+=max(0,h)$, $h^-=max(0,-h)$

I have that $$\nu(h) = \nu(h^+) - \nu(h^-)$$ and $\nu(h^-)<\infty$, $\nu(h^+)\infty$ where by $\nu(h)$ I denote the integral w.r.t. measure $\nu()$, etc.

I have difficulty however seeing how one transitions from one measure to another.

Answer 1

You're given a measure space $(S,\Sigma,\mu)$ and a non-negative measurable function $f\colon S \to [0,\infty]$.

Consider $\nu$ such that $\displaystyle\nu(E)=\int_E f\, d\mu \quad\forall E\in\Sigma$. It's a measure on $(S,\Sigma)$ and $$ \int_E h\,d\nu = \int_E hf\,d\mu $$ for every measurable function $h$.

This also tells you that $h\in\mathcal{L}^1(S,\Sigma,\nu)$ iff $hf\in\mathcal{L}^1(S,\Sigma,\mu)$.

To prove the equivalence of the integrals use the "standard machine" (prove it for simple functions, then non-negative ones via Monotone Convergence Theorem, lastly for real valued functions decomposing them into positive and negative part). Pay attention to put the correct sign when decomposing $h$: $$ h=h^+\color{red}{-}h^-. $$

Answer 2

Let $h(x)=1$ and $f(x)=\frac{1}{x}$ for $0<x\le 1$, $f(0)=0$. Both are measurable, $f$ is nonnegative, $h$ is integrable on $[0,1]$. But the integral $\displaystyle\int_0^1 (fg) \, dx$ diverges.