For some Lebesgue Measurable/Integrable function $g$ such that $f \in$ [a,b], for finite $a,b$, then show that $\exists$ $k \in [a,b]$, such that $\mathbb E^{\mathbb P}[f|g|] = k \mathbb E^{\mathbb P}[|g|]$, where $\mathbb P$ is a probability measure on a measurable space.
Firstly, does $\mathbb E^{\mathbb P}(g)$ simply mean the expectation of a function with respect to the measure $P$?
If so then I am thinking to write $\mathbb E^{\mathbb P}(g)$ as an integral but then I come a bit unstuck on how to proceed, perhaps using Holder's Inequality/Tonelli's Theorem? Any advice is welcome, thank you. $$ \int_a^b f(x)|g(x)|d\mathbb P = ... ? $$
For fixed $f$ and $g$ and $g\neq 0$ consider the map $m:k\mapsto \mathbb{E}^{\mathbb{P}}[f|g|]-k\mathbb{E}^{\mathbb{P}}[|g|]$. $m$ is continuous being just linear in $k$. So, if we assume that $\forall k\in[a,b]$ the inequality $\mathbb{E}^{\mathbb{P}}[f|g|]\neq k\mathbb{E}^{\mathbb{P}}[|g|]$ holds, we obtain that $\forall k\in[a,b]$ either $\mathbb{E}^{\mathbb{P}}[f|g|]>k\mathbb{E}^{\mathbb{P}}[|g|]$ (case (a) ) or $\mathbb{E}^{\mathbb{P}}[f|g|]< k\mathbb{E}^{\mathbb{P}}[|g|]$ (case (b) ) since otherwise we would find a $0$ of $m$ which would solve the equality we want to show.
In case (a) we choose $k=\sup_{x} f(x)$ and in case (b) we choose $k=\inf_{x} f(x)$, which leads in both cases to the contradiction $k<k$. Hence there is a constant $k\in[a,b]$ such that $\mathbb{E}^{\mathbb{P}}[f|g|]= k\mathbb{E}^{\mathbb{P}}[|g|]$.