Show there exists an $f $ s.t. $ \int_{B} f d\mathbb{P} = \int_{B} 1-f d\mathbb{Q} $

Question

Given finite measures $\mathbb{P}$ and $\mathbb{Q}$ on the measureable space $(X,A)$, show that there exists a measurable function $f:X\to [0,1] $ such that $$ \int_{B} f d\mathbb{P} = \int_{B}( 1-f) d\mathbb{Q} $$ for all $B\in A $

** My thoughts **

So I don't know even where to start. But i have some ideas on what $f $ should look like...

1) $f $ could maybe exploit the finiteness of the measures to keep the image in $[0,1] $

2) There should be some symmetry in f

3) Use some results from theory of finite signed measures??

Answer 1

Let $\mathbb{P} = \nu_0 + \nu_1$ be the Lebesgue-decomposition of $\mathbb{P}$ with respect of $\mathbb{Q}$, i.ei. $\nu_0 \ll \mathbb{Q}$ and $\nu_1 \perp \mathbb{Q}$. Thus $$\nu_0(A \cap N^c) = \int_A g \, d \mathbb{Q}$$ and $\mathbb{Q}(N) =0$, $\nu_1(N^c)=0$.

1.) Since $\mathbb{Q}(N) =0$, we need so set $f = 0$ on $N$.

2.) On $N^c$ we set $f = 1/(1+g)$.

Then we have \begin{align*} \int f \, d \mathbb{P} = \int_{N^c} f \, d \nu_0 &= \int_{N^c} \frac{g}{g+1} \, d \mathbb{Q} \\ &= \int_{N^c} (1- f) \, d \mathbb{Q} \\ &= \int (1- f) \, d \mathbb{Q}. \end{align*}

Answer 2

Hint:

$\mathbb Q\ll\mathbb P+\mathbb Q$ so that $d\mathbb Q=g\;d(\mathbb P+\mathbb Q)$ for some nonnegative measurable $g$.

It can be shown that also $d\mathbb Q=f\;d(\mathbb P+\mathbb Q)$ where $f=\min(1,g)$.

Then $\int f\;d\mathbb P+\int f\;d\mathbb Q=\int_Bf\;d(\mathbb P+\mathbb Q)=\int_B\;d\mathbb Q$ for every measurable $B$.