Derivative of a probability measure
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $f: (\Omega, \mathcal{F}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ a measurable non-negative function s.t. $\int_\Omega f d\mathbb{P}=1$. Define for $A\in\mathcal{F} :$ $$\tilde{\mathbb{P}}(A)=\int_A f d\mathbb{P}$$ Show, using 'standard machinery', that for $g\in\mathcal{L_1(\Omega)}$ that: $$\int_\Omega g d\tilde{\mathbb{P}}=\int_\Omega gf d\mathbb{P}$$
This question seems trivial to me since it follows from working out the differential:
$$d\tilde{\mathbb{P}}=fd\mathbb{P}$$
I however assume that is too easy, so I started reading something about the Radon-Nikodym-theorem. It seems like this answer is a direct consequence of that theorem, given that $\tilde{\mathbb{P}}$ is a probabilty measure. Am I correct?
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$\begingroup$By using Radon-Nikodym, this follows directly, but you can do it using 'standard' techniques (namely following the construction of the integral). We want to prove $\def\P{\mathbb P}$ $$ \tag 1 \int_\Omega g \,d\tilde\P = \int_\Omega fg\,d\P$$ for all $g\in L^1(\tilde\P)$. (1) holds by definition of $\tilde P$ for characteristic functions, as for $g = \chi_A$ we have $$ \int_\Omega g \,d\tilde\P = \tilde P(A) = \int_A f \,d\P = \int_\Omega fg\,d\P $$ By linearity, (1) holds for simple functions, i. e. linear combinations of characteristic functions. If now $g$ is non-negative measurable, choose simple $g_n$ such that $g_n \nearrow g$. Then, we have $fg_n \nearrow fg$, hence $$ \int_\Omega g \,d\tilde\P = \lim\int_\Omega g_n \,d\tilde\P = \lim\int_\Omega fg_n\,d\P ) \int_\Omega fg_n\,d\P $$ For general $g$, now write $g = g^+ - g^-$ and use linearity again.
$\endgroup$ $\begingroup$By the standard machinery, they mean prove it first for indicator functions (the identity follows in this case by definition), then for simple functions by linearity, then for non-negative functions by the monotone convergence theorem, and finally for general $L^1$ functions by writing $g=g^+-g^{-}$ and using linearity again.
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