3
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Yes, it is a consequence of Radon-Nikodym Theorem. Note that $\tilde{\mathbb{P}}<<\mathbb{P}$. $\endgroup$
    – sinbadh
    Feb 29 '16 at 18:22
  • $\begingroup$ It looks quite elementary. So it seems not unthinkable that it actually "precedes" the Radon-Nikodym theorem. $\endgroup$
    – drhab
    Feb 29 '16 at 18:33
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.