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$\mathbb{P}_0$ is Lebesgue measure, $\mathbb{P}_1$ is the probability measure given by $\mathbb{P}_1([a,b])=\int_a^b (4\omega-1) \, d\mathbb{P}_0(w)$ $\Omega$- is the interval $[0,1]$.

I need to find the Radon-Nikodym derivative $\frac{d\mathbb{P}_1}{d\mathbb{P}_0}$

What is $\mathbb{P}_0$ here?

I decided that it is $\int_a^b 1 \, dw $ - is it right?

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  • $\begingroup$ I decided that it is not. ${}$ $\endgroup$ – user251257 Mar 8 '16 at 17:29
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    $\begingroup$ Are you sure that $\Bbb P_1$ is a probability measure? For example, $\Bbb P_1([0,1/4]) = -1/8$ if we use your recipe for $\Bbb P_1$. $\endgroup$ – John Dawkins Mar 8 '16 at 17:39
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    $\begingroup$ Where you wrote $\int_a^b (4\omega-1) \, d\mathbb{P}_0(w)$, might you have meant either $\int_a^b (4\omega-1) \, d\mathbb{P}_0(\omega)$ or $\int_a^b (4w-1) \, d\mathbb{P}_0(w)$? $\qquad$ $\endgroup$ – Michael Hardy Mar 8 '16 at 18:23
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Where you wrote $\displaystyle\int_a^b (4\omega-1) \, d\mathbb{P}_0(w)$, I suspect you meant either $\displaystyle\int_a^b (4\omega-1) \, d\mathbb{P}_0(\omega)$ or $\displaystyle\int_a^b (4w-1) \, d\mathbb{P}_0(w)$.

If $\displaystyle\mathbb P_1([a,b]) = \int_{[a,b]} (4w-1)\, d\mathbb P_0(w)$, then $\dfrac{d\mathbb P_1}{d\mathbb P_0}(w) = 4w-1$. That is just an instance of the definition of the concept of a Radon–Nikodym derivative.

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