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Let $ \mathcal{M}$ be the set of $\sigma$-algebra of Lebesgue measurable subsets of $ \mathbb{R} $. Let $f:\mathbb{R} \rightarrow (0,+\infty)$ be measurable and $g \in L^1(\mathbb{R}) $. For $E \in \mathcal{M}$, define \begin{equation*} \mu(E)= \int_E f(x)dx \text{ and } \lambda(E)= \int_E g(x)dx \end{equation*} whenever $ E \in \mathcal{M} $. Show $ \lambda << \mu $. Find $ d\lambda/d\mu $.

if we write f= $\frac{d\mu}{dx} $ and g= $\frac{d\lambda}{dx} $ then $\frac{g}{f} = \dfrac{\frac{d\lambda}{dx}}{\frac{d\mu}{dx}} = \frac{d\lambda}{d\mu}$. So the Radon-Nikodym derivative of $ \lambda $ with respect to $ \mu $ is $\frac{d\lambda}{d\mu}=\dfrac{g(x)}{f(x)}$.

I need some help to show that $ \lambda << \mu $. Given $E \in \mathcal{M}$ such that $\mu(E)=0$, we want to show that $\lambda(E)=0$

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If $\mu(E) = 0$, then $f > 0$ implies that $m(E) = 0$, from which it follows that $\lambda(E) = 0$. If you're not comfortable with the differential pushing, you can find the Radon-Nikodym derivative by finding a function $h$ such that $$\lambda(E) = \int_E h \;\mathrm{d}\mu.$$ By uniqueness, such a function is necessarily the Radon-Nikodym derivative. Finding such a function is straightforward: $$\lambda(E) = \int_E g\; \mathrm{d}x = \int_E \frac{g}{f} \cdot f \;\mathrm{d}x = \int_E \frac{g}{f} \;\mathrm{d}\mu.$$ The change of variables in the last step may be justified by approximating with simple functions.

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  • $\begingroup$ (What is differential pushing?) $\endgroup$ – user66081 May 16 '17 at 7:03
  • $\begingroup$ An ad-hoc term for how OP found the Radon-Nikodym derivative, which involved some shady "cancelling" of d$x$'s that may not be allowed at an introductory level. $\endgroup$ – user217285 May 16 '17 at 7:04

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