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I have a doubt in the solution of an exercise.

Exercise: Suppose that $\mu$ and $\nu$ are $\sigma$-finite measures on $(X,M)$ such that $μ <<\nu$ and $\nu << \mu$. Prove that

$\dfrac{d\nu}{d\mu} \neq 0$ a.e. and $\dfrac{\mu}{\nu} = \dfrac{1}{\frac{\nu}{\mu}}$.

My answer

Let $Z = \displaystyle\{x \in X| \dfrac{d\nu}{d\mu}(x) = 0 \}$ we have that $\nu(Z) = \displaystyle \int_{Z} \dfrac{d\nu}{d\mu} d\mu = \int_{Z} 0 d\mu = 0$ then $\mu(Z) = 0$. Therefore $\dfrac{d\nu}{d\mu} \neq 0$ a.e.

Moreover, $\mu(X) =\displaystyle \int\dfrac{d\mu}{d\nu}d\nu = \int \dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu} d\mu$; Let $\{E_{i}\}_{i = 1}^{\infty}$ a countable family of measurable sets such that $X = \bigcup_{i = 1}^{\infty} E_i$ and $\mu(E_i) < \infty$ $\forall i$. (For each $i$ $\displaystyle\int_{E_i} \dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu}d\mu = \int_{E_i} 1 d\mu$ and by the uniqueness of Radon derivative $\dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu} = 1$ a.e. for each $i$)*.

My doubt:

Is The passage * between parentheses valid?

1 Answers 1

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I'm not sure if this will answer your question, but a sketch of the "standard" proof is as follows:

Wlog $\mu,\nu \ge 0$ and consider the equation $\int fd\nu=\int f\frac{d\nu}{d\mu}d\mu .$ It is true for characterisitc functions of measurable sets $f=\chi_E,$ and so by linearity true for all simple functions. Then, an application of MCT shows it is true for all $f\ge 0$. The result follows in general because $f=f^+-f^-$.

Now, set $f=\chi_E\cdot \frac{d\mu}{d\nu}.$ Then, $\int_Ed\nu =\nu(E)=\int_E \frac{d\nu}{d\mu}d\mu=\int _E\frac{d\mu}{d\nu}\frac{d\nu}{d\mu}d\mu.$ Uniqueness od the R-N derivative now implies that $\frac{d\mu}{d\nu}\frac{d\nu}{d\mu}=1$ a.e.-$\nu$. This also shows that $\frac{d\nu}{d\mu} \neq 0$ a.e.-$\mu$.

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    would not $\nu(E) = \int_{E} \dfrac{d\nu}{d\mu} d\mu = \int_{E}\dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu} d\nu$? Note that i did the same thing on each $E_{i}$ where i know that the constant function $1 \in L^{1}(\mu)$ and then the uniqueness of Radon-Nikodym theorem is valid. Then, what is wrong? – 2017-02-16
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    Well, if you already have the formula $\nu(E) = \int_{E} \dfrac{d\nu}{d\mu} d\mu = \int_{E}\dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu} d\nu$ the result is almost trivial. I thought you were asked to prove the formula, at least in the special case you are dealing with. – 2017-02-16
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    excuse me by the mistake, but i already have the formula. My doubt is my solution of exercise using the formula. – 2017-02-16
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    The $\sigma-$ finiteness is a condition for the R-N Theorem to hold, so once you have that, you can just use your formula $\mu(X) =\displaystyle \int\dfrac{d\mu}{d\nu}d\nu = \int \dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu} d\mu$ and the result is immediate. Or maybe I do not understand the question--- – 2017-02-17
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    I know, but the uniqueness is ensure for functions in $L^1(\mu)$, and the constante function 1 not necessarily is in $L^1(\mu)$ whereas $X$ as space. On the other hand taking $(E_i, E_{i} \cap M)$ as measurable space 1 is in $L^1(\mu)$ and $\dfrac{d\mu}{d\nu}\dfrac{d\nu}{d\mu}$ necessarily is in $L^1(\mu)$, then we apply the uniqueness in each $E_i$. This thought seems fine? – 2017-02-17