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Let $m$ be Lebesgue measure on $\mathbb R_+=(0,\infty)$ and $\mathcal A = \sigma\left(( \frac 1{n+1} , \frac 1n ]:n=1,2,...\right)$. Define a new measure $\lambda$ on $\mathcal A$, for each $E \in \mathcal A$, by $\lambda(E)= \int_E fdm $, where $f(x)=2x^2$ . Find the Radon-Nikodym derivative $\frac{d\lambda}{dm}$.

It is clear that $\lambda $ is absolutely continuous with respect to $m$ by definition, and Lebesgue measure is $\sigma$-finite in $(\mathbb R_+ , \mathfrak M_+)$, where $\mathfrak M_+$ is the collection of all Lebesgue measurable subsets of $\mathbb R_+$, so $m$ is $\sigma$-finite in $(\mathbb R_+, \mathcal A)$ since every element of $\mathcal A$ is also Lebesgue measurable.

However, to apply Radon-Nikodym theorem, $\lambda$ must be a finite measure, but $\lambda(E)= \int_E fdm = \infty$ if $E=\mathbb R_+ - ( \frac 12,1]$, so we cannot apply that theorem. Is there any other approach to this problem?

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  • $\begingroup$ I don't believe you say what $f$ is. (For instance if it's an integrable function, then $\lambda$ is a finite measure.) $\endgroup$ – spaceisdarkgreen Feb 24 '17 at 2:17
  • $\begingroup$ Oh, $f(x)=2x^2$. I'll add it. $\endgroup$ – bellcircle Feb 24 '17 at 2:32
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There is no need for the Radon-Nikodým theorem.

By the very definition of the Radon-Nikodým derivative, we are looking for a function $g: (0,\infty) \to [0,\infty)$ which is measurable with respect to $\mathcal{A}$ and satisfies $$\lambda(E) = \int_E g \, dm, \qquad E \in \mathcal{A}. \tag{1}$$ Note that $g(x) := f(x) :=2x^2$ is not measurable with respect to $\mathcal{A}$ and therefore we cannot simply choose $g=f$.

If we define

$$E_n := \begin{cases} \bigg( \frac{1}{n+1}, \frac{1}{n} \bigg], & n \in \mathbb{N}, \\ (1,\infty), & n = 0 \end{cases}$$

then $\mathcal{A} = \sigma(E_n; n \in \mathbb{N}_0)$. Since the intervals $E_n$, $n \in \mathbb{N}_0$, are disjoint and cover $(0,\infty)$, equation $(1)$ is equivalent to

$$\lambda (E_n) = \int_{E_n} g \, dm \qquad \text{for all} \, \, n \in \mathbb{N}_0. \tag{2}$$

Moreover, any $\mathcal{A}$-measurable function $g$ is of the form

$$g(x) = \sum_{n \in \mathbb{N}_0} c_n 1_{E_n}(x) \tag{3}$$

for constants $c_n \in \mathbb{R}$. The only thing which we have to do is to choose the constants $c_n \geq 0$ such that $(2)$ holds. To this end, we plug our candidate $(3)$ into $(2)$ and find

$$\lambda(E_n) \stackrel{!}{=} \int_{E_n} g \, dm \stackrel{(3)}{=} c_n \int_{E_n} \, dm= c_n m(E_n) = c_n \left( \frac{1}{n}-\frac{1}{n+1} \right)$$

which implies

$$c_n = \lambda(E_n) n (n+1)$$

for all $n \in \mathbb{N}$. $\lambda(E_n)$ can be calculated explicitly using the very definition of $\lambda$; I leaves this to you. For $n=0$ we get $$\lambda(E_0) = \infty \stackrel{!}{=} c_0 m(E_0) = c_0 \infty,$$ i.e. we can choose $c_0 := 1$. Hence,

$$g(x) = 1_{E_0}(x)+ \sum_{n \geq 1} \lambda(E_n) n (n+1) 1_{E_n}(x)$$

is a non-negative $\mathcal{A}$-measurable function which satisfies $(2)$ (hence, $(1)$), i.e.

$$g = \frac{d\lambda}{dm}.$$

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  • $\begingroup$ I think $(1, \infty)$ should be added because the $\sigma$-algebra also contains complements. $\endgroup$ – bellcircle Feb 25 '17 at 0:28
  • $\begingroup$ @bellcircle Ah right, thanks... $\endgroup$ – saz Feb 25 '17 at 7:14
  • $\begingroup$ @saz I think for this problem: math.stackexchange.com/questions/2524383/… I need to do something similar. Could you please help me with it? Thank you. $\endgroup$ – ALannister Nov 17 '17 at 11:50

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