MHB Finding Lebesgue Decomposition for a Measure Defined by an Integral

[joypav]

These are both problems from Royden's Real Analysis. Trying to get through the problems from the Radon-Nikodym Theorem section of the book.

Problem 1:
Let $(\mu_n)$ be a sequence of measures on a measurable space $(X, M)$ for which there is a constant $c>0$ such that $\mu_n(X)<c, \forall n$.
Define $\mu: M \rightarrow [0, \infty]$ by,

$\mu = \sum_{n=1}^{\infty} \frac{\mu_n}{2^n}$.

Show that $\mu$ is a measure on $M$ and that each $\mu_n$ is absolutely continuous with respect to $\mu$.

Solution:
Let $\lambda_n = \frac{\mu_n}{2^n}$.
Claim: $\lambda_n$ is a measure for every $n$
$\lambda_n(\emptyset) = \frac{\mu_n(\emptyset)}{2^n} = \frac{0}{2^n} = 0$

and

$\lambda_n(\cup_{i=1}^{\infty}A_i) = \frac{\mu_n(\cup_{i=1}^{\infty}A_i)}{2^n} = \frac{\sum_{i=1}^{\infty}\mu_n(A_i)}{2^n} = \sum_{i=1}^{\infty} \frac{\mu_n(A_i)}{2^n} = \sum_{i=1}^{\infty} \lambda_n(A_i)$
$\implies \lambda_n$ is a measure for every $n$

Then,
$\mu = \sum_{n=1}^{\infty} \lambda_n$

$\lambda_n(\emptyset) = 0, \forall n \implies \mu(\emptyset) = \sum_{n=1}^{\infty} \lambda_n(\emptyset) = \sum_{n=1}^{\infty} 0 = 0$

and

$\mu(\cup_{i=1}^{\infty} A_i) = \sum_{n=1}^{\infty} \lambda_n(\cup_{i=1}^{\infty} A_i) = \sum_{n=1}^{\infty}\sum_{i=1}^{\infty} \lambda_n(A_i) = \sum_{i=1}^{\infty}\sum_{n=1}^{\infty} \lambda_n(A_i) = \sum_{i=1}^{\infty} \mu(A_i)$
(I was thinking we could switch the order of the summations because each $\mu_n(A_i)<c$ so it's finite?)

I also need to show absolute continuity of each $\mu_n$ with respect to $\mu$.
For $A \in M$,
$\mu(A) = 0 \implies \sum_{n=1}^{\infty} \frac{\mu_n(A)}{2^n} = 0 \implies \mu_n(A) = 0, \forall n$
Because if any are nonzero then the sum would be $>0$?

Problem 2:
Let $(X, M, \mu)$ be a measure space and $f$ a non-negative function that is integrable over $X$ with respect to $\mu$. Find the Lebesgue decomposition with respect to $\mu$ of the measure $\nu$ defined by

$\nu(E) = \int_{E} f d\mu$, for $E \in M$.

Alright... this problem I really have no idea about. I was thinking at first to split $f$ into its positive and negative parts, but $f$ is non-negative.

Also, isn't $f$ actually $\frac{d\nu}{d\mu}$?

 

[Euge]

Interchanging double series of nonnegative terms is justified by the monotone convergence theorem applied the counting measure on $\Bbb N$. The condition $\mu_n(X) < c$ for each $n$ is unnecessary unless one is to show that $\mu$ is a finite positive measure on $X$.

For the second problem, it's true that $f = \frac{d\nu}{d\mu}$ a.e. $[\mu]$, but you can simply note $\nu << \mu$, so that the absolutely continuous part of $\nu$ with respect to $\mu$ is $\nu$ itself; $\nu$ has no singular part with respect to $\mu$.

 

[joypav]

Interchanging double series of nonnegative terms is justified by the monotone convergence theorem applied the counting measure on $\Bbb N$. The condition $\mu_n(X) < c$ for each $n$ is unnecessary unless one is to show that $\mu$ is a finite positive measure on $X$.

For the second problem, it's true that $f = \frac{d\nu}{d\mu}$ a.e. $[\mu]$, but you can simply note $\nu << \mu$, so that the absolutely continuous part of $\nu$ with respect to $\mu$ is $\mu$ itself; $\nu$ has no singular part with respect to $\mu$.

Ahh okay, then it is as simple as $\nu = \nu_1 + \nu_2$, where $\nu_1=\nu$ and $\nu_2 \equiv 0$?

 

[Euge]

Yes, that's correct.