4
$\begingroup$

Theorem: Let $\lambda, \mu$ be $\sigma$-finite measures defined on the $\sigma$-algebra $\mathcal{A}$ of the space $X$. Then,

a) Lebesgue decomposition: $\lambda=\lambda_a+\lambda_s$ where $\lambda_a \ll \mu$ and $\lambda_s \perp \mu$

b) $\exists h \in L^1(\mu)$ such that $\lambda_a = \int h d\mu$

Proof:

Define $\phi=\lambda + \mu \space \phi(X)<\infty, \space f \in L^2(\phi) $. Then $f \to \int f d\lambda $ is a continuous linear functional on $L^2(\phi)$ Because: $$\int_X |f| d\lambda \le \int |f| d\phi \le \Bigg( \int_X |f|^2 d\phi \Bigg)^{\frac12}\Bigg( \int_X d\phi \Bigg)^{\frac12} $$

Q1: How does this prove the aforementioned claim? Don't we already have $f \to \int f d\lambda $ is a continuous linear functional on $L^2(\phi)$ from the fact that $f \in L^2(\phi) $?

Moving on, from the Riesz Representation theorem we know, that $\exists ! \space g \in L^2(\phi) $ such that $$\int_X f d\lambda = \int_X fg d\phi $$.

Choosing $f=\chi_E, \space E \in \mathcal{A}, \space \phi(E)<\infty$ we obtain $\lambda(E)=\int_E g d\phi$ and I'm told, that

$$\lambda(E)=\int_E g d\phi \Rightarrow 0 \le \frac{1}{\phi(E)}\int_E g d\phi = \frac{\lambda(E)}{\phi(E)} \le 1$$

Q2 Why does the above implication hold?

From the above equality we know, that $g(x) \in [0,1]$, an absolutely crucial fact for the rest of the proof. We define $A=\{x:g(x)<1\} \text{ and } B=\{x: g(x)=1\}$. Now, $$\lambda(E) = \lambda(E\cap A) + \lambda(E \cap B) = \lambda_a (E)+\lambda_s(E)$$

$$\int_X f d\lambda = \int_X fg d (\lambda+\mu) \Rightarrow \int_X f(1-g) d \lambda = \int_X fg d \mu $$.

Choosing $f=\chi_B$ we obtain $\mu(B)=0 \Rightarrow \mu \perp \lambda_s$

Choosing $f=\chi_E \sum_{k=0}^n g^k$ we obtain

$$\int_E (1-g^{n+1}) d \lambda = \int_E \sum_{k=1}^n g^k d \mu \\ \lim_{n \to \infty} \int_E (1-g^{n+1}) d \lambda = \lim_{n \to \infty} \int_E \sum_{k=1}^n g^k d \mu$$.

and since $(1-g^{n+1}) \nearrow 1$ and $\sum_{k=1}^n g^k \nearrow h$ we can use the monotone convergence theorem to get

$$\lambda(E\cap A)=\lambda_a(E) = \int_E h d \mu $$

Remark: for $x \in B$ we get 0 measure because $g(x)=1$.

$\endgroup$
4
$\begingroup$

Question 1: We define a functional $\ell\colon L^2(\phi) \to \mathbb K$ by $$ \ell(f) = \int_X f \, d\lambda, \quad f \in L^2(\phi). $$ So to each $f \in L^2(\phi)$ we associate a number, namely the integral over $X$ with respect to $\lambda$. To know that $f$ is linear and continuous, we have to prove it. Linearity is easy, so we concentrate on continuity, we have for all $f \in L^2(\phi)\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$ \begin{align*} \abs{\ell(f)} &= \abs{\int_X f\, d\lambda}\\ &\le \int_X \abs f\, d\lambda\\ &\le \int_X \abs f \, d(\lambda + \mu) \text{ note that $\int_X \abs f \, d\mu \ge 0$}\\ &= \int_X \abs f\cdot 1 \, d\phi\\ &\le \left(\int_X \abs{f}^2\, d\phi\right)^{1/2} \left(\int_X 1\, d\phi\right)^{1/2} \text{ this is Cauchy-Schwarz}\\ &= \phi(X)^{1/2} \cdot \norm f_{L^2(\phi)} \end{align*} So $\ell$ is a bounded, hence continuous linear functional on $L^2(\phi)$.

Question 2: As $\lambda$ and $\phi$ are positive measures, we obviously have $\lambda(E)$, $\phi(E)\ge 0$, moreover note, that by defintion $$ \phi(E) = \lambda(E) + \mu(E) \ge \lambda(E) $$ So for $\phi(E) > 0$, we have $$ 0 \le \frac{\lambda(E)}{\phi(E)} = \frac{\lambda(E)}{\lambda(E) + \mu(E)} \le 1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.