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 $\lambda_a = \int h d\mu$


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 $g(x)=1$.


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 $$ 0 \le \frac{\lambda(E)}{\phi(E)} = \frac{\lambda(E)}{\lambda(E) + \mu(E)} \le 1. $$


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