I'm trying to prove Lebesgue's decomposition theorem. I took the ideas for the existence part from here, and those for the uniqueness part from here
Lebesgue's decomposition theorem: Let $\mu, \nu$ be $\sigma$-finite measures on a measurable space $(X, \Sigma)$. Then there exist a unique pair $(\nu_0, \nu_1)$ of $\sigma$-finite measures such that:
- $\nu=\nu_0+\nu_1$,
- $\nu_0 \ll \mu$, i.e., $\nu_0$ is absolutely continuous with respect to $\mu$,
- $\nu_1 \perp \mu$, i.e., $\nu_1$ and $\mu$ are singular.
Could you check if my proof of the part $\color{blue}{\nu (A) = \nu (A \cap B)}$ is fine? Thank you so much!
My attempt:
- Existence:
Let $\lambda = \mu + \nu$. Then $\lambda$ is $\sigma$-finite and $\mu \ll \lambda$. By Radon–Nikodym theorem, the Radon–Nikodym derivative $f:X \to [0, \infty)$ exists, i.e., $$ f = \frac{\mathrm d \mu}{\mathrm d \lambda}. $$
Let $B := \{f = 0\} \in \Sigma$. We define non-negative measures $\nu_0, \nu_1$ by $$ \nu_0 (A) := \nu(A \cap B^c) \quad \text{and} \quad \nu_1 (A) := \nu(A \cap B) \quad \forall A \in \Sigma. $$
Clearly, $\nu=\nu_0+\nu_1$. Because $B$ is a $\mu$-null set, we get $\nu_1 \perp \mu$. Let's prove that $\nu_0 \ll \mu$. Assume $A \in \Sigma$ such that $\mu(A) = 0$. Then $$ 0=\mu (A) = \int_A f \mathrm d [\mu + \nu] = \int_A f \mathrm d \mu + \int_A f \mathrm d \nu = \int_A f \mathrm d \nu. $$
It follows that $f (x)=0$ for $\nu$-a.e. $x \in A$. We want to prove $\nu (A \cap B^c) = 0$. It suffices to prove $\color{blue}{\nu (A) = \nu (A \cap B)}$. This indeed true because $$ \nu (A) = \nu (\{x \in A \mid f(x) = 0\}) = \nu (\{x \in A \mid x \in B\}) = \nu (A \cap B). $$
- Uniqueness:
Let $(\lambda_0, \lambda_1)$ be another pair that satisfies the required properties. Then $$ \nu_0 - \lambda_0 = \lambda_1 - \nu_1. $$
Because $\mu \perp \nu_1$, there is $\Omega_1 \in \Sigma$ with the following properties
- $\mu (A) = 0$ for all $A \in \Sigma$ such that $A \subset \Omega_1$.
- $\nu_1 (A) = 0$ for all $A \in \Sigma$ such that $A \subset \Omega_1^c$.
Because $\mu \perp \lambda_1$, there is $\Omega_2 \in \Sigma$ with the following properties
- $\mu (A) = 0$ for all $A \in \Sigma$ such that $A \subset \Omega_2$.
- $\lambda_1 (A) = 0$ for all $A \in \Sigma$ such that $A \subset \Omega_2^c$.
Clearly, $\nu_0 - \lambda_0 \ll \mu$. Let's prove that $\lambda_1 - \nu_1 \perp \mu$. Let $\Omega := \Omega_1 \cup \Omega_2$. Then
- $\mu (A) = 0$ for all $A \in \Sigma$ such that $A \subset \Omega$. This is because $\mu (\Omega) \le \mu (\Omega_1) + \mu (\Omega_2) = 0$.
- $(\lambda_1 - \nu_1) (A) = 0$ for all $A \in \Sigma$ such that $A \subset \Omega^c$. This is because $(\lambda_1 - \nu_1) (A) = \lambda_1 (A) - \nu_1 (A)=0$.
We have $$ \begin{align} (\nu_0 - \lambda_0) (A) &= (\nu_0 - \lambda_0) (A \cap \Omega) + (\nu_0 - \lambda_0) (A \cap \Omega^c) \\ &= (\nu_0 - \lambda_0) (A \cap \Omega) + (\lambda_1 - \nu_1) (A \cap \Omega^c) \\ &=0+0\\ &=0. \end{align} $$
This completes the proof.
