16 — Radon–Nikodym, Lebesgue Decomposition, and Kakutani
Goal: Prove the Radon–Nikodym theorem in the simple setting of measures on
$(0,1], \mathcal{B}$, following Durrett §5.5.3, p. 206.
We work with two finite measures $\mu$ and $\gamma$, and define \(\rho = \frac{\mu + \gamma}{2}.\)
1. Dyadic σ–fields and the martingale $X_n$
Define the dyadic intervals \(I_{n,k} = \left( \frac{k-1}{2^n}, \frac{k}{2^n} \right],\qquad k=1,\dots,2^n.\)
Let \(\mathcal{F}_n = \sigma\{ I_{n,k} : 1 \le k \le 2^n\} \quad\text{and}\quad X_n(t)= \frac{\mu(I_{n,k})}{\rho(I_{n,k})},\ t\in I_{n,k}.\)
Facts:
- ${X_n,\mathcal{F}_n}$ is a martingale in the probability space $((0,1],\mathcal{B},\rho)$.
- $0\le X_n \le 2$.
- By MGCT, \(X_n \to X \quad \rho\text{-a.s.}\)
2. Key identity relating $\mu$, $\gamma$, and $\rho$
Since $\rho = (\mu+\gamma)/2$, for any Borel $A$, \(\int_A X_n\, d\rho = \int_A X_n\, \frac{d\mu + d\gamma}{2} = \frac12\left[\int_A X_n\, d\mu + \int_A X_n\, d\gamma\right].\)
But by definition of $X_n$, \(\int_A X_n\, d\rho = \mu(A).\)
Therefore, \(\mu(A) = \frac12\left[\int_A X_n\, d\mu + \int_A X_n\, d\gamma\right].\)
Solve for the term with $d\mu$: \(\int_A (2-X_n)\, d\mu = \int_A X_n\, d\gamma.\)
3. Goal (intuition of Radon–Nikodym)
We want
\((2-X)\, d\mu = X\, d\gamma,\)
which gives formally
\(d\mu = \frac{X}{2-X}\, d\gamma,\)
whenever $X<2$.
If $X=2$ on a set, $\mu$ may have a singular component there.
Thus the RN derivative should be \(\frac{d\mu}{d\gamma} = \frac{X}{2-X} \quad\text{on } \{X<2\}.\)
4. Dominated Convergence and Simple Functions
Let $Y_n = X_n/(2-X_n)$.
On ${X<2}$, $Y_n\to Y := X/(2-X)$ a.s.
Then for any $A\in\mathcal{B}$, \(\mu(A) = \lim_{n\to\infty} \int_A Y_n\, d\gamma = \int_A Y\, d\gamma.\)
This gives the Radon–Nikodym derivative: \(\frac{d\mu}{d\gamma} = Y.\)
5. Lebesgue Decomposition via $X$
Define \(M_r(A) = \int_A Y\, d\gamma, \qquad M_s(A) = \mu(A) - M_r(A).\)
Then:
- $M_r \ll \gamma$,
- $M_s \perp \gamma$,
- $\mu = M_r + M_s$.
The singular part is supported on ${X=2}$, since $X=2$ is precisely where the formula for the density breaks down.
6. Example: Pure Point Mass
Example in notes:
Let $\mu = \delta_{1/2}$.
Then $X_n(t) = \mu(I_{n,k}) / \rho(I_{n,k})$ is only nonzero for intervals containing $1/2$.
Limit $X = 2\cdot 1_{{1/2}}$.
Thus all mass is singular: $M_s = \mu$, $M_r = 0$.
7. Kakutani Product Measure Criterion
Consider product measures: \(\mu = \bigotimes_{k=1}^\infty \mu_k, \quad P = \bigotimes_{k=1}^\infty P_k.\)
Define the finite–dimensional Radon–Nikodym approximations: \(Y_n = \prod_{k=1}^n \frac{d\mu_k}{dP_k}.\)
Then $Y_n\to Y$ a.s., and exactly one of the following holds:
-
$E_P(Y)=0$: then
\(\mu \perp P.\) -
$E_P(Y)=1$: then
\(\mu \ll P.\)
Kakutani’s criterion:
-
Mutual absolute continuity iff
\(\prod_{k=1}^\infty E_P\left(\sqrt{\frac{d\mu_k}{dP_k}}\right) > 0.\) -
Mutual singularity iff
\(\prod_{k=1}^\infty E_P\left(\sqrt{\frac{d\mu_k}{dP_k}}\right) = 0.\)
This is exactly the statement on the last page of your notes.
Summary
- Construct a martingale $X_n$ by dyadic partitioning.
- Use MGCT to get limit $X$.
- Show $(2-X)\, d\mu = X\, d\gamma$.
- Obtain Radon–Nikodym derivative $d\mu/d\gamma = X/(2-X)$ on ${X<2}$.
- Singular mass sits where $X=2$.
- Kakutani’s theorem characterizes absolute continuity vs. singularity for infinite product measures.
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