# Lebesgue Decomposition Theorem only true for Borel sets?

In Evan's book "Geometric Measure Theory and Fine Properties of Functions", we have the following two theorems:

Differentiation Theorem for Radon measures. Let $\nu, \mu: \mathcal P(\Bbb R^n) \to [0, \infty]$ be two Radon measures with $\nu \ll \mu$ (which means that for each $A \subset \Bbb R^n$ with $\mu(A) = 0$ we always have $\nu(A) = 0$). Then $$\nu(A) = \int_A \mathrm D_{\mu} \nu \, \mathrm d\mu \quad \text{for each \mu-measurable A \subset \Bbb R^n} \; .$$

and

Lebesgue Decomposition Theorem. Let $\nu, \mu: \mathcal P(\Bbb R^n) \to [0, \infty]$ be two Radon measures. Then

1. There exists two Radon measures $\nu_{ac}, \nu_s: \mathcal P(\Bbb R^n) \to [0, \infty]$ with $\nu_{ac} \ll \mu$ and $\nu_s \perp \mu$ (which means that there is a Borel set $B \subset \Bbb R^n$, such that $\nu_s(B) = \mu(\Bbb R^n \backslash B) = 0$), such that $$\nu = \nu_{ac} + \nu_s \; .$$

2. It holds that $$\mathrm D_{\mu} \nu = \mathrm D_{\mu} \nu_{ac} \quad \text{and} \quad \mathrm D_{\mu} \nu_s = 0 \quad \text{\mu-a.e.} \; ,$$ and consequently $$\nu(A) = \int_A \mathrm D_{\mu} \nu \, \mathrm d\mu + \nu_s(A) \quad \text{for each Borel set A \subset \Bbb R^n} \; .$$

Now my question is, doesn't the last equality in Lebesgue's Decomposition Theorem hold for each $\mu$-measurable set $A \subset \Bbb R^n$, and not only for the Borel sets?

Because if $A \subset \Bbb R^n$ is $\mu$-measurable, then by property 1. of Lebesgue's Decomposition Theorem and the Differentiation Theorem we have $$\nu(A) = \nu_{ac}(A) + \nu_s(A) = \int_A \mathrm D_{\mu} \nu_{ac} \, \mathrm d\mu + \nu_s(A) = \int_A \mathrm D_{\mu} \nu \, \mathrm d\mu + \nu_s(A) \; .$$ Did I overlook somehting? Note that $\mu$ und $\nu$ are what some authors call an outer measure.

• I'm familiar with this being proved as a single theorem (Lebesgue-Radon-Nikodym Theorem) from Rudin's Real and Complex Analysis. However, I don't recall the restriction $A \subset \mathbb{R}^n$ is Borel in order for equality to hold. I would look it up, but I don't have my copy of Rudin on hand at the moment. Perhaps you know someone who happens to have it? – Robert Short May 27 '15 at 16:47