Integral Measures: Identification

Question

Problem

Given a Borel space $\Omega$.

Consider a Borel measure: $$\mu:\mathcal{B}(\Omega)\to\overline{\mathbb{R}}:\quad\mu\geq0$$

Regard a Borel measure: $$\rho\geq0:\quad\mu_\rho(A):=\int_A\rho\,\mathrm{d}\mu$$

Denote its Borel support: $$\sigma_\rho:=\left({\bigcup}_{U\in\mathcal{T}(\Omega):\mu_{\rho}(U)=0}U\right)^\complement$$

Then isometrically: $$\mathcal{L}(\sigma_\rho;\mu_\rho)\cong\mathcal{L}(\sigma_\rho;\mu)$$

How can I prove this?

Attempt

Define the map: $$U_\rho:\mathcal{L}(\sigma_\rho;\mu_\rho)\to\mathcal{L}(\sigma_\rho;\mu):\eta\mapsto\eta\rho$$

It is isometric since: $$\|U_\rho\eta\|=\int|\omega|\rho\mathrm{d}\mu=\int|\omega|\mathrm{d}\mu_\rho=\|\eta\|$$

So also well-defined: $$\|\omega\|=0\implies\|U_\rho\omega\|=0$$

But why surjective?

Answer 1

The assertion is wrong:

Given the interval $[-1,1]$.

Consider the Borel measure: $$\mu:\mathcal{B}([-1,1])\to\overline{\mathbb{R}}:\quad\mu:=\lambda+\delta$$

Regard the Borel measure: $$\rho:=1-1_{\{0\}}:\quad\mu_\rho(A):=\int_{[-1,1]}\rho\,\mathrm{d}\mu=\lambda$$

It has Borel support: $$\sigma_\rho=\sigma(\lambda)=[-1,1]$$

But it is not surjective: $$\vartheta:=1_{\{0\}}:\quad\vartheta\notin\mathcal{R}U_\rho$$

Concluding counterexample.