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Let me turn my comment into an answer.

I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.

To see this, fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that for $r\leq \delta$, $$ \left|\int f(z)d\mu(z)- \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w)\right|\leq $$ $$ \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.

Let me turn my comment into an answer.

I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.

To see this, fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that for $r\leq \delta$, $$ \left|\int f(z)d\mu(z)- \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w)\right|=$$ $$ \left|\int d\mu(z)\int_{|w-z|\leq r} d\lambda(w)\frac{f(z)}{\pi r^2} - \int d\lambda(w)\int_{|z-w|\leq r} d\mu(z)\frac{f(w)}{\pi r^2} \right| =$$ $$\left| \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{f(z)-f(w)}{\pi r^2}\right|\leq$$ $$ \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.

Let me turn my comment into an answer.

I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.

To see this, fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that for $r\leq \delta$, $$ \left|\int f(z)d\mu(z)- \frac{1}{\pi r^2} \mu_w([0,r))f(w) d\lambda(w)\right|\leq $$ $$ \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.

Let me turn my comment into an answer.

I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.

To see this, fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that for $r\leq \delta$, $$ \left|\int f(z)d\mu(z)- \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w)\right|\leq $$ $$ \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.

Let me turn my comment into an answer.

I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.

To see this, fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that $$ \left|\int f(z)d\mu(z)- \frac{1}{\pi \delta^2} \mu_w([0,\delta))f(w) d\lambda(w)\right|\leq $$ $$ \int d\mu(z)\int_{|w-z|\leq \delta}d\lambda(w) \frac{|f(z)-f(w)|}{\pi \delta^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.

Let me turn my comment into an answer.

I claim that for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$, $$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$ where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.

To see this, fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that for $r\leq \delta$, $$ \left|\int f(z)d\mu(z)- \frac{1}{\pi r^2} \mu_w([0,r))f(w) d\lambda(w)\right|\leq $$ $$ \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.