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Sep 22, 2022 at 19:27 vote accept Ibra
Sep 22, 2022 at 18:05 answer added Dominik Kutek timeline score: 1
Sep 22, 2022 at 17:47 comment added Ibra I honestly could not conclude. Application of monotone class theorem requires the set of Balls to be stable by intersection which is not the case. I don't know whether I can proceed differently. Thanks!
Sep 22, 2022 at 16:19 comment added Dominik Kutek So I would suggest defining $\phi_{k,n} = 1_{B(x_k,\frac{1}{n})}$. Then $\int_Y \phi_{k,n}d\nu_1 = \int_{Y}\phi_{k,n} d\nu_2$ means $\nu_1(B(x_k,\frac{1}{n})) = \nu_2(B(x_k,\frac{1}{n}))$. Try to show that this implies $\nu_1(U) = \nu_2(U)$ for any open set $U$.
Sep 22, 2022 at 16:15 comment added Ibra Yes, I know a similar result
Sep 22, 2022 at 16:08 comment added Dominik Kutek Do you know, that any open set in a separable metric space can be written as countable union of open balls? Precisely, if $\{x_k\}_k$ is a countable dense subset in $Y$, then any open set $U$ can be written as $U = \bigcup_{x_k \in U, n \in \mathbb N} B(x_k,\frac{1}{n})$, where $B(x_k,\frac{1}{n})$ is an open ball centered at $x_k$ with radius $\frac{1}{n}$.
Sep 22, 2022 at 16:04 comment added Ibra Sorry for my lack of precision. It is indeed the Borel sigma-field.
Sep 22, 2022 at 15:34 comment added Dominik Kutek Is $\mathcal Y$ arbitrary $\sigma-$field on $Y$, or Borel $\sigma-$field?
Sep 22, 2022 at 15:14 history edited Ibra CC BY-SA 4.0
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Sep 22, 2022 at 14:59 history asked Ibra CC BY-SA 4.0