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A POVM is typically defined as a collection of operators $\{\mu(a)\}_{a\in\Sigma}$ with $\mu(a)\in\mathrm{Pos}(\mathcal X)$ positive operators such that $\sum_{a\in\Sigma}\mu(a)=I$, where I take here $\Sigma$ to be some finite set and $\mathcal X$ some vector space (using the notation from Watrous' TQI).

On the other hand, when discussing *observables* in a measure-theoretic framework, given a nonempty set $\Omega$, a $\sigma$-algebra $\mathcal F$ on $\Omega$ (in other words, given a measurable space $(\Omega,\mathcal F)$), and denoting with $\mathcal E(\mathcal X)$ the set of *effects* on the space $\mathcal X$, that is, the set of Hermitian operators $E$ such that $0\le E\le I$, we say that a mapping $\mathrm A:\mathcal F\to\mathcal E(\mathcal X)$ is an *observable* if, for any state $\psi\in\mathcal X$, the function
$$\mathcal F\ni X\mapsto \langle \psi|\mathrm A(X)\psi\rangle\in\mathbb R$$
is a probability measure. Here I'm taking definition and notation from Heinosaari et al. (2008).
In other words, $\mathrm A$ is an observable iff, for all $\psi\in\mathcal X$, the triple $(\Omega,\mathcal F,\mathrm A_\psi)$ is a probability space, with $\mathrm A_\psi$ defined as $\mathrm A_\psi(X)=\langle \psi|\mathrm A(X)\psi\rangle$.

I'm trying to get a better understanding of how these two different formalisms match.
In particular, an *observable* as thus defined is closer to a POVM than an "observable" as usually defined in physics (which is just a Hermitian operator), right?

Are these *observables* equivalent to POVMs? That is, does any such *observable* correspond to a POVM, and *vice versa*?

I can see that a POVM can be thought of as/is the map $\mu:\Sigma\to\mathrm{Pos}(\mathcal X)$, which then extends to a map $\tilde\mu:2^{\Sigma}\to\mathrm{Pos}(\mathcal X)$ such that $\tilde\mu(2^\Sigma)=1$, which is then an *observable*.
However, I'm not sure whether any *observable* also corresponds to such a POVM.

As far as i understand, wikipedia's POVM definition and this definition of an observable are the same. – tsgeorgios – 2021-01-21T21:46:23.967

well, in one case you only impose $\mu(a)\ge0$ and $\sum_a \mu(a)=I$, while in the other you ask for $F$ s.t. $F(\Omega)=I$ and $\mathcal F\ni E\mapsto \langle F(E)\psi|\psi\rangle$ is a non-negative countably additive measure on $\mathcal F$ for all $\psi$. I guess the question is pretty much why these are equivalent – glS – 2021-01-21T22:02:33.877