PhyslibAlpha.ProbabilisticTheory.JordanOrderUnit.ProjectionResolution

Every measurable projection resolution $P$ on a measurable space $\Omega$ with values in a space $E$ can be viewed as an effect-valued measure. This allows a projection resolution to be used in any context where an effect-valued measure is expected.

For a measurable projection resolution $P$ over a sample space $\Omega$ with values in an effect space $E$, this instance allows $P$ to be treated as a function that maps any measurable set $s \subseteq \Omega$ to its corresponding effect $P(s) \in E$.

Let $\Omega$ be a measurable space and $E$ be a type representing a system of effects (typically within a JB algebra). Let $P$ and $Q$ be measurable projection resolutions on $\Omega$ taking values in $E$. If for every measurable set $s \subseteq \Omega$, the projections $P(s)$ and $Q(s)$ are equal as elements of $E$, then the projection resolutions $P$ and $Q$ are equal.

For any measurable projection resolution $P$ on a measurable space $\Omega$ with values in the effects $E$, the value of $P$ on the empty set $\emptyset$ (representing the impossible event) is the zero effect $0$.

For any measurable projection resolution $P$ on a space $\Omega$ taking values in a Jordan algebra $E$, the value assigned to the universal set $\Omega$ is the unit element $1$ of $E$.

Let $\Omega$ be a measurable space and $E$ be a Jordan algebra. If $P$ is a measurable projection resolution from $\Omega$ to $E$, then for any measurable set $s \subseteq \Omega$, the value $P(s) \in E$ is a Jordan projection.

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ taking values in a Jordan algebra $E$. For any measurable sets $s, t \subseteq \Omega$, the Jordan product of the projections $P(s)$ and $P(t)$ is equal to the projection of their intersection $s \cap t$: $$P(s) \cdot P(t) = P(s \cap t)$$

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. For any two measurable sets $s, t \subseteq \Omega$, if $s$ and $t$ are disjoint (i.e., $s \cap t = \emptyset$), then the corresponding projections $P(s)$ and $P(t)$ are Jordan-orthogonal in $E$.

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in an ordered Jordan algebra $E$. For any sequence of pairwise disjoint measurable sets $\{s_n\}_{n \in \mathbb{N}}$ in $\Omega$, the projection of their union $P\left(\bigcup_{n \in \mathbb{N}} s_n\right)$ is the least upper bound (supremum) of the set of finite partial sums $\left\{ \sum_{n=0}^{N-1} P(s_n) \mid N \in \mathbb{N} \right\}$.

Given a measurable projection resolution $P$ on a measurable space $\Omega$ with values in a Jordan algebra $E$, and a measurable function $f: \Omega \to \mathbb{R}$ that is bounded by a constant $M$ (i.e., $|f(x)| \le M$ for all $x \in \Omega$), the bounded Borel calculus is the element in $E$ defined by the integral of $f$ with respect to the projection resolution $P$, denoted as $\int_\Omega f \, dP$.

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. Let $f: \Omega \to \mathbb{R}$ be a measurable function that is bounded by a constant $M$ (i.e., $|f(x)| \leq M$ for all $x \in \Omega$). If $f$ is non-negative ($f(x) \geq 0$ for all $x \in \Omega$), then its bounded Borel functional calculus under $P$, denoted as $\int_\Omega f \, dP$, is a non-negative element in $E$ (i.e., $0 \leq \int_\Omega f \, dP$).

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. Let $f: \Omega \to \mathbb{R}$ be a measurable function. Suppose $M$ and $M'$ are two real constants such that $|f(x)| \le M$ and $|f(x)| \le M'$ for all $x \in \Omega$. Then the value of the bounded Borel functional calculus $\int_{\Omega} f \, dP$ is independent of the choice of bound, i.e., the integral calculated using $M$ is equal to the integral calculated using $M'$.

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ taking values in a Jordan algebra $E$. Let $f, g: \Omega \to \mathbb{R}$ be measurable functions such that $f$ is bounded by $M$ ($|f(x)| \le M$) and $g$ is bounded by $M'$ ($|g(x)| \le M'$). If the sum $f + g$ is measurable and bounded by $M + M'$, then the bounded Borel functional calculus is additive: $$\int_{\Omega} (f + g) \, dP = \int_{\Omega} f \, dP + \int_{\Omega} g \, dP$$

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. Let $f: \Omega \to \mathbb{R}$ be a measurable function such that $|f(x)| \le M$ for some $M \in \mathbb{R}$. For any scalar $c \in \mathbb{R}$, the bounded Borel functional calculus (denoted by $\int f \, dP$) is real-homogeneous, satisfying: $$\int_{\Omega} (c \cdot f) \, dP = c \cdot \int_{\Omega} f \, dP$$

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. For any measurable set $s \subseteq \Omega$, let $\mathbf{1}_s: \Omega \to \mathbb{R}$ denote the indicator function of $s$ (where $\mathbf{1}_s(x) = 1$ if $x \in s$ and $0$ otherwise). The bounded Borel functional calculus of $\mathbf{1}_s$ with respect to $P$ is equal to the projection $P(s)$, i.e., $$\int_\Omega \mathbf{1}_s \, dP = P(s).$$

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. The bounded Borel functional calculus of the constant function $1$ (defined as $f(x) = 1$ for all $x \in \Omega$) is equal to the order unit (identity) $1$ of the algebra $E$. This represents the normalization of the projection resolution, expressed as: $$\int_{\Omega} 1 \, dP = 1$$

Given a measurable projection resolution $P$ on a measurable space $\Omega$ with values in a Jordan algebra $E$, and a measurable function $f: \Omega \to \mathbb{R}$ such that $0 \le f(x) \le 1$ for all $x \in \Omega$, this definition represents the bounded Borel functional calculus $\int_\Omega f \, dP$ as an **effect** in $E$. An effect is an element of $E$ belonging to the order interval $[0, 1]$. This definition packages the properties that the integral of a $[0, 1]$-valued function is positive ($0 \le \int f \, dP$) and normalized ($\int f \, dP \le 1$).

Let $P$ be a measurable projection resolution on a measurable space $\Omega$ with values in a Jordan algebra $E$. Let $f: \Omega \to \mathbb{R}$ be a measurable function such that $0 \le f(x) \le 1$ for all $x \in \Omega$. Then the effect-valued bounded Borel functional calculus (the integral $\int_\Omega f \, dP$ viewed as an element of the order interval $[0, 1] \subseteq E$) is equal to the standard bounded Borel functional calculus $\int_\Omega f \, dP$ viewed as an element of $E$. In this context, coercing the effect-valued integral into the algebra $E$ simply forgets the established interval bounds $[0, 1]$.

Let $\Omega$ be a measurable space and $E$ be a Jordan algebra. Let $P$ and $Q$ be measurable projection resolutions on $\Omega$ with values in $E$. If for every measurable function $f: \Omega \to \mathbb{R}$ that is bounded by some constant $M \in \mathbb{R}$ (i.e., $|f(x)| \leq M$ for all $x \in \Omega$), the bounded Borel functional calculus results are equal, $\int_\Omega f \, dP = \int_\Omega f \, dQ$, then the projection resolutions themselves are equal, $P = Q$.