PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.SpectralPointMass

Given a measure $\mu$ and a pair of endpoints $p = (a, b) \in \mathbb{R}^2$ representing the closed interval $[a, b]$, this function computes the midpoint $m = \frac{a + b}{2}$. It returns the pair $(a, m)$ if the measure of the left sub-interval $\mu([a, m]) = 1$; otherwise, it returns the pair $(m, b)$.

Given a measure $\mu$ and initial endpoints $a, b \in \mathbb{R}$, this function defines a sequence of pairs $(a_n, b_n) \in \mathbb{R} \times \mathbb{R}$ representing the endpoints of nested intervals. The sequence is defined recursively by $(a_0, b_0) = (a, b)$, and the $(n+1)$-th interval is obtained by applying a bisection step `bisectStep` to the $n$-th interval. At each step $n$, the width of the interval is halved, such that $b_n - a_n = \frac{b-a}{2^n}$.

Let $\mu$ be a measure on $\mathbb{R}$. For any pair of real numbers $p = (a, b)$ representing an interval such that $a \le b$, the pair $(a', b') = \text{bisectStep}(\mu, p)$ resulting from the bisection step also satisfies $a' \le b'$.

For any pair of real numbers $p = (a, b)$ representing an interval, the width of the interval resulting from one bisection step $\text{bisectStep}(\mu, p)$ is equal to half the width of the original interval. That is, if $(a', b') = \text{bisectStep}(\mu, p)$, then $b' - a' = \frac{b - a}{2}$.

Let $\mu$ be a measure on $\mathbb{R}$. For any pair $p = (a, b) \in \mathbb{R}^2$ representing an interval $[a, b]$ such that $a \le b$, the first component (left endpoint) of the pair produced by the bisection step $\text{bisectStep}(\mu, p)$ is greater than or equal to $a$.

Let $\mu$ be a measure on $\mathbb{R}$. For any pair of real numbers $p = (a, b)$ representing the closed interval $[a, b]$ such that $a \le b$, the second component (right endpoint) of the pair produced by the bisection step $\text{bisectStep}(\mu, p)$ is less than or equal to the original right endpoint $b$.

Let $p = (a, b)$ be a pair of real numbers representing the closed interval $[a, b]$ such that $a \le b$. Let $(a', b') = \text{bisectStep}(\mu, p)$ be the pair of endpoints resulting from one step of the bisection process with respect to the measure $\mu$. Then the closed interval $[a', b']$ is a subset of the original closed interval $[a, b]$, i.e., $[a', b'] \subseteq [a, b]$.

Let $a, b \in \mathbb{R}$ be real numbers such that $a \le b$. For any natural number $n$, let $(a_n, b_n)$ be the $n$-th bisection interval starting from $(a, b)$ with respect to a measure $\mu$. Then, the left endpoint is less than or equal to the right endpoint, i.e., $a_n \le b_n$.

Let $(a_n, b_n)$ be the $n$-th interval in the bisection sequence starting from $(a, b)$ relative to a spectral measure $\mu$. For any $a, b \in \mathbb{R}$ and any $n \in \mathbb{N}$, the width of the $n$-th interval satisfies $b_n - a_n = \frac{b - a}{2^n}$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ and let $[a, b]$ be an initial interval with $a \le b$. Let $(a_n, b_n)$ denote the endpoints of the $n$-th interval produced by the bisection process (which recursively selects the half-interval where the measure $\mu$ is 1). Then, for any natural number $n$, the left endpoint of the $n$-th interval is less than or equal to the left endpoint of the $(n+1)$-th interval, i.e., $a_n \le a_{n+1}$.

Let $\mu$ be a measure on $\mathbb{R}$. For any real numbers $a$ and $b$ such that $a \le b$, let $(a_n, b_n)$ denote the $n$-th interval produced by the bisection process starting from initial endpoints $a$ and $b$. Then for any natural number $n$, the right endpoint of the $(n+1)$-th interval is less than or equal to the right endpoint of the $n$-th interval, i.e., $b_{n+1} \le b_n$.

Let $\mu$ be a measure on $\mathbb{R}$ and $a, b \in \mathbb{R}$ such that $a \leq b$. Let $(a_n, b_n)$ be the $n$-th bisection interval obtained from the initial interval $[a, b]$ with respect to $\mu$. Then for any $n \in \mathbb{N}$, the $(n+1)$-th interval is contained within the $n$-th interval, i.e., $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$.

Let $\mu$ be a measure on $\mathbb{R}$ such that for every measurable set $E \subseteq \mathbb{R}$, $\mu(E)$ is either $0$ or $1$. Let $p = (a, b)$ be a pair of real numbers with $a \le b$ such that the measure of the closed interval is $\mu([a, b]) = 1$. Then the interval resulting from a single bisection step, $[a', b'] = \text{bisectStep}(\mu, p)$, also has measure $\mu([a', b']) = 1$. (Note: The bisection step $\text{bisectStep}(\mu, p)$ returns the left sub-interval $[a, \frac{a+b}{2}]$ if its measure is 1, and the right sub-interval $[\frac{a+b}{2}, b]$ otherwise.)

Let $\mu$ be a measure on $\mathbb{R}$ that only takes the values $0$ and $1$. For any real numbers $a \le b$ such that the measure of the closed interval $[a, b]$ is $\mu([a, b]) = 1$, the sequence of intervals $[a_n, b_n]$ produced by the bisection process starting from $[a, b]$ satisfies $\mu([a_n, b_n]) = 1$ for all $n \in \mathbb{N}$.

Let $\mu$ be a measure on $\mathbb{R}$ such that for every measurable set $E$, $\mu(E)$ is either $0$ or $1$. Let $[a, b]$ be an interval with $a \le b$ such that $\mu([a, b]) = 1$. If $I_n = [a_n, b_n]$ denotes the $n$-th interval in the bisection sequence starting from $[a, b]$, then for any $n \in \mathbb{N}$, the measure of the set difference between the $n$-th and $(n+1)$-th intervals is zero, i.e., $\mu(I_n \setminus I_{n+1}) = 0$.

Let $\mu$ be a measure on $\mathbb{R}$ and let $a, b \in \mathbb{R}$ be such that $a \leq b$. Let $(a_n, b_n)$ be the sequence of intervals defined by the bisection process starting from $[a, b]$ with respect to $\mu$. Then the sequence of left endpoints $n \mapsto a_n$ is monotone (non-decreasing).

Let $\mu$ be a measure on $\mathbb{R}$ and let $a, b$ be real numbers such that $a \leq b$. Let $(a_n, b_n)$ denote the endpoints of the $n$-th interval produced by the bisection process $\text{bisect}(\mu, a, b, n)$. The sequence of right endpoints $b_n$ is antitone; that is, for any $n, m \in \mathbb{N}$, if $n \leq m$, then $b_m \leq b_n$.

Let $\mu$ be a measure and $a, b \in \mathbb{R}$ such that $a \leq b$. Let $(a_n, b_n)$ be the $n$-th bisection interval starting from $(a, b)$. For any natural number $n$, the right endpoint $b_n$ of the $n$-th interval is less than or equal to the initial right endpoint $b$.

Let $\mu$ be a spectral measure and let $a, b \in \mathbb{R}$ be such that $a \le b$. If $(a_n, b_n)$ denotes the endpoints of the $n$-th bisection interval starting from $[a, b]$, then for any natural number $n$, the left endpoint $a_n$ satisfies $a \le a_n$.

Let $\mu$ be a spectral measure and let $a$ and $b$ be real numbers such that $a \leq b$. Consider the sequence of intervals $[a_n, b_n]$ generated by the bisection process starting from $[a, b]$ (defined by `bisect μ a b n`). Then the set of left endpoints $\{a_n : n \in \mathbb{N}\}$ is bounded above.

Let $a, b \in \mathbb{R}$ be such that $a \leq b$. For a measure $\mu$, let $(a_n, b_n)$ denote the $n$-th interval produced by the bisection process starting from $[a, b]$. Then the set of right endpoints $\{b_n : n \in \mathbb{N}\}$ is bounded from below.

Given a measure $\mu$ and an initial interval $[a, b]$ with $a \leq b$, let $(a_n, b_n)$ be the sequence of nested intervals produced by the bisection procedure (where each step selects the sub-interval of measure 1). This value is defined as the supremum of the left endpoints, $\sup_{n \in \mathbb{N}} a_n$, which represents the unique limit point of the nested intervals.

Let $a, b \in \mathbb{R}$ be such that $a \leq b$. For a measure $\mu$, let $(a_n, b_n)$ denote the $n$-th interval generated by the bisection process starting from $[a, b]$, where at each step the sub-interval of measure $1$ is selected. Let $r$ be the limit point of these intervals, defined as $r = \sup_{n \in \mathbb{N}} a_n$. Then the sequence of left endpoints $a_n$ converges to $r$ as $n \to \infty$.

For any real numbers $a$ and $b$, let $(a_n, b_n)$ denote the $n$-th interval generated by the bisection process starting from initial endpoints $a$ and $b$. Then the width of these intervals, given by the difference $b_n - a_n$, converges to $0$ as $n \to \infty$.

Given a spectral measure $\mu$ and initial endpoints $a, b \in \mathbb{R}$ with $a \leq b$, let $(a_n, b_n)$ be the sequence of nested intervals produced by the bisection process. Then the sequence of right endpoints $b_n$ converges to the bisection limit point $r = \sup_{n \in \mathbb{N}} a_n$ as $n \to \infty$.

Let $\mu$ be a spectral measure taking values in $\{0, 1\}$. For an initial interval $[a, b]$ with $a \leq b$, let $(a_n, b_n)$ be the sequence of nested intervals produced by recursively bisecting the interval and selecting the sub-interval with measure $1$. Let $x^*$ be the limit point of these intervals, defined as the supremum of the left endpoints, $x^* = \sup_{k \in \mathbb{N}} a_k$. Then for any $n \in \mathbb{N}$, the left endpoint of the $n$-th interval satisfies $a_n \leq x^*$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ that takes values in $\{0, 1\}$. For an initial interval $[a, b]$ with $a \le b$, let $(a_n, b_n)$ be the sequence of nested intervals produced by recursively bisecting the interval and selecting the sub-interval with measure $1$. Let $x^*$ be the limit point of these intervals, defined as the supremum of the left endpoints, $x^* = \sup_{k \in \mathbb{N}} a_k$. Then for any $n \in \mathbb{N}$, the bisection point $x^*$ satisfies $x^* \le b_n$.

Let $a, b \in \mathbb{R}$ be real numbers such that $a \leq b$. Let $(a_n, b_n)$ denote the sequence of endpoints of the nested intervals produced by the bisection procedure starting from the interval $[a, b]$. Then the intersection of all the closed intervals $[a_n, b_n]$ for $n \in \mathbb{N}$ is the singleton set $\{r\}$, where $r$ is the bisection limit point defined as $r = \sup_{n \in \mathbb{N}} a_n$. That is, $$\bigcap_{n \in \mathbb{N}} [a_n, b_n] = \{r\}.$$

Let $\mu$ be a measure on $\mathbb{R}$ such that for every measurable set $E \subseteq \mathbb{R}$, $\mu(E) \in \{0, 1\}$. If $a \le b$ and the measure of the closed interval $[a, b]$ is $\mu([a, b]) = 1$, then $\mu(\{x^*\}) = 1$, where $x^*$ is the bisection point (the unique limit point of the sequence of nested sub-intervals of $[a, b]$ that each have measure $1$).

Let $\mu$ be a measure on the real numbers $\mathbb{R}$ such that for every measurable set $E \subseteq \mathbb{R}$, the measure $\mu(E)$ is either $0$ or $1$. Suppose there exists a closed interval $[a, b]$ with $a \le b$ such that $\mu([a, b]) = 1$. Then there exists a point $r \in \mathbb{R}$ such that $\mu(\{r\}) = 1$, and for any measurable set $E$, if $r \notin E$, then $\mu(E) = 0$.