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For the left shift, ${\sigma }^p(s)=s$ precisely when the binary sequence is obtained by repeating its first $p$ symbols. There are exactly $2^p$ such sequences, although some have smaller least period. Hence the set of all periodic sequences is a countable union of finite sets and is at most countable. Passing from periodic points to periodic orbits cannot increase this cardinality.

To prove infinitude, for every integer $p\ge 1$ take the repeated word $(0^{p-1}1{)}^{\infty }$. Its least period is $p$, because its ones are separated by exactly $p$ positions; a smaller positive period would carry a one to a zero. Distinct least periods give distinct periodic orbits. Thus there are countably infinitely many periodic orbits.

Let $E_n$ be null. Given $\epsilon >0$, cover $E_n$ by intervals of total length less than $\epsilon 2^{-n}$, for $n\ge 1$. The union of all these interval families is countable, covers ${\bigcup }_n{E}_n$, and has total length less than ${\sum }_n\epsilon 2^{-n}=\epsilon$. Thus the countable union is null.

For each $N\ge 1$, put $C_N={\bigcup }_{n\ge N}{A}_n$. Closure of a sigma-algebra under countable unions and countable intersections givesThe events $C_N$ decrease to $A$. Continuity of probability from above applies because $\mathbb{P}(C_1)\le 1<\infty$. Also, $A_n\subseteq C_N$ whenever $n\ge N$, soLetting $N\to \infty$ provesThis reverse Fatou inequality for events requires no independence.

For $B_n=\{X_n>a_n\}$, the definition of the common cumulative distribution function givesThe strict exceedance matters here: an atom of a measure at $a_n$ belongs to $\{X_n\le a_n\}$, so it is already included in $F(a_n)$. The events $B_n$ are mutually independent events, since each depends on a separate member of the independent random variables $X_n$.

If ${\sum }_n(1-F(a_n))<\infty$, countable subadditivity gives, for every $N$,Continuity of probability then gives $\mathbb{P}({\mathrm{limsup}}_n{B}_n)=0$. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and $1-u\le e^{-u}$ give, for every fixed $N$,Taking $M\to \infty$ shows that the probability of no exceedance after $N$ is zero. A countable union over $N$ shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves

The product measure $\mu \otimes \nu$ is the unique probability measure on the sigma-algebra generated by measurable rectangles for whichHere the printed product notation $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ denotes the generated product sigma-algebra, conventionally written $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$; it is not just the collection of rectangles. On ${\mathbb{R}}^2$ it equals $\mathcal{B}({\mathbb{R}}^2)$: coordinate projections show that rectangles are Borel sets, while open rectangles with rational endpoints give every open set as a countable union and hence generate the Borel sigma-algebra.

Existence is the product measure theorem for finite measures, which are in particular sigma-finite measures. Uniqueness follows because the rectangles form a pi-system generating the product sigma-algebra, and two finite measures agreeing there agree everywhere by the Dynkin lemma.^[ref]

For each integer $m\ge 1$, letIf $D_m$ had $m+1$ distinct members, finite additivity on their disjoint singleton Borel sets would give total measure at least $(m+1)/m>1$, contradicting $\mu (\mathbb{R})=1$. Thus $|D_m|\le m$.

Every positive singleton measure is at least $1/m$ for some sufficiently large $m$. Consequently the set of point atoms of a measure isA countable union of finite sets is at most countable. This proves the claim, including the possibility that the set is empty. The same argument gives the countability of point atoms of a finite measure, with $|D_m|\le m\mu (\mathbb{R})$ in place of $m$.

Apply part (i) for every positive integer $K$ and discard the countable union of its exceptional null sets. A sample path bounded below never reaches $-K$ for some such $K$, and must therefore converge to a finite limit of a sequence.

Apply the same argument to the martingale $(-M_n)$. Outside one more null set, every sample path bounded above also converges finitely. Thus a sample path which does not converge finitely must be unbounded in both directions. Each finite initial segment has finite values, so removing any such segment preserves unboundedness. It follows that the limit superior is $+\infty$ and the limit inferior is $-\infty$. ConsequentlyThis proves the bounded-increment martingale convergence-or-oscillation dichotomy; in particular one-sided divergence to infinity is impossible with positive probability.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.

Let $\mathcal{G}=T^{-1}(\mathcal{F})=\{T^{-1}(A):A\in \mathcal{F}\}$. Inverse images commute with complements and countable unions, so this is a sub-sigma-algebra of $\mathcal{F}$. The random variable $X\circ T$ is $\mathcal{G}$-measurable.

The assumption $\mathbb{P}\circ T^{-1}=\mathbb{P}$ says that $T$ is a measure-preserving transformation. Consequently $X\circ T$ has the same probability distribution as $X$ and is integrable. It also implies $\mathbb{E}[W\circ T]=\mathbb{E}[W]$ for every integrable random variable $W$, first for indicator functions, then for simple functions, and then by positive and negative approximation.

To identify a conditional expectation, take any $A\in \mathcal{F}$. The assumed identity with the bounded random variable $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.

For a set $M$ of finite measure apply the preceding result to $f=1_M$. Since its local average minus its point value is bounded in absolute value by its mean absolute oscillation, it givesFor a general measurable $M$, apply the same argument to $1_{M\cap [-j,j]}\in L^1(\mathbb{R})$. At every interior point of $(-j,j)$, sufficiently short averages agree with those of $1_M$. The countable union of these interiors covers the line, and a countable union of null exceptional sets remains null. This proves the Lebesgue density theorem without a finite-measure assumption on $M$. The source's $\chi$ is the indicator of $M$.

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.

A sigma-algebra on a set $\Omega$ is a collection $\mathcal{F}$ of subsets such that $\Omega \in \mathcal{F}$, $A\in \mathcal{F}$ implies $\Omega \setminus A\in \mathcal{F}$, and $A_n\in \mathcal{F}$ for all $n$ implies ${\bigcup }_n{A}_n\in \mathcal{F}$. Thus it contains the empty set and is closed under countable intersections as well. This is precisely the closure needed to apply countable additivity without leaving the domain of a measure.

For $\Omega =\{a,b,c\}$,is a sigma-algebra. It describes observations that distinguish $a$ from the pair $\{b,c\}$, but cannot distinguish $b$ from $c$. The power set provides a finer sigma-algebra with all eight subsets. Both are valid; choosing a sigma-algebra records which questions the model can answer.

On the real numbers, the Borel sigma-algebra contains all open sets and everything generated from them by these operations. The Lebesgue measurable sets form a larger sigma-algebra, including all subsets of Borel null sets. Closure under countable unions accommodates constructions involving an entire sequence of observations or approximations.

For a real random variable $X$ on a probability space, $\sigma (X)$ is the smallest sigma-algebra making $X$ a measurable function into the Borel sigma-algebra on $\mathbb{R}$. The collection of all inverse images of Borel sets is already a sigma-algebra, since inverse images preserve complements and countable unions; this proves the displayed formula without completing any measure.

The tail-sum formula expresses an expected value as the area under a survival function. For every nonnegative random variable $Y$,This Lebesgue integral and the expected value may both be $+\infty$; a probability density function is unnecessary. For a random variable $N$ taking values in the nonnegative integers, the corresponding sum isThe events $\{N>k\}$ and $\{N\ge k+1\}$ coincide because $N$ takes integer values. Thus the strict sum starts at zero, while the inclusive sum starts at one.^[ref]

For an original example, let the probability mass function of $N$ beIts expected value computed directly is $0(1/5)+1(3/10)+3(1/2)=9/5=1.8$. The strict tail probabilities at thresholds $0,1,2,3$ are $4/5,1/2,1/2,0$: their sum is again $1.8$. There is no probability mass at two, yet the event $\{N>2\}$ still has probability $1/2$. The survival function has heights $4/5,1/2,1/2$ on the successive intervals $[0,1),[1,2),[2,3)$; each interval has length one, so its integral is the same sum.

The indexing matters numerically. Starting the inclusive sum at zero adds $\mathbb{P}(N\ge 0)=1$, giving $2.8$ for this example. Starting the strict sum at one omits $\mathbb{P}(N>0)=4/5$, giving $1$. For a general nonnegative random variable, replacing $>$ with $\ge$ inside the integral is harmless: the difference is $\mathbb{P}(Y=t)$, nonzero at only a countable set of thresholds. For each positive integer $m$, at most $m$ thresholds can have probability mass at least $1/m$; their countable union contains all thresholds with positive probability mass and has Lebesgue measure zero. This reasoning concerns an integral, not equality of the two survival function conventions at every point.

For a continuous original example, take the uniform distribution on the interval $[0,4]$. Its survival function is $S(t)=1-t/4$ for $0\le t\le 4$ and zero for $t>4$. The area is a triangle of base four and height one:Directly integrating against the probability density function gives the same expected value, ${\int }_0^{4}y(1/4)dy=2$. The integral of the probability density function alone is one; it is the survival function whose area is the expected value.

The proof counts layers of each outcome. For a realized value $y\ge 0$, the indicator function $1_{\{y>t\}}$ equals one over an interval of length $y$, soTonelli theorem permits the interchange because the indicator function is nonnegative; no finite expected value is assumed. Similarly, an outcome $N=n$ contributes one indicator random variable for each of the $n$ events $\{N\ge k\}$ that occur:Again Tonelli theorem applies to the nonnegative sum, or equivalently the monotone convergence theorem applies to its partial sums. This exchanges counting layers within each outcome with averaging across outcomes. The existing integer-valued hypothesis in the tail-sum formula explains what changes when the values are not integers; the tail integral formula for moments weights the layers to obtain higher moments.

For an integrable random variable $X$ that can have either sign, apply the nonnegative formula separately to its positive part $X_{+}=\max (X,0)$ and negative part $X_{-}=\max (-X,0)$:Integrability makes both integrals finite. More generally the difference defines an extended expected value when at least one is finite; if both are infinite, subtraction does not define an expected value.^[ref]

Tail probabilities are often easier to measure or bound than a whole probability density function. For a device's survival time, the survival function records the probability of still operating at each threshold; its integral gives the mean operating time. A finite observation window naturally gives the expected value of a capped nonnegative random variable. For a mathematical bound, $\mathbb{P}(Y>t)\le e^{-t/2}$ for all $t\ge 0$ implies $\mathbb{E}[Y]\le {\int }_0^{\infty }{e}^{-t/2}dt=2$. The tail-sum formula turns control of a whole family of events into control of an average.

For a measurable space $(\Omega ,\mathcal{F})$ and any subset $E\subseteq \Omega$, the trace sigma-algebra is the displayed family of subsets of $E$. It contains $E=E\cap \Omega$, is closed under relative complements since $E\setminus (E\cap A)=E\cap (\Omega \setminus A)$, and is closed under countable unions. If $E\in \mathcal{F}$, it equals $\{B\in \mathcal{F}:B\subseteq E\}$. Without that measurability assumption, the traces need not be measurable sets in the original sigma-algebra.