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Because each increment is $\pm 1$, the event at time $k$ defining $T$ is exactly $\{X_{k-1}=X_k=1\}$. Therefore, for $n\ge 2$,and for $n<2$ this event is empty. Thus $T$ is a stopping time for the natural filtration.

On the other hand, $\{U\le 0\}=\{T=2\}=\{X_1=X_2=1\}$. This event has probability $1/4$, whereas the initial sigma-algebra ${\mathcal{F}}_0$ is trivial, even after completion up to null sets. It therefore cannot belong to ${\mathcal{F}}_0$. Hence $U=T-2$ is not a stopping time. Subtracting a deterministic delay from a stopping time can require information from the future.

Define $U(x)=U_0({\delta }_x)$, where ${\delta }_x$ is the Dirac measure at $x$. For the full sigma-algebra of subsets of a finite $E$, affineness directly gives $U_0(\lambda )={\sum }_{x\in E}\lambda (\{x\})U(x)$.

The statement also holds for an arbitrary sigma-algebra on a finite set. List each atom of a sigma-algebra as $A_1,\dots ,A_r$ and choose $x_j\in A_j$. Points in one atom have identical Dirac measures on this sigma-algebra, so $U$ is constant on each atom and is measurable. Every probability measure decomposes as $\lambda ={\sum }_j\lambda (A_j){\delta }_{x_j}$, whenceThis is the expected utility representation on a finite measurable space. It does not silently assume that each singleton is measurable.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.

SetThen $B\in \Sigma$ and $\mu (B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative measurable functions by approximation with simple functions and the monotone convergence theorem. Consequently,and, for every $A$,HenceThis proves Lebesgue decomposition from a sum-measure density. The argument includes the cases of zero measures; there is no division on $B$, where the denominator vanishes.

Absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If $\nu \ll \mu$, then $\nu (B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu (A)={\int }_{A}fd\mu$ with a nonnegative integrable $f$, that integral is zero on every $\mu$-null set. ThusThis is the finite-measure Radon-Nikodym theorem as a consequence of the proved construction.

Finally restrict the base measure to the sub-sigma-algebra ${\Sigma }_0$. For a real $f\in L^1(\mu )$, write $f=f^{+}-f^{-}$, with both parts nonnegative and integrable. On ${\Sigma }_0$ define finite positive measuresThey are absolutely continuous with respect to $\mu {|}_{{\Sigma }_0}$. The result just proved gives nonnegative, ${\Sigma }_0$-measurable densities $g_{\pm }$ satisfyingSet $f_0=g_{+}-g_{-}$. Since $\int (g_{+}+g_{-})d\mu =\int |f|d\mu <\infty$, this is in $L^1(\Omega ,{\Sigma }_0,\mu )$, and subtraction givesFor complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is conditional expectation from finite-measure densities. The resulting conditional expectation is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in ${\Sigma }_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.

A probability space consists of a sample space $\Omega$, a sigma-algebra $\mathcal{F}$ of events, and a probability measure $\mathbb{P}$ satisfying:

For arbitrary events, Boole's inequality is $\mathbb{P}({\bigcup }_j{A}_j)\le {\sum }_j\mathbb{P}(A_j)$. To prove it, replace each $A_j$ by $D_j=A_j\setminus {\bigcup }_{i<j}{A}_i$. The $D_j$ are disjoint, have the same union, and satisfy $D_j\subseteq A_j$. Countable additivity and monotonicity give $\mathbb{P}({\bigcup }_j{A}_j)={\sum }_j\mathbb{P}(D_j)\le {\sum }_j\mathbb{P}(A_j)$. Finite unions are included by taking subsequent events empty.

Let $H_i$ be the $i$th heads event. For every $m$, the event of infinitely many heads lies inside ${\bigcup }_{i\ge m}{H}_i$. By the union bound, its probability is at most ${\sum }_{i\ge m}{p}_i$, which tends to zero. Thus the probability of infinitely many heads is $0$. This is the First Borel-Cantelli lemma, and requires no independence of the coin tosses.

For the dice experiment there are $4$ ordered outcomes giving a sum of $5$ and $6$ giving $7$, out of $36$. Each trial therefore has $p_5=1/9$, $p_7=1/6$ and irrelevant-outcome probability $13/18$. By independence between trials, summing over the number of initial irrelevant outcomes givesThe geometric series also shows that one of the two relevant sums eventually appears almost surely. Equivalently, first-step analysis gives $p=p_5+(13/18)p$.

An independence graph of events, also called a dependency graph of events, has one vertex for each event $E_i$, with $E_i$ independent of the entire family of events indexed by its nonneighbors. Precisely, $E_i$ is independent of the sigma-algebra generated by those events. Pairwise independence alone is insufficient.

The asymmetric Lovász local lemma says that if numbers $0\le x_i<1$ satisfythen $\mathbb{P}({\bigcap }_i{E}_i^c)\ge {\prod }_i(1-x_i)>0$ for a finite family. In the commonly used symmetric Lovász local lemma, if each probability is at most $q$ and the maximum graph degree is at most $d$, the sufficient condition is

The conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ is a $\mathcal{G}$-measurable integrable random variable $Z$ such thatIt is defined up to almost sure equality. Measurability with respect to the sub-sigma-algebra and equality of these integrals are both essential: the first expresses that only the information in $\mathcal{G}$ is retained, and the second preserves all averages visible through that information. Existence follows from the Radon-Nikodym theorem; uniqueness is up to sets of probability zero.

On a fixed underlying set, a pi-system is a nonempty family closed under finite intersections. A Dynkin system, also called a d-system, contains the whole underlying set, is closed under complements relative to it, and under countable disjoint unions. Equivalently it is closed under differences of nested members and increasing countable unions. A sigma-algebra contains the whole set, is closed under complements and arbitrary countable unions, hence also countable intersections. A Dynkin system need not be a sigma-algebra unless additional intersection closure is available.