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The uniqueness theorem for measures says that two measures agreeing on a generating pi-system agree on its generated sigma-algebra, provided they are finite with the same total mass, or there is a common increasing exhaustion by pi-system sets of finite measure.

For the finite case let $\mathcal{D}=\{A:\mu (A)=\nu (A)\}$. Equality of total masses gives closure under complements, and countable additivity gives closure under disjoint countable unions. Thus $\mathcal{D}$ is a Dynkin system containing the generating pi-system. The pi-lambda theorem puts the entire generated sigma-algebra inside $\mathcal{D}$.

For the exhaustion version, let $E_n\uparrow E$ with $E_n$ in the pi-system and $\mu (E_n)=\nu (E_n)<\infty$. The finite measures $A\mapsto \mu (A\cap E_n)$ and $A\mapsto \nu (A\cap E_n)$ agree on the generating pi-system, since it is intersection-closed. The finite proof applies, and continuity from below of a measure as $n\to \infty$ gives $\mu (A)=\nu (A)$ for every measurable $A$. A finiteness hypothesis is essential to this standard extension-uniqueness statement.

Let $D_n=B_n\setminus B_{n+1}$. These sets are pairwise disjoint andBy countable additivity, ${\sum }_n\mu (D_n)=\mu (B_1)-\mu (B)<\infty$. The tails of this convergent nonnegative series tend to zero, soThis proves continuity from above of a measure and explains the finite-measure assumption.

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.

No. On the real line, the disjoint bounded Borel sets $[n,n+1)$, $n\ge 0$, each have assigned mass zero, while their union $[0,\infty )$ has assigned mass one. This contradicts countable additivity, so the set function is not a Borel measure.

If $\theta$ is measure-preserving, equality holds on every measurable set and hence on the specified $\pi$-system. Conversely, let $\mathcal{D}=\{A\in \mathcal{E}:{\mu }_{\theta }(A)=\mu (A)\}$. Both finite measures have equal total mass. Therefore $E\in \mathcal{D}$, complements of members remain in $\mathcal{D}$ by subtracting from the finite total mass, and disjoint countable unions remain in $\mathcal{D}$ by countable additivity. Thus $\mathcal{D}$ is a Dynkin system.

It contains the generating pi-system $\mathcal{A}$, so Dynkin lemma gives $\mathcal{E}=\sigma (\mathcal{A})\subseteq \mathcal{D}$. ConsequentlyThis is exactly the measure-preserving property.

The Kolmogorov axioms for a probability measure $\mathbb{P}$ are nonnegativity, normalization $\mathbb{P}(\Omega )=1$, and countable additivity on pairwise disjoint events. For $\mathbb{P}(B)>0$, conditional probability is defined byBecause the events $A\cap B_i$ are pairwise disjoint and their union is $A$, countable additivity gives the law of total probabilityFor any $j$ with $\mathbb{P}(A)>0$, the definition of conditional probability and this identity give Bayes' theorem:

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.

The integers form a countable set, and the events $\{X=k\}$ are disjoint. By countable additivity and part ii,Countability is essential: one cannot use the same sum argument over an arbitrary uncountable collection of zero-probability events.

For each nonnegative integer $z$, the events $\{X=z-y,Y=y\}$, $0\le y\le z$, are pairwise disjoint and their union is $\{X+Y=z\}$. By countable additivity and then independence,This is the discrete convolution of the two probability mass functions. If the convention for natural numbers excludes zero, set both probability mass functions equal to zero there; the formula remains valid. Also $p_Z(z)=0$ for negative $z$.

A sigma-algebra $\mathcal{F}$ on $\Omega$ contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. Consequently it contains the empty set and is closed under countable intersections. A measure space is a triple $(\Omega ,\mathcal{F},\mu )$ with a map $\mu :\mathcal{F}\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$ and countable additivity on pairwise disjoint measurable sets.

For the increasing sequence, set $C_1=A_1$ and $C_n=A_n\setminus A_{n-1}$ for $n\ge 2$. These measurable sets are disjoint, their union is $A$, and $A_N={\bigcup }_{n=1}^N{C}_n$. Thus countable additivity and finite additivity giveThe identity permits the value infinity; it uses no subtraction of infinite measures. This is continuity from below of a measure.

For a decreasing sequence, the analogous assertion is not always true. On $\Omega =\mathbb{N}$ with the counting measure, take $B_n=\{n,n+1,\dots \}$. Then ${\bigcap }_n{B}_n=\varnothing$, so $\mu (B)=0$, whereas $\mu (B_n)=\infty$ for every $n$ and their limit is infinity.

A sufficient extra condition for continuity from above of a measure is $\mu (B_1)<\infty$. Indeed $B_1\setminus B_n$ increases to $B_1\setminus B$, and the increasing-sequence result givesMore generally it suffices that one of the $B_n$ has finite measure. The missing finiteness assumption is exactly what the counterexample violates.

Set $H={\int }_0^{\pi }|h(t)|dt<\infty$. By periodicity and countable additivity of the Lebesgue integral of a nonnegative function,Thus $f$ is Lebesgue integrable for every $\beta >0$.

Fix $A\in {\mathcal{A}}_1$ and defineThis is a Dynkin system: it contains $\Omega$; taking complements preserves the equality because $\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)$; and countable additivity preserves it under disjoint unions. The hypothesis says that ${\mathcal{D}}_A$ contains the pi-system ${\mathcal{A}}_2$. Hence Dynkin lemma gives ${\mathcal{D}}_A={\mathcal{G}}_2$.

Now fix any $B\in {\mathcal{G}}_2$ and putThe same complement and disjoint-union calculations make this a Dynkin system. The first step shows that it contains ${\mathcal{A}}_1$, so Dynkin lemma gives ${\mathcal{E}}_B={\mathcal{G}}_1$. ThusThis is precisely the definition of independent sigma-algebras, and proves independence extended from generating pi-systems.

Start with $R_0=A$. Having chosen disjoint $A_1,\dots ,A_i$, put $R_i=A\setminus {\bigcup }_{j\le i}{A}_j$ andThe empty set shows $s_i\ge 0$, and finite total variation bounds it above. If $s_i>0$, choose $A_{i+1}\subseteq R_i$ with $\mu (A_{i+1})\ge s_i/2>0$; if $s_i=0$, choose $A_{i+1}=\varnothing$. This is the required recursion, including the initial step $i=0$.

Disjointness and nonnegativity give ${\sum }_i\mu (A_i)\le |\mu |(A)<\infty$, hence $\mu (A_i)\to 0$. Write $R=A\setminus {\bigcup }_i{A}_i$. For every measurable $B\subseteq R$,Thus every measurable subset of $R$ has nonpositive measure. Each finite partition of $R$ consequently has variation sum exactly $-\mu (R)$, and therefore $|\mu |(R)=-\mu (R)$. It follows that ${\mu }^{+}(R)=0$. Finally countable additivity gives

From now on $|\mu |(X)<\infty$, so all signed quantities below are finite. Since $|\mu |(E)\ge |\mu (E)|$, the set functionis nonnegative. Countable additivity follows from that of the two terms, so ${\mu }^{+}$ is a finite positive measure.

Set $s=\sup \{|\mu |(M):{\mu }^{+}(M)=0\}$. This is finite. Choose measurable $M_j$ with ${\mu }^{+}(M_j)=0$ and $|\mu |(M_j)\to s$. Thenis ${\mu }^{+}$-null by countable subadditivity, and monotonicity gives $|\mu |(M_{-})\ge s$. The definition of $s$ gives the reverse inequality. Hence $M_{-}$ attains the maximum, rather than merely approximating it.

For a signed measure define its total variation measure byThe value at the empty set is zero. For disjoint measurable sets $A_j$ with union $A$, any finite partition $E_r$ of $A$ givesTaking the supremum proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus {\bigcup }_{j\le k}{A}_j$ givesIf one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\epsilon \downarrow 0$ and $k\to \infty$ proves countable additivity, including infinite values. Thus $|\mu |$ is a positive measure. This uses only countable additivity of $\mu$, without assuming a Hahn decomposition.

Write $J(A)$ for the partition supremum. Empty cells are permitted, so $J\ge {\mu }_i$ for each $i$, and $J(A)\le {\sum }_i{\mu }_i(A)$ is finite. Also $J(\varnothing )=0$.

Let $A={⨆}_{j\ge 1}{B}_j$. Any measurable partition $A={⨆}_i{A}_i$ yieldsTaking its supremum gives one inequality. Conversely, for any $\epsilon >0$ choose for each $j$ a partition $B_j={⨆}_i{B}_{ji}$ whose sum exceeds $J(B_j)-\epsilon 2^{-j}$. The sets $A_i={⨆}_j{B}_{ji}$ partition $A$, and countable additivity of each ${\mu }_i$ gives $J(A)\ge {\sum }_{j}J(B_j)-\epsilon$. Let $\epsilon \downarrow 0$. Thus $J$ is a finite measure.

If another measure $\nu$ dominates every ${\mu }_i$, then for every such partition ${\sum }_i{\mu }_i(A_i)\le {\sum }_i\nu (A_i)=\nu (A)$. Consequently $J\le \nu$. Hence $J={\bigvee }_i{\mu }_i$ is the least dominating measure, the finite supremum of positive measures. It generally differs from the pointwise maximum of the numbers ${\mu }_i(A)$: the maximizing index may differ between pieces of $A$.

Define $\mathbb{Q}$ on ${\mathcal{F}}_t$ by density $Z_t$. Nonnegativity and countable additivity of the integral make it a measure, and $\mathbb{Q}(\Omega )={\mathbb{E}}_P{Z}_t=1$. Strict positivity of $Z_t$ also makes the two measures equivalent. The Radon-Nikodym derivative identity gives ${\mathbb{E}}_Q[1_{A}Y]={\mathbb{E}}_P[1_{A}Y{Z}_t]$, first for simple variables and then for every Q-integrable $Y$ by approximation.

Write $U={\mathbb{E}}_Q[Y\mid {\mathcal{F}}_s]$. The density on ${\mathcal{F}}_s$ is $Z_s={\mathbb{E}}_P[Z_t\mid {\mathcal{F}}_s]$. For every $A\in {\mathcal{F}}_s$,Moreover $Z_{s}U$ is P-integrable because ${\mathbb{E}}_P[Z_s|U|]={\mathbb{E}}_Q|U|<\infty$. Uniqueness in the definition of conditional expectation provesThis is conditional expectation under a martingale density.

The partition theorem is the law of total probability. If $B_1,B_2,\dots$ form a finite or countable partition of a sample space into pairwise disjoint events with positive probabilities, then for any event $A$,Indeed, $A$ is the disjoint union of the events $A\cap B_j$, so countable additivity gives the formula after applying the definition of conditional probability. Zero-probability members can be omitted.

Let $(B_i)$ be a finite or countable partition of a sample space into events of positive probability. The events $A\cap B_i$ are disjoint and their union is $A$. Countable additivity and the definition of conditional probability therefore give the law of total probabilityCells of zero probability contribute zero through $\mathbb{P}(A\cap B_i)$ and can be omitted rather than assigning an undefined conditional probability.