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Let $M=\Vert f{\Vert }_{\infty }$, the essential supremum of $|f|$. If $M=0$, $f=0$ almost everywhere and every Lp norm is zero. Suppose $M>0$. Since $|f|\le M$ almost everywhere and $m((0,1))=1$,For any $a$ with $0<a<M$, the measurable set $A_a=\{x\in (0,1):|f(x)|>a\}$ has positive Lebesgue measure; otherwise $a$ would be an essential upper bound. HenceFor fixed $a$, $m(A_a{)}^{1/p}\to 1$. Thus ${\mathrm{liminf}}_{p\to \infty }\Vert f{\Vert }_p\ge a$. Letting $a\uparrow M$ and combining with the upper bound provesThe positive-measure superlevel sets explain why the Lp norms converge to the supremum norm even when the essential maximum is never attained.

Equip $S$ with a sigma-algebra $\mathcal{S}$. A Poisson random measure with intensity measure of a point process $\mu$ is a random counting measure $\Pi$ such that, for every measurable $A$ with $\mu (A)<\infty$,and the counts on every finite collection of pairwise disjoint measurable sets are independent. Thus $\mathbb{E}\Pi (A)=\mu (A)$. For a sigma-finite measure $\mu$, sets of infinite intensity contain infinitely many points almost surely: exhaust such a set by increasing finite-intensity sets and use the Poisson count tails as their means tend to infinity. The count is a measurable random variable for every measurable $A$, which is part of saying that $\Pi$ is a random measure.

Under the usual point representation write $\Pi ={\sum }_j{\delta }_{Z_j}$, counting multiplicities. If $\mu$ is a non-atomic measure, this is a simple Poisson point process; if $\mu$ has atoms, several point occurrences can have the same location, and each occurrence is counted separately. In a topological setting, local finiteness requires that the intensity be finite on compact sets.

Sufficient existence conditions are that $S$ is a standard Borel space and $\mu$ is a sigma-finite measure. This gives existence of a Poisson random measure on a standard Borel space. If a simple locally finite configuration is desired, additionally take a non-atomic, locally finite intensity on a locally compact second-countable Hausdorff space. These conditions are sufficient, not claimed necessary. In particular sigma-finiteness suffices for a countable family of occurrences that can each be independently coloured.

Let $R$ and $G$ be the red and green counting measures. The marking assumption means that, conditional on $\Pi$, the colours of all point occurrences are independent, with the same red probability $r$. For a measurable set $A$ of finite intensity $m=\mu (A)$, condition on its total count. To obtain $a$ red and $b$ green points the total must be $a+b$, so direct multiplication gives the joint red and green Poisson count formulaThus the two counts are independent Poisson variables with respective means $rm$ and $(1-r)m$. This computation includes zero intensity with the usual convention for the probability of zero counts.

For disjoint finite-intensity sets $A_1,\dots ,A_k$, the original counts are independent. Conditional marking in different sets uses disjoint collections of independent colours. Repeating the preceding calculation therefore givesThis proves the disjoint-count definition for each coloured Poisson point process and independence of their two count vectors.

To obtain independence of the whole processes, take any finite list of red test sets and any finite list of green test sets, initially all of finite intensity. Refine their union into the disjoint cells given by their membership patterns. Every test-set count is the sum of the counts of its cells. The entire red cell vector is independent of the entire green cell vector by the product formula, hence so are the corresponding test-set count vectors. For arbitrary measurable test sets, intersect first with a finite-intensity exhaustion of $S$ and pass to the increasing count limits. The count cylinder events generate the two evaluation sigma-algebras, and independence extended from generating pi-systems gives independence of those sigma-algebras. ConsequentlyThis is a direct count proof of the Poisson thinning theorem; no general marking or thinning result has been used to deduce it.

Represent the raindrops by a Poisson random measure on the space of location, impact time and volume. Physical volume is nonnegative, so take the intensity density to vanish for $v<0$, or equivalently work on ${\mathbb{R}}^2\times \mathbb{R}\times [0,\infty )$ and extend it by zero to ${\mathbb{R}}^4$. Assume the radius function is measurable and nonnegative for elapsed times at least zero. Its stated monotonicity means patches grow rather than dry away.

A drop contributes to wetting $\xi$ at time $\tau$ precisely if its impact time is no later than $\tau$ and its current patch contains $\xi$. Thus define the deterministic measurable setThe number of covering drops is $\Pi (A_{\xi ,\tau })$. Its intensity measure of a point process isSince the point is dry if and only if that count is zero, the Poisson count definition gives the void probability for growing Poisson rain patchesHere $e^{-\infty }=0$: when the intensity is infinite, exhaust the covering set by increasing finite-intensity sets; their decreasing zero-count events have probabilities tending to zero. This also covers storms with infinitely many drops that can reach the point. The formula is valid for spatially inhomogeneous intensity. Only if the intensity is spatially constant may the spatial integral be replaced by $\pi r(\tau -t,v{)}^2$ times that intensity. Whether the circular boundary is included makes no difference for a density with respect to Lebesgue measure, since each such boundary has zero spatial area. This construction is a time-dependent Poisson germ-grain model.

For total rainfall, the required general result is the Campbell first-moment formula: if $\Pi$ is a Poisson random measure with intensity $\mu$ and $h$ is any nonnegative measurable function, thenwith infinite values permitted. For a signed $h$, it holds with finite absolute expectation whenever $\int |h|d\mu <\infty$. No square-integrability hypothesis is required for this first-moment identity. The result is stated rather than proved, as allowed here.

The total rainfall volume is $V_{\mathrm{total}}={\sum }_{(X,T,V)\in \Pi }V=\int v\Pi (dxdtdv)$, counting every drop once, independently of overlaps between its patch and other patches. Apply the stated formula with $h(x,t,v)=v$ on the physical mark space. Extending by zero to negative volume givesIf the displayed nonnegative integral is finite, the total volume is integrable and finite almost surely. If it is infinite, its expectation is infinite; this alone does not imply that the random total is infinite almost surely. The radius affects wet-area coverage but does not change this total-volume expectation.

A measure space consists of a set $\Omega$, a sigma-algebra $\mathcal{A}$ of subsets, and a countably additive map $\mu :\mathcal{A}\to [0,\infty ]$ with $\mu (\varnothing )=0$. A sigma-algebra contains $\Omega$ and is closed under complements and countable unions. Countable additivity means $\mu ({\bigcup }_n{A}_n)={\sum }_n\mu (A_n)$ for pairwise disjoint measurable sets. Such sets are called measurable; a real function is measurable when the inverse image of every Borel set belongs to $\mathcal{A}$.

The Radon-Nikodym theorem for positive measures says: if $\mu$ and $\nu$ are sigma-finite measures on the same measurable space, and $\nu$ is absolutely continuous with respect to $\mu$, then there is a nonnegative measurable function $g$, unique $\mu$-almost everywhere, such thatHere sigma-finiteness means that the space is a countable union of measurable sets of finite measure; absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (E)=0$ implies $\nu (E)=0$. The function $g=d\nu /d\mu$ is the Radon-Nikodym derivative. For a finite signed or complex measure $\nu$ of finite total variation norm of a measure, absolutely continuous with respect to a sigma-finite $\mu$, the corresponding density belongs to $L^1(\mu )$. This follows by applying the positive theorem to the positive and negative parts of the real and imaginary parts of $\nu$.

For every measure space and $1<p<\infty$, $(L^p{)}^{\ast }$ is isometrically $L^q$, where $1/p+1/q=1$. We use the complex-linear pairingWith the convention $\int h\overline{g}$, the same identification is conjugate-linear in $g$. By Hölder's inequality, ${\Phi }_g$ is a bounded linear functional with $\Vert {\Phi }_g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, setinterpreting the numerator as zero where $g=0$. The identity $(q-1)p=q$ gives $\Vert h{\Vert }_p=1$ and ${\Phi }_g(h)=\Vert g{\Vert }_q$. ThusIt remains to represent an arbitrary $\Phi \in (L^p{)}^{\ast }$, rather than merely produce functionals from $L^q$.

We prove Lp duality on an arbitrary measure space without imposing sigma-finiteness on $\mu$. Write $C=\Vert \Phi \Vert$. For every measurable set $E$ with $\mu (E)<\infty$, define a complex measure on $E$ byIt is countably additive: for disjoint $F_j\subseteq E$, the partial sums of their indicator functions tend in $L^p$ to $1_{\bigcup F_j}$, because the measure of the omitted tail tends to zero. It is absolutely continuous with respect to $\mu {|}_E$, since indicator functions of null sets represent zero in $L^p$.

Its total variation norm of a measure is finite. For any finite measurable partition $E={\bigcup }_j{E}_j$, choose scalars $c_j$ of modulus $1$ with $c_j{\nu }_E(E_j)=|{\nu }_E(E_j)|$. ThenTaking the supremum over partitions gives the variation bound. Since $\mu {|}_E$ is finite, the Radon-Nikodym theorem supplies $g_E\in L^1(E)$ with ${\nu }_E(F)={\int }_F{g}_{E}d\mu$. By uniform approximation with simple functions,for every bounded measurable $h$ supported in $E$.

To improve $g_E$ from $L^1$ to $L^q$, test with the bounded functionIf $I_N={\int }_{E\cap \{|g_E|\le N\}}|g_E{|}^{q}d\mu$, thenThus $I_N^{1/q}\le C$ when $I_N>0$, and the same bound is trivial when it is zero. The monotone convergence theorem givesIf $E,F$ both have finite measure, the densities $g_E,g_F$ agree almost everywhere on $E\cap F$: their integrals over every measurable subset of the intersection equal the same functional value. This is uniqueness in the Radon-Nikodym theorem.

We now perform support localization of an Lp functional. SetChoose finite-measure sets $E_n$ whose displayed integrals tend to $M$, and let $D_n=E_1\cup \cdots \cup E_n$, $S={\bigcup }_n{D}_n$. If $M=0$, take $E_n=\varnothing$. Compatibility allows us to define a measurable function $g$ on $S$ by taking $g_{D_n}$ on the disjoint measurable sets $D_n\setminus D_{n-1}$, and put $g=0$ off $S$. It agrees almost everywhere with $g_{D_n}$ on every $D_n$. Moreover,Indeed each integral is at most $M$, and it is at least the integral over $E_n$, which tends to $M$.

For any finite-measure set $F\subseteq \Omega \setminus S$, compatibility on the disjoint union $D_n\cup F$ givesLetting $n\to \infty$ forces $g_F=0$ almost everywhere. For an arbitrary finite-measure set $E$, compatibility on $E\cap D_n$ and the preceding conclusion on $E\setminus S$ show that $g_E=g$ almost everywhere on $E$. Consequently $\Phi (h)=\int hgd\mu$ for every simple function supported on a finite-measure set.

Those simple functions are dense in $L^p$ even for this arbitrary measure space. To see the needed finite-support property, for $h\in L^p$ the sets $\{|h|>1/n\}$ have finite measure, bounded by $n^p\Vert h{\Vert }_p^p$. First truncate $h$ to such sets and to bounded values, then approximate by simple functions; the discarded $L^p$ integral tends to zero. Continuity of $\Phi$ and Hölder's inequality therefore extend the representation to every $h\in L^p$.

We have constructed $g\in L^q$ with $\Phi ={\Phi }_g$, and the previously proved norm identity gives $\Vert g{\Vert }_q=\Vert \Phi \Vert$. It also proves uniqueness: if ${\Phi }_g={\Phi }_{\tilde{g}}$, then $\Vert g-\tilde{g}{\Vert }_q=0$. This completes the isometric duality of Lp spaces.

The dominated sequence actually converges to zero in norm, and hence weakly. The assumptions give $|f_n{|}^p\le |f{|}^p\in L^1$ and $|f_n{|}^p\to 0$ almost everywhere. The dominated convergence theorem yieldsFor every $\Phi \in (L^p{)}^{\ast }$, $|\Phi (f_n)|\le \Vert \Phi \Vert \Vert f_n{\Vert }_p\to 0$. Therefore

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.

If $S\subseteq M_{-}$, then ${\mu }^{+}(S)=0$, so $\mu (S)=-|\mu |(S)\le 0$. Suppose instead that $S\cap M_{-}=\varnothing$ but $\mu (S)<0$. The preceding construction gives $R\subseteq S$ with ${\mu }^{+}(R)=0$ and $|\mu |(R)=-\mu (R)>0$. Then $M_{-}\cup R$ is ${\mu }^{+}$-null with variation strictly larger than that of $M_{-}$, a contradiction. Hence $\mu (S)\ge 0$ on the complement of $M_{-}$. This is a Hahn decomposition.

On $M_{+}=X\setminus M_{-}$ every partition has nonnegative signed measures, so $|\mu |(E)=\mu (E)={\mu }^{+}(E)$ for $E\subseteq M_{+}$. On $M_{-}$, ${\mu }^{+}$ vanishes. Splitting an arbitrary measurable set between these two sets givesIn particular ${\mu }^{+}\ll |\mu |$, and the exhibited Radon-Nikodym derivative verifies the Radon-Nikodym theorem for this pair directly.

An outer measure on $X$ is a function $\mu :\mathcal{P}(X)\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$, monotonicity, and countable subadditivity:Its measurable sets are those satisfying Caratheodory measurability criterion:The test runs over all subsets $E$, including nonmeasurable ones.

For $E\subseteq {\mathbb{R}}^n$, define the Lebesgue outer measure bywhere the boxes are bounded and $\mathrm{vol}(Q_j)={\prod }_r(b_{jr}-a_{jr})$. The empty covering handles the empty set. Monotonicity follows by inclusion of admissible covers. Taking covers within $\epsilon 2^{-j}$ of the outer measures of a sequence of sets proves countable subadditivity; when the sum is infinite the inequality is automatic. Lebesgue measure is the restriction of ${\mathcal{L}}^{\ast }$ to its Caratheodory-measurable sets.

If $d(A,B)=\delta >0$, subdivide each box of any countable cover of $A\cup B$ into finitely many half-open boxes of diameter smaller than $\delta$, preserving its total volume. No small box can meet both $A$ and $B$. The boxes meeting $A$ cover $A$, and the disjoint family meeting $B$ covers $B$, so the original cover costs at least ${\mathcal{L}}^{\ast }(A)+{\mathcal{L}}^{\ast }(B)$. Taking the infimum and then using subadditivity givesThus the outer measure underlying Lebesgue measure is a metric outer measure, and in particular every Borel set is Lebesgue measurable.

The restriction of a measure to a measurable subset is $(\mu {|}_M)(A)=\mu (A\cap M)$ for every measurable $A$. Absolute continuity of measures, written $\mu \ll \kappa$, means $\kappa (A)=0⟹\mu (A)=0$. The measures are mutually singular when a measurable $S$ satisfies $\kappa (S)=0$ and $\mu (X\setminus S)=0$. Thus singularity means concentration on a null set of the other measure; it does not assert that every measurable set is null for one of them.

Choose measurable sets $S_i$ carrying ${\mu }_i^s$ with $\kappa (S_i)=0$, and put $S={\bigcup }_i{S}_i$. Every absolutely continuous component vanishes on $S$, and every singular component vanishes on its complement. The given restriction identity therefore givesThe first is singular with respect to $\kappa$. The second is absolutely continuous because it is bounded above by ${\sum }_i{\mu }_i^a$. They sum to $\mu$, so uniqueness identifies them as its Lebesgue components.

Let $h_i=d{\mu }_i^a/d\kappa$ and $h={\max }_i{h}_i$, using measurable representatives. Each partition sum is at most ${\int }_{A}hd\kappa$. For the reverse inequality, partition $A$ measurably by assigning a point to the smallest index attaining this finite maximum. On its assigned cell $h_i=h$, and the partition sum is exactly ${\int }_{A}hd\kappa$. ThusThe maximum of finitely many integrable nonnegative densities is integrable, since it is at most their sum.

A set $H$ is Lebesgue measurable when the Caratheodory measurability criterion holds:Subadditivity always gives the inequality with left side at most right side. For $H=\{x_i>t\}$, split each covering box $Q_j$ across the plane $x_i=t$. Its two pieces have total volume $|Q_j|$. The piece with a closed cut face can be enlarged by arbitrarily little into an open box; hence its outer measure is at most its corresponding volume, just as for the other piece. Subadditivity applied separately on the two sides consequently givesFor finite outer measure take a covering sum arbitrarily close to ${\mathcal{L}}^{\ast }(E)$; infinite outer measure follows already from subadditivity. Thus every coordinate half space is measurable.

Carathéodory's theorem says that the measurable sets form a sigma-algebra and the outer measure restricted to it is a complete measure. Coordinate half spaces generate all open boxes and then all open sets, using countably many rational-endpoint boxes. They therefore generate the Borel sigma-algebra. Hence every Borel set is Lebesgue measurable.

Write $s={\sigma }^2\ge 1$. The two multivariate normal distribution probability density functions have ratioHere $\Vert x\Vert$ is the Euclidean norm. Consequently the smallest rejection sampling envelope constant, and its acceptance probability, areThe supremum is attained at $x=0$; hence no smaller constant can dominate the ratio. Draw $Y\sim N(0,s{I}_d)$ and an independent $U$ from the uniform distribution on $(0,1)$. Accept $Y$ whenotherwise repeat with fresh independent random variables. One can generate $Y=\sqrt{s}G$ using a random vector $G$ with independent standard normal distribution coordinates.

For any measurable set $B$, the joint acceptance probability is ${\int }_{B}q(y)p(y)/(M_{\ast }q(y))dy=M_{\ast }^{-1}{\int }_{B}p(y)dy$. Dividing by $1/M_{\ast }$ proves that the returned random vector has probability density function $p$. The number of proposals has a geometric distribution on the positive integers, with expected value $M_{\ast }$. When $s=1$, every proposal is accepted.

Integrating the envelope $p\le Mq$ shows $M\ge 1$. One trial of rejection sampling draws $Y\sim q$ and an independent $U$ from the uniform distribution on $(0,1)$, and accepts $Y$ if $U\le p(Y)/(Mq(Y))$. Repeat with fresh independent random variables after each rejection. Ratios need only be evaluated where $q>0$; the envelope makes the target zero wherever $q=0$.

For any measurable set $B$, integration of the conditional probability of acceptance givesTaking the whole domain gives acceptance probability $1/M$, and dividing gives $\mathbb{P}(Y\in B\mid \mathrm{accept})={\int }_{B}p(y)dy$. Fresh trials therefore supply independent and identically distributed random variables $X_1,\dots ,X_m$ with target probability density function $p$. The corresponding Monte Carlo estimator isThe proposal count $T$ up to and including one acceptance has the geometric distributionFor example, the tail-sum formula gives $\mathbb{E}T={\sum }_{t\ge 0}(1-1/M{)}^t=M$. Thus the expected cost is $M$ proposal draws per accepted target draw.

For a measurable set $C$, (i) gives the joint acceptance probabilityDivide by $Z_p/(M{Z}_q)$ from (ii). The resulting conditional probability isThus every accepted value has target probability density function $p$. Repeating rejected trials with fresh independent random variables does not change this law: the contribution from acceptance on trial $j$ is multiplied by $(1-Z_p/(M{Z}_q){)}^{j-1}$, and summing that geometric series gives the same conditional law. Independent repetitions produce independent target draws.

A Markov chain with transition kernel $P$ is $\phi$-irreducible if $\phi$ is a nonzero sigma-finite measure and every measurable set $A$ with $\phi (A)>0$ satisfies ${\sum }_{n\ge 1}{P}^n(x,A)>0$ from every starting state $x$. This is positive reachability, not certain eventual hitting.

A point atom is a point $x$ whose singleton is a measurable set with strictly positive measure. That singleton is an atom of a measure. This is more specific than a general atom of a measure, which is defined as a positive-measure set that cannot be divided into two measurable sets of positive measure.

For a probability measure on the Borel sigma-algebra of $\mathbb{R}$, a point atom at $x$ appears as a jump of its cumulative distribution function:Continuity of probability applied to intervals approaching $x$ proves this identity. Keeping such jumps is essential when deciding whether an endpoint belongs to an integration region.

A Poisson point process is a point process with independent Poisson distribution counts on disjoint measurable sets. Its intensity measure $\mu$ specifies those counts: for $\mu (A)<\infty$,Thus both the expected value and variance of $N(A)$ are $\mu (A)$. The independence requirement concerns disjoint test sets; overlapping sets can share points and have dependent counts.^[ref]

For a homogeneous Poisson point process in Euclidean space, $\mu (A)=\lambda |A|$, with constant intensity $\lambda$ and volume $|A|$. For example, intensity $2$ on the real line gives expected count $2$ and empty-interval probability $e^{-2}$ in a unit interval. Counts in two disjoint unit intervals are independent.

The counting measure formulation records multiplicity. A non-atomic intensity measure gives a simple point process, with no coincident points; an atom of a measure can instead carry a random multiplicity. Poisson random measures make that counting formulation explicit, while Poisson thinning theorem and the Superposition theorem for Poisson point processes explain useful ways of constructing related processes.

A Poisson random measure $N$ with intensity measure $\mu$ assigns independent Poisson random variables to disjoint measurable sets, with $N(A)$ having mean $\mu (A)$. Its compensated version is $\tilde{N}=N-\mu$ on sets of finite intensity.

A Harris recurrent chain visits every measurable set of positive irreducibility measure almost surely. Positive Harris recurrence additionally supplies an invariant Probability measure. The ergodic theorem for a positive Harris recurrent Markov chain gives consistency of time averages of integrable functions.