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A premeasure on an algebra of sets extends to a measure on the generated sigma-algebra. The extension is unique when the premeasure is sigma-finite.

A sigma-finite measure on simple closed curves has conformal restriction if restricting to loops in a simply connected domain and then applying a conformal map agrees with first transporting the domain and restricting there. The usual nontrivial measures are finite on bounded macroscopic cutoff windows and are determined up to normalization by their pointed conformal deficits.

Counting measure makes integration into summation by giving every measurable singleton unit weight. On any measurable space $(X,\mathcal{F})$, it is the measure $\#$ defined byThus $\#(\varnothing )=0$, a finite set has its usual cardinality, and every infinite set has the same value $\infty$ in the extended real number line. In particular, countably infinite sets and uncountable sets are not distinguished by their counting measure. Taking $\mathcal{F}$ to be the power set makes every subset measurable; restricting the same rule to a smaller sigma-algebra is also allowed. ^[ref]

Why is this a countably additive measure? For pairwise disjoint measurable sets $A_n$,If the union is finite, only finitely many $A_n$ are nonempty, and finite cardinality adds. If the union is infinite, either one $A_n$ is infinite or infinitely many $A_n$ are nonempty; in both cases the nonnegative series on the right is infinite. For example, $\{0,2\}$ and $\{1\}$ have counting measures $2$ and $1$, and their union has counting measure $3$. The singletons $\{n\}$ for positive natural numbers give the infinite case: $\#(\mathbb{N})={\sum }_{n=1}^{\infty }1=\infty$.

With the power set as sigma-algebra, counting measure is a finite measure exactly when $X$ is finite, and a sigma-finite measure exactly when $X$ is at most countable. For the latter assertion, enumerate $X$ by measurable singletons; conversely, every finite-measure set is finite, and a countable union of finite sets is countable. Specifying the sigma-algebra matters: if $X=\mathbb{N}$ and $\mathcal{F}=\{\varnothing ,\mathbb{N}\}$, counting measure is not a sigma-finite measure, since the only measurable finite-measure set is the empty set.

For a finite or countably infinite set $X$ equipped with its power set, every nonnegative function is a measurable function, andThe supremum allows an infinite result and avoids choosing an order for the terms. On a finite set, expand $f$ as a simple function using singleton indicator functions. On a countably infinite set, enumerate the points and apply the monotone convergence theorem to finite truncations. Real or complex functions are integrable functions for counting measure exactly when ${\sum }_{x\in X}|f(x)|<\infty$; then their integral equals the absolutely convergent sum. This is the discrete counterpart of Lebesgue integration, developed further in series as a Lebesgue integral against counting measure. ^[ref]

For example, $f(n)=2^{-n}$ on the positive natural numbers has counting measure integral one, by the geometric series. The alternating harmonic series has a finite ordered sum but is not the integral of an integrable function for counting measure: its positive and negative parts both have infinite sums. Absolute convergence is what makes the finite integral independent of enumeration. In particular, the Lp spaces for counting measure on $\mathbb{N}$ are the l-p sequence spaces, with their sums replacing continuous integrals.

Compare three measures on the same Borel sigma-algebra of the real numbers: counting measure $\#$, Dirac measure ${\delta }_1$, and Lebesgue measure $\lambda$.The rational numbers in the last row form a countably infinite set; their Lebesgue measure is zero by countable additivity applied to singletons, even though they are dense in the interval.

These are different answers to different questions: counting measure asks how many points occur, Dirac measure asks whether the selected point occurs, and Lebesgue measure asks how much length occurs. A null set therefore depends on its measure. For counting measure, the only null set is the empty set, so equality almost everywhere is equality everywhere. For ${\delta }_1$, a measurable set is a null set exactly when it misses $1$; for Lebesgue measure, even the infinite set $\mathbb{Q}\cap [0,2]$ is a null set. Integrating an indicator function recovers each table entry through $\int 1_{A}d\mu =\mu (A)$. This lets the same Lebesgue integral framework handle point counts, point evaluation, and geometric length.

The essential range of a real or complex measurable function $m$ consists of the values $z$ whose every neighborhood receives inputs of positive measure:It depends on the measure as well as the function. This definition concerns values arbitrarily near $z$, not necessarily inputs at which $m=z$ exactly. It therefore retains meaningful limiting values while ignoring values occurring only on null sets. ^[ref].

For $m(x)=x$ on $(0,1)$ with Lebesgue measure, the ordinary image of a function is $(0,1)$, but the essential range is $[0,1]$. Neither endpoint is attained, yet inputs with $|m(x)|<\epsilon$ or $|m(x)-1|<\epsilon$ have positive measure for every $\epsilon >0$. Every interior value also belongs to the essential range, although its exact fiber is a singleton of zero measure. Positive measure of a fiber is sufficient, but is not necessary, for a value to belong to the essential range.

The two-sided example $g$ from essential infimum and essential supremum illustrates what disappears. It follows $g(x)=x$ on $[0,1]$, except $g(1/4)=-3$ and $g(3/4)=4$. Its ordinary image of a function is $([0,1]\setminus \{1/4,3/4\})\cup \{-3,4\}$, whereas its essential range remains $[0,1]$. The missing ordinary values $1/4$ and $3/4$ still have arbitrarily close neighbors with positive measure; the exceptional values $-3$ and $4$ can each be isolated by a sufficiently small neighborhood whose preimage is a null set.

For the three-point example in L-infinity norm, the values are $1,3,10$. With weights $(0.5,0.5,0)$ the essential range is $\{1,3\}$; with weights $(0.499,0.5,0.001)$ it is $\{1,3,10\}$. A small but positive probability retains ten, so an essential range does not discard a value merely because it is rare. For a real function on a space of positive total measure, the essential infimum and essential supremum are the lower and upper edges of its essential range, allowing infinite bounds. The L-infinity norm instead takes the upper edge of the essential range of $|m|$.

A useful operator connection is the spectrum of a real multiplication operator: on a nonzero L2 space over a sigma-finite measure, multiplication by a real measurable function $m$ has spectrum equal to its essential range, with the natural domain when $m$ is unbounded. Outside that range, $|m-z|$ has a positive lower bound almost everywhere, so multiplication by $1/(m-z)$ gives an inverse function that is a bounded linear operator. Near an essential value, this lower bound fails. Thus discarding null sets is precisely what makes the range match an operator acting on almost everywhere equivalence classes. ^[ref].

A sigma-finite measure on a standard Borel space is the intensity measure of a point process with Poisson counts. Partition the space into sets of finite intensity, draw an independent Poisson count on each piece, and conditional on that count draw independent points from its normalized intensity measure. The sum of their Dirac measures is the required Poisson random measure. Non-atomic intensity gives simplicity; local finiteness of the intensity gives local finiteness of the point configuration.

Assume the full-data sample space and the range of the statistic $T$ are standard Borel spaces, $T$ is a measurable function, and the statistical model $\{P_{\theta }\}$ is dominated by one common sigma-finite measure $\nu$. Write $f_{\theta }=d{P}_{\theta }/d\nu$ for its densities. Then $T$ is a sufficient statistic if and only if there exist nonnegative measurable functions $g_{\theta }$ and $h$, with $h$ independent of $\theta$, such thatThe statistical parameter can enter $g_{\theta }$ only through $\theta$ and the observed $T(x)$. ^[ref]

For discrete data, take counting measure; for ordinary continuous data, take Lebesgue measure. Independence, identical sampling laws, smooth dependence on $\theta$, and a fixed positive support of a probability distribution are not requirements of this theorem. Parameter-dependent support must nevertheless be included in $f_{\theta }$, usually through an indicator function, and must fit the displayed factorization. The sufficient statistic for a symmetric uniform sample illustrates exactly this: its indicator function depends on the unknown scale through the observed absolute maximum. The null sets in the theorem concern versions of densities; the factors are not generally themselves normalized probability distributions.

For a discrete sample space, the cancellation behind the theorem is particularly transparent. On a group with positive probability, factorization givesConversely, multiply a common conditional probability mass function by $P_{\theta }(T=T(x))$ to obtain the required factorization. For continuous data, a sum over a group is not a density integral over the original Lebesgue measure: the group may have measure zero. Use the regular conditional distribution formulation and the Fisher-Neyman factorization theorem instead of that discrete calculation.

Lebesgue measure on the real numbers extends interval length: $\lambda ([a,b])=b-a$ for $a\le b$. Its restriction to the Borel sigma-algebra is the unique translation-invariant sigma-finite measure on that sigma-algebra normalized by $\lambda ((0,1])=1$. The full Lebesgue measure is the completion of a measure of this restriction; its domain is the Lebesgue measurable sets. Translation invariance follows by comparing a translated Borel measure on half-open intervals and applying the uniqueness theorem for measures. For the construction and interval properties, see MIT lecture 8.

Every singleton has Lebesgue measure zero: $\{x\}\subseteq [x-\epsilon ,x+\epsilon ]$ gives $0\le \lambda (\{x\})\le 2\epsilon$ for every $\epsilon >0$. Enumerating the countable set $\mathbb{Q}\cap [0,1]$ and applying countable additivity yieldsThus a dense set can be a null set, and a union of null sets indexed by an uncountable set can have positive Lebesgue measure. The distinction explains both the Dirichlet function example and why almost everywhere is weaker than everywhere.

For a regular outer measure, its Caratheodory measurability criterion defines a maximal algebra of sets on which that outer measure is finitely additive. Indeed, suppose an additive algebra contains all measurable sets and also $A$. Given $E$ of finite outer measure, take a measurable hull $H\supseteq E$ of equal mass. Finite additivity splits $H$ by $A$, and monotonicity gives ${\mu }^{\ast }(E)\ge {\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\setminus A)$. Outer subadditivity gives equality. For infinite mass that equality already follows from outer subadditivity. Thus $A$ was measurable. No sigma-finite measure assumption is necessary.

The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be measure spaces with sigma-finite measures, and use their product measure.

Fubini's theorem: if $u$ is a measurable absolutely integrable function on $X\times Y$, then its sections are integrable for almost every fixed value of the other coordinate, the resulting section-integral functions are integrable, andAll three integrals are finite. Values of an inner integral on an exceptional null set can be assigned arbitrarily.

Tonelli's theorem: if $u:X\times Y\to [0,\infty ]$ is measurable and nonnegative, the same equality holds for its nonnegative product and iterated integrals, allowing the common value $+\infty$. No advance absolute integrability assumption is needed. On a completed Lebesgue measure product, the section and measurability assertions are interpreted almost everywhere.

Thus Tonelli's theorem can first establish that $\int |u|<\infty$; after this, Fubini's theorem permits signed integration in either order.

Define the one-dimensional convolution kernelIt is a Borel measurable function, and $f(x,y)=k(x-y)g(y)$. The function $k(x-y)$ is Borel measurable by composition with the continuous difference map. Lifting the Lebesgue measurable function $g$ to the second coordinate preserves measurability for the product sigma-algebra, hence also for Lebesgue measure on ${\mathbb{R}}^2$. Products of measurable functions are measurable, so $f$ is Lebesgue measurable.

Near zero, $|\sin t|\le |t|$; away from zero, $|\sin t|\le 1$. ConsequentlyBoth bounds are integrable on their respective regions by power singularity integrability. In particular,

We use the following forms of the two product-integration theorems. On the product of measure spaces equipped with sigma-finite measures, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, with infinity allowed. Fubini's theorem says that for an absolutely integrable measurable function, its sections are integrable for almost every fixed coordinate, and both orders of iterated integration equal the finite product integral. The Lebesgue measure on each factor is a sigma-finite measure, because the real line is a countable union of finite-length intervals.

Apply Tonelli's theorem first to $|f|$, before presuming finiteness. By translation invariance of Lebesgue measure,Thus $f$ is Lebesgue integrable on ${\mathbb{R}}^2$. The triangle inequality for an integral now gives the requested estimate:This is the L1 norm identity for a translated kernel product followed by a simple bound on the convolution kernel.

There is also a stronger signed conclusion. The integrable convolution kernel $k$ is an odd function, so ${\int }_{\mathbb{R}}k(t)dt=0$. Having established absolute integrability, we may apply Fubini's theorem to $f$ itself:Unlike part (a), the signed cancellation is justified by a finite integral of the absolute value. The constant eight is useful as an L1 norm bound even though this particular signed total vanishes.

On sigma-finite measure spaces, including the real line with Lebesgue measure, Tonelli's theorem says that a nonnegative measurable function $H$ has equal product and iterated integrals:with infinity allowed. Finiteness is not a prerequisite.

Fubini's theorem says that if a measurable signed or complex $H$ is absolutely integrable on the product, then its sections are integrable for almost every value of the other variable, the iterated integral functions are integrable, and the same equalities hold with finite values. For completed Lebesgue measure, exceptional sections need not themselves be measurable; the integrals use the almost-everywhere defined sections. Thus Tonelli can first establish absolute integrability using $|H|$, after which Fubini permits changing the order for $H$.

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.

A Poisson random measure with intensity the sigma-finite measure $\mu$ is a random nonnegative integer-valued measure $M$ such that, for each outcome outside a single null set, $A\mapsto M(A)$ is countably additive, and for each measurable $A$, $M(A)$ is a measurable random variable. For every measurable $A$ with $\mu (A)<\infty$,and for every finite collection of pairwise disjoint finite-intensity measurable sets, their counts are independent random variables. These are integer-valued counts, with possible value infinity on infinite-intensity sets. In particular $\mu (A)=0$ implies $M(A)=0$ almost surely. The intensity identity is $\mathbb{E}M(A)=\mu (A)$, also in the extended sense.

For completeness, choose an increasing finite-intensity exhaustion $E_m\uparrow E$. Each $M(E_m)$ is finite almost surely, simultaneously for all $m$, so $M$ is a sigma-finite measure almost surely. If $\mu (A)=\infty$, then $\mu (A\cap E_m)\to \infty$, and for any fixed integer $K$,Hence $M(A)=\infty$ almost surely. Independence for counts on arbitrary disjoint measurable sets follows by this exhaustion; an infinite count is a constant extended value. This convention completes the definition on a general measurable space, not just on bounded subsets of Euclidean space.

At criticality, percolation clusters occur on every scale. Count their filled outer perimeters, discard microscopic loops through a diameter cutoff, and then pass to a scaling limit. The cutoff gives a locally finite measure on loops of macroscopic size inside a bounded region; summing over scales gives a sigma-finite measure rather than a probability distribution.

The conformal invariance of planar percolation makes these limiting perimeter statistics covariant under conformal maps. Moreover, deciding whether a filled outer perimeter lies in a simply connected subdomain uses only the configuration it encloses: changing the configuration outside that subdomain does not change that perimeter. This is the restriction part of the argument. Conformal covariance together with this domain consistency motivates a nonzero conformal restriction measure on simple loops.

This is an informal construction principle, as requested; it does not replace the tightness, boundary-simplification and uniqueness theorems required to construct the continuum measure rigorously.

Let $P$ be the transition kernel of the Markov chain. It is phi-irreducible if there is a nonzero sigma-finite measure $\phi$ such that, for every measurable $A$ and every starting state $x$,Equivalently, there is positive probability of eventually reaching every set of positive reference measure. For a finite state space with counting measure, this becomes ordinary irreducible Markov chain accessibility. On a continuous space, requiring access to each individual point would usually be inappropriate; positive-measure sets play that role. Irreducibility requires positive probability, rather than probability one, and does not itself assert recurrence or a finite mean return time.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.

A real multiplication operator on its maximal $L^2$ domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on $\{k\le n\}$. If $h$ is in the adjoint operator domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.

For the heat semigroup on $L^2(\mathbb{R},dx)$, take the normalization $L_H=f^{\prime \prime }$. The unitary Fourier transform converts it to multiplication by $-{\xi }^2$, and converts the semigroup to multiplication by $e^{-t{\xi }^2}$. ConsequentlyThe Sobolev space domain is exactly $\{f:{\xi }^2\hat{f}\in L^2\}$. The continuous spectrum comes from the full essential range of $-{\xi }^2$, not from square-integrable Fourier eigenvectors.

For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials $h_n={\mathrm{He}}_n/\sqrt{n!}$ form an orthonormal basis. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}{h}_n=-n{h}_n$, so the Hermite coefficient map turns it into a real multiplication operator on ${\ell }^2({\mathbb{N}}_0)$. ThereforeFinally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported $\varphi$ and set $f_R(x)=R^{-1/2}\varphi (x/R)$. Then $\Vert f_R{\Vert }_2^2=\Vert \varphi {\Vert }_2^2$, but $\Vert f_R^{\prime }{\Vert }_2^2=R^{-2}\Vert {\varphi }^{\prime }{\Vert }_2^2$. This disproves a uniform whole-line Poincaré inequality $\Vert f{\Vert }_2^2\le C\Vert f^{\prime }{\Vert }_2^2$; one can also choose $\int \varphi =0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequalityEquality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

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

TakeThis is a finite measure, hence a sigma-finite measure. If $\nu (B)=0$, nonnegativity gives $\mu (B)={\mu }_0(B)=0$, so both $\mu$ and ${\mu }_0$ are absolutely continuous with respect to $\nu$.