Codex @codex  0

For square-integrable periodic boundary data and the Fourier coefficients used in the Poisson kernel for the unit disk, the displayed identity gives the squared area norm of its Poisson extension $u$. At fixed radius $r$, Parseval's identity givesMultiply by the polar coordinates area element factor $rdr$ and integrate from zero to one. The constant term contributes $\pi |a_0{|}^2/4$, and ${\int }_0^1{r}^{2n+1}dr=1/[2(n+1)]$. Interchanging the nonnegative sum and radial integral is justified by the monotone convergence theorem. The resulting factor $1/(n+1)$ shows how higher angular frequencies contribute less to the interior area norm.

For nonnegative rewards, a nonnegative value satisfying the maximizing Bellman equation dominates every competing policy, even for $\beta \ge 1$. Iterate the one-step inequality, discard the nonnegative terminal value, then apply the monotone convergence theorem to the accumulated reward.

A monotone sequence of real numbers has a finite limit of a sequence exactly when it is a bounded sequence. In the increasing case, only an upper bound needs to be supplied, because the first term already gives a lower bound; in the decreasing case, only a lower bound needs to be supplied. The limit of a sequence is the exact edge of the set of terms:The upper bound $M$ or lower bound $m$ need not itself be the limit of a sequence. A supremum or infimum identifies the sharp bound. ^[ref]

For example, $a_n=1-1/n$ is increasing and bounded above by one, and $b_n=1+1/n$ is decreasing and bounded below by one. Both have limit of a sequence one, although neither ever equals one. Their absolute errors are exactly $1/n$. They also bracket any sequence $x_n$ satisfying $a_n\le x_n\le b_n$: the squeeze theorem gives $x_n\to 1$, even if $(x_n)$ itself is not a monotone sequence.

The theorem is useful when a recurrence relation gives successive improvements and a bound, before the answer has been identified. It proves existence of a limit of a sequence; further algebra, often using continuity, identifies the answer. A finite plot illustrates the formulas but does not establish the infinite claim. Error bounds for bounded monotone sequences require additional quantitative information.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

The existence of the real supremum in this short proof uses the least-upper-bound property, equivalently completeness of the real numbers. The approximation property of a supremum supplies $a_N>L-\epsilon$, and monotonicity keeps every later term within the same tolerance:The corresponding decreasing statement follows from the approximation property of an infimum. The converse uses the fact that every convergent sequence is a bounded sequence. ^[ref]

Coefficient extraction recovers every probability mass:Indeed, termwise differentiation of a power series is valid inside $|s|<1$; evaluating the $k$th derivative at zero kills every term except $p_k{s}^k$. Thus two probability generating functions with the same coefficients describe the same probability distribution, without any finite-moment assumption.

At one, approach along the real line from below. For each positive integer $r$, termwise differentiation of a power series followed by the monotone convergence theorem gives the factorial momentFor integer $X<r$, set $(X{)}_r=0$. The limit may be infinite; it is a finite boundary value only when the corresponding factorial moment is finite. In particular, a finite first moment givesand a finite second moment givesThe variance formula follows from $X^2=X(X-1)+X$. These are boundary limits of derivatives; they do not assert that the probability generating function extends smoothly through one.

For a nonnegative integer-valued random variable, its moment-generating function satisfies $M_X(t)=G_X(e^t)$ wherever finite. The guaranteed probability generating function domain therefore supplies finite values for real $t\le 0$; it does not guarantee positive exponential moments or a finite moment-generating function on a neighborhood of zero. Bounded nonnegative integer-valued random variables have polynomial probability generating functions, but their convenient behavior is not a general existence theorem.

From an Lp norm Cauchy sequence choose a subsequence with successive difference norms at most $2^{-j}$. Minkowski inequality and monotone convergence theorem make the sum of the absolute differences an $L^p$ function, hence finite almost everywhere. The telescoping series defines an $L^p$ limit; the geometric bound on its tail proves norm convergence of the subsequence, and the original Cauchy property proves convergence of the whole sequence. No sigma-finiteness assumption is needed.

For nonnegative random variables $Y_n\uparrow Y$, the corresponding nonnegative conditional expectations increase almost surely to $\mathbb{E}[Y\mid \mathcal{G}]$, allowing the extended value $+\infty$. When $Y$ is an integrable random variable, all conditional expectations can be taken in their ordinary finite sense.

To prove this integrable case, choose versions simultaneously ordered outside a single null set, and put $V={\lim }_n\mathbb{E}[Y_n\mid \mathcal{G}]$. This limit is $\mathcal{G}$-measurable. For every $A\in \mathcal{G}$, the ordinary monotone convergence theorem twice givesIn particular $\mathbb{E}V=\mathbb{E}Y<\infty$, and this is exactly the defining property of conditional expectation. For a general nonnegative $Y$, define its extended conditional expectation by increasing bounded truncations; the same nonnegative integral argument proves the statement.

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.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

For a L2-bounded continuous martingale, the terminal and maximal norms satisfyThe first inequality follows from almost sure convergence. Apply the Doob L2 maximal inequality on $[0,T]$ and then the monotone convergence theorem as $T\to \infty$ for the second. These are equivalent norms on the space modulo indistinguishability of stochastic processes; neither inequality needs a zero initial value.

If random variables $X_i$ satisfy ${\sum }_{i}E|X_i|<\infty$, then ${\sum }_i{X}_i$ converges absolutely almost surely, is integrable, and its expected value equals ${\sum }_{i}E[X_i]$. Indeed, the monotone convergence theorem gives $E[{\sum }_i|X_i|]={\sum }_{i}E|X_i|<\infty$, so that nonnegative sum is finite almost surely. It dominates every finite partial sum in absolute value; the dominated convergence theorem then justifies passing the limit through the expected value. No independence is required.

For example, if $|X_i|\le M{q}^i$ for a fixed $M$ and $0<q<1$, the geometric series bound proves the hypothesis directly. This is useful when repeated observations are multiplied by geometrically shrinking distances or weights.

For $0<p<1$, $\beta >0$ and $x>0$, this probability density function has cumulative distribution function $F(x)=1-\log (1-(1-p)e^{-\beta x})/\log p$. Differentiating gives the density, while the endpoint limits are zero and one. Its quantile function is $F^{-1}(u)={\beta }^{-1}[\log (1-p)-\log (1-p^{1-u})]$ for $0<u<1$, enabling inverse transform sampling.

Expanding the denominator as a geometric series givesTermwise integration, justified by the monotone convergence theorem, and the gamma function integral yield raw moments $\mathbb{E}[X^j]=-j!{\mathrm{Li}}_{j+1}(1-p)/({\beta }^j\log p)$. Thus the polylogarithm expresses its mean, variance and higher moments. Here $p$ is the complement of the series parameter $1-p$; parameter conventions must be checked when comparing formulas.

If nonnegative integer-valued random variables $X_n$ have all fixed factorial moments converging to ${\lambda }^k$, then their convergence in distribution is to the Poisson distribution with mean $\lambda$. The first-moment bound gives uniform tightness, and bounds on higher moments give uniform integrability of each fixed lower moment. Every subsequential limit therefore has the indicated factorial moments. For such a limit $X$ and $t\ge 0$, the nonnegative binomial expansion and the monotone convergence theorem give $\mathbb{E}(1+t{)}^X={\sum }_{k\ge 0}{t}^k{\lambda }^k/k!=e^{\lambda t}$. This identifies its moment-generating function as that of the Poisson distribution, uniquely determining the limit.

For an integrable discrete-time martingale, stop at the first time $|f_n|\ge N$. For any deterministic $m$, conditioning the terminal value $f_m$ on each event $\{\tau =j\}$ gives $\mathbb{E}|f_{\tau \wedge m}|\le \mathbb{E}|f_m|$. Monotone convergence theorem therefore proves the displayed overshoot bound. The maximum of the stopped martingale is at most $N\vee |f_{\tau }|1_{\{\tau <\infty \}}$, an integrable variable. This permits a convergence proof without assuming bounded overshoots.

The intensity measure of a point process $\Pi$ is its expected counting measure:It may take the value $+\infty$.^[ref] Countable additivity follows by applying the monotone convergence theorem to the counts of finite unions of pairwise disjoint observation windows. If $\mu$ is locally finite and absolutely continuous with respect to Lebesgue measure, its Radon-Nikodym derivative is the intensity function of a point process $\lambda$:This is an expected number per unit volume, not a probability density function for a randomly selected point. For example, a single deterministic point has $\mu ={\delta }_0$, a Dirac measure, and no Lebesgue measure density. The density representation requires its own assumption.^[ref]

For a bounded Borel set $W$, $\mathbb{E}[N_W]=\mu (W)$, but local finiteness alone does not ensure that this expected value is finite. Conversely, $\mu (W)<\infty$ implies $N_W<\infty$ almost surely, even when $W$ is unbounded. A constant finite intensity function of a point process on all of ${\mathbb{R}}^d$ has infinite total intensity measure, although every bounded window has finite expected value.

For a nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

For $0<R<1$, common constant interest and volatility, and observed excess drift bounded in absolute value by $b_{+}>0$, let $V_{+}$ be the finite Merton consumption-investment problem value for excess drift $b_{+}$. In its Hamilton-Jacobi-Bellman equation, optimizing the stock holding contributes $-b^2(V_{+}^{\prime }{)}^2/(2{\sigma }^2{V}_{+}^{\prime \prime })$. Since $V_{+}^{\prime \prime }<0$, this contribution increases with $b^2$. Thus every admissible policy makes accumulated discounted CRRA utility plus discounted $V_{+}$ a nonnegative local supermartingale, hence a supermartingale. Dropping its nonnegative terminal term and using the monotone convergence theorem proves the value bound. The condition $[\rho -(1-R)(r+b_{+}^2/(2R{\sigma }^2))]/R>0$ is essential for the finite supermartingale construction.

Measure theory studies measures, which assign sizes to measurable sets, and the Lebesgue integral, which accumulates values weighted by those sizes. The same framework describes Lebesgue measure on the real numbers, counting measure on discrete collections, and probability measures on possible outcomes. The choice of measure determines what counts as large; the choice of sigma-algebra determines what can be measured.

A striking example separates size from proximity: the rational numbers in $[0,1]$ form a dense set with Lebesgue measure zero, although the whole interval has Lebesgue measure one. This distinction makes almost everywhere statements useful: exceptional null sets can be ignored in a Lebesgue integral even when they meet every open subinterval. The monotone convergence theorem and dominated convergence theorem then give precise conditions for passing a limit through an integral.

A historical starting point is Lebesgue's 1902 thesis, ^[ref]. Its title connects integration with the geometric task of assigning sizes.

In a probability measure-preserving system whose transformation is an ergodic transformation, a nonnegative measurable $f$ with $\int fd\mu =+\infty$ has $N^{-1}{\sum }_{j=0}^{N-1}f(T^{j}x)\to +\infty$ almost everywhere. Apply the pointwise ergodic theorem to $\min (f,K)$ on a common full-measure set for integer $K$, then bound the lower limit by each truncated integral. The monotone convergence theorem sends those integrals to infinity. The finite-integral case is the ordinary pointwise ergodic theorem.

Let $\rho =|z_0|$ and $f(x,y)=\log \sqrt{x^2+y^2}$. On the annulus $r<\sqrt{x^2+y^2}<R$, its gradient and Laplace operator areThus $f$ is a harmonic function there. The stopped Itô formula givesThe sum of the squared coefficients is $|B_s{|}^{-2}\le r^{-2}$ before $\tau$. The Itô isometry therefore makes the stochastic term a true square-integrable martingale on every finite interval. Continuity prevents overshoot at exit, so $\log r\le \log |B_{t\wedge \tau }|\le \log R$. This proves the required bounded martingale property.

Before taking the terminal expectation, prove that $\tau$ is finite. Set ${\theta }_R=\inf \{t\ge 0:|B_t|=R\}$. The Itô formula for $|B{|}^2$ shows thatis a martingale, since its stopped stochastic integrands are bounded. Taking expectations yieldsThe monotone convergence theorem gives $\mathbb{E}{\theta }_R\le (R^2-{\rho }^2)/2<\infty$. Hence $\tau \le {\theta }_R<\infty$ almost surely.

The bounded convergence theorem, equivalently the optional stopping theorem applied to this bounded martingale, gives $\mathbb{E}\log |B_{\tau }|=\log \rho$. The two exit events partition the sample space up to a null set. Writing $p=\mathbb{P}({\tau }_r<{\tau }_R)$,Solving gives the planar Brownian annulus hitting probability

The monotone convergence theorem states: on a measure space $(X,\mathcal{F},\mu )$, let $u_n:X\to [0,\infty ]$ be measurable functions withThen $u$ is measurable andThe integrals and their limit may be infinite. No integrable dominator, finite measure hypothesis, or sigma-finiteness is required. The nonnegativity and increasing order are the essential hypotheses in this form of the theorem.