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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.

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

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]

The essential infimum and essential supremum describe the lower and upper bounds of a measurable function after disregarding null sets for a specified measure. A pointwise infimum or supremum sees every input; an essential bound sees whether values below or above a proposed threshold occur on a set of positive measure. Thus "essential" means insensitive to changes on a null set, rather than insensitive to unusual values. This distinction makes essential bounds compatible with Lebesgue integrals and the equivalence classes used in Lp spaces. ^[ref].

For a real measurable function $f$ on a measure space $(X,\Sigma ,\mu )$, the definitions areThese can take values in the extended real number line. On a space of positive total measure, finite essential bounds obey ${\mathrm{essinf}}_{\mu }f\le f\le {\mathrm{esssup}}_{\mu }f$ almost everywhere. If $s={\mathrm{esssup}}_{\mu }f$ is finite, every threshold $b<s$ has $\mu (\{f>b\})>0$; the corresponding lower-edge statement is $\mu (\{f<a\})>0$ for every $a>{\mathrm{essinf}}_{\mu }f$. These are the measure-based counterparts of the approximation property of a supremum and approximation property of an infimum.

Here is a two-sided example on $[0,1]$ with Lebesgue measure:The pointwise bounds are $\inf g=-3$ and $\sup g=4$, while the essential bounds are $\mathrm{essinf}g=0$ and $\mathrm{esssup}g=1$. Both exceptional singletons are null sets. But for $0<a<1$, the inputs where $g<a$ still have measure $a$, and those where $g>a$ have measure $1-a$: deleting or adding two individual inputs cannot change those lengths. Hence neither essential bound can be moved inward.

The choice of measure is part of the answer. The same $g$ has essential bounds $-3$ and $4$ under counting measure, because every nonempty set has positive measure. Under the Dirac measure at $1/4$, both essential bounds equal $-3$. A null set for one measure need not be a null set for another.

Nor must the exceptional inputs be isolated. The Dirichlet function, equal to one at rational numbers and zero at irrational numbers, has pointwise supremum one on $[0,1]$, but essential supremum zero for Lebesgue measure: the rational numbers form a countable set that is a null set, even though they are dense. Conversely, a value ten on an interval of length $0.001$ affects the essential supremum just as decisively as it would on half the domain. Small positive measure is not zero measure.

The essential supremum is the upper bound for a measurable function after disregarding null sets for a specified measure $\mu$. Its definition iswith values allowed in the extended real number line. A supremum controls every value; an essential supremum controls values almost everywhere. It therefore depends on the chosen measure, as well as on the function. ^[ref].

For a worked example, use Lebesgue measure on $[0,1]$ and putIts supremum is ten, attained at $1/2$. But its essential supremum is one: $f\le 1$ outside the singleton $\{1/2\}$, which is a null set. Every candidate $M<1$ fails, because the interval $(\max (0,M),1)$ has positive Lebesgue measure and includes values larger than $M$; changing one point cannot remove that interval. In particular, $\mu (\{f>0.8\})=0.2$, while $\mu (\{f>1\})=0$.

Why is this useful? Lebesgue integrals ignore changes on null sets, and Lp spaces identify functions that agree almost everywhere. The L-infinity norm uses $\mathrm{esssup}|f|$ so it respects that identification. For this nonnegative example the supremum norm is ten and the L-infinity norm is one; replacing $f(1/2)$ by $1/2$ changes the former while leaving the latter unchanged. ^[ref].

The measure qualification is essential. For $\mu ={\delta }_{1/2}$, the Dirac measure concentrated at the exceptional input, that singleton has measure one and this function has essential supremum ten. A set ignored by one measure need not be ignored by another. The lower counterpart is the essential infimum, with $\mathrm{essinf}f=-\mathrm{esssup}(-f)$.

An expected value is the probability-weighted arithmetic mean of a real-valued random variable. On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, its general definition is the Lebesgue integralwhere ${\mu }_X$ is the probability distribution of $X$. For an integrable random variable, this is a finite real number. With a probability mass function $p_X$, use ${\sum }_{x}x{p}_X(x)$; with a probability density function $f_X$, use ${\int }_{\mathbb{R}}x{f}_X(x)dx$. These are two ways of computing the same weighted average, rather than different definitions for different experiments.

For example, suppose $X$ takes the values $-1,2,8$ with probabilities $1/2,3/10,1/5$. ThenThe most likely outcome is $-1$, and $1.7$ is not a possible outcome at all. The expected value summarizes all outcomes and their weights; it need not predict the result of one trial. Geometrically, placing masses equal to the probability masses at the outcome positions gives a center of mass at $1.7$, since ${\sum }_x{p}_X(x)(x-1.7)=0$. This balance-point interpretation also explains the connection to a weighted arithmetic mean. ^[ref].

For the dimensionless random variable $Y$ with probability density function $f_Y(y)=2y$ on $[0,1]$, the expected value is $\mathbb{E}[Y]={\int }_0^{1}y(2y)dy=2/3$. Integrating the probability density function itself gives one; to compute the expected value, multiply by the outcome first. ^[ref].

Why is this useful? Linearity of expectation lets us compute an average total by adding average contributions, even for dependent random variables. And for independent and identically distributed integrable random variables, the strong law of large numbers says their sample mean approaches their common expected value almost surely. The theorem supplies the link to repeated observations; the weighted-average definition itself does not promise that any particular finite run will be close to the expected value.

There is a useful historical connection to valuing uncertain rewards. In his 1657 De ratiociniis in ludo aleae, Christiaan Huygens assigns the value $(a+b)/2$ to equally likely rewards $a,b$, and $(pa+qb)/(p+q)$ when they have $p,q$ equally likely chances. These are the finite weighted-average formulas for expected value. His opening discussion asks how to divide the stakes when a game stops before completion, connecting probability to a concrete question about an unfinished contest. ^[ref].

A real or complex random variable $X$ is integrable when $\mathbb{E}|X|<\infty$. This is integrability with respect to its probability measure, and ensures that its expected value is finite. A bounded random variable is always an integrable random variable; so is the example taking $-1,2,8$, for which $\mathbb{E}|X|=2.7$.

For a real random variable, put $X_{+}=\max (X,0)$ and $X_{-}=\max (-X,0)$, its positive part and negative part. Each has a nonnegative Lebesgue integral, possibly infinite. The expression $\mathbb{E}[X]=\mathbb{E}[X_{+}]-\mathbb{E}[X_{-}]$ is defined when at least one of these two expected values is finite; if both are infinite, the expected value is undefined. Integrability makes both finite. A nonnegative random variable can have expected value $+\infty$, even though it is not an integrable random variable.

The standard Cauchy distribution is a useful warning. With its probability density function $f(x)=1/[\pi (1+x^2)]$, the positive contribution is ${\int }_0^{R}xf(x)dx=\log (1+R^2)/(2\pi )$, which tends to infinity; the negative contribution has the same magnitude. Integrating symmetrically over $[-R,R]$ gives zero for each $R$, but this cancellation does not define an expected value. The Cauchy sample mean has an unchanged distribution, showing why the integrability hypothesis in the strong law of large numbers matters.

For an integrable random variable, conditional expectation preserves integrability, and linearity of expectation can safely combine finite averages without forming $\infty -\infty$.

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 integrable real-valued functions $f,g$ on a measure space $(X,\mu )$ and fixed real numbers $a,b$, their linear combination $af+bg$ is integrable, and the Lebesgue integral satisfiesThe same rule applies to any finite summation: ${\int }_X{\sum }_{j=1}^n{a}_j{f}_{j}d\mu ={\sum }_{j=1}^n{a}_j{\int }_X{f}_{j}d\mu$ for integrable functions $f_j$ and fixed scalars $a_j$. Finite signed values matter: integrability ensures all these integrals are finite, so no undefined $\infty -\infty$ occurs.

This is the distributive property of finite summation carried into integration: a fixed factor can be pulled out, and a finite sum can be integrated term by term. For simple functions, distribute the finite sums of value times measure; approximation extends additivity to nonnegative measurable functions, and splitting into a positive part and negative part gives the integrable signed case. See Lecture 12: Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem (Proposition 122: linearity for integrable functions). For example, ${\int }_0^1(2x-3)dx=2{\int }_0^{1}xdx-3{\int }_0^{1}1dx=-2$. With a probability measure, linearity of integration gives linearity of expectation.

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.

The negative part of a real-valued function $f$ is the nonnegative function $f_{-}(x)=\max (0,-f(x))$. It records the magnitude of the negative values, rather than retaining their sign: if $f(x)=-3$, then $f_{-}(x)=3$. Together with the positive part, it gives $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$. This decomposition defines the Lebesgue integral of a signed function and makes the integrability condition explicit.

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.

A simple function is a measurable function taking finitely many values, expressible as $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ for disjoint measurable sets. For nonnegative coefficients define $\int sd\mu ={\sum }_j{a}_j\mu (A_j)$. The Lebesgue integral of a nonnegative measurable $f$ isSuppose $0\le g_n\uparrow f$ with each $g_n$ simple. Their integrals increase and are bounded above by $\int f$. For the reverse bound, fix any simple $0\le s\le f$ and $0<c<1$. The sets $E_n=\{g_n\ge cs\}$ increase to a set containing $\{s>0\}$. Hence $\int g_n\ge c\int s{1}_{E_n}$, whose right side tends to $c\int s$ by continuity from below of the measure on the finitely many level sets of $s$. Let $c\uparrow 1$ and then take the supremum over $s$. This proves $\int g_n\uparrow \int f$, including infinite limiting integrals.

The monotone convergence theorem states the same conclusion for any nonnegative measurable sequence $f_n\uparrow f$, without simplicity. The preceding lower-bound argument works unchanged: $E_n=\{f_n\ge cs\}$ and monotonicity of the integral give $\int f_n\ge c\int s{1}_{E_n}$. Taking limits and the same two suprema provesThe upper bound follows from $f_n\le f$. Finiteness of the measure makes all bounded simple integrals finite in this proof; the theorem in fact extends to arbitrary measure spaces.

A simple function is a measurable function with finitely many values, written $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ using disjoint measurable sets. For nonnegative $s$ its Lebesgue integral is $\mu (s)={\sum }_j{a}_j\mu (A_j)$, with $0\cdot \infty =0$. An integrable signed simple function has ${\sum }_j|a_j|\mu (A_j)<\infty$ and the same signed sum defines its integral.

For a nonnegative measurable $f$ define $\mu (f)=\sup \{\mu (s):s\text{ simple},0\le s\le f\}$. A real measurable $f$ is Lebesgue integrable when $\mu (|f|)<\infty$; then $\mu (f)=\mu (f^{+})-\mu (f^{-})$, both terms finite. The monotone convergence theorem states that if $0\le f_n\uparrow f$ pointwise, or almost everywhere, then $\mu (f_n)\uparrow \mu (f)$, allowing infinity.

For nonnegative $f,g$, choose nonnegative simple approximations $s_n\uparrow f$, $t_n\uparrow g$. Their sums increase to $f+g$; simple-function additivity and monotone convergence give $\mu (f+g)=\mu (f)+\mu (g)$. The same approximation proves $\mu (cf)=c\mu (f)$ for $c\ge 0$.

For integrable signed $f,g$, $|f+g|\le |f|+|g|$ makes their sum integrable. Write $u=f^{+}+g^{+}$, $v=f^{-}+g^{-}$. The identity $u+(f+g{)}^{-}=v+(f+g{)}^{+}$ and nonnegative additivity, with finite integrals, give $\mu (f+g)=\mu (f)+\mu (g)$. Also $(cf{)}^{\pm }=c{f}^{\pm }$ when $c\ge 0$, while negative $c$ interchanges the two parts, so $\mu (cf)=c\mu (f)$ for every real $c$. Integrals do not change upon alteration on a null set, hence this descends to Lebesgue space. The indicated map is linear.

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.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

The Beppo Levi theorem, or monotone convergence theorem, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a measure space andpointwise almost everywhere, then $f$ is measurable andwhere either side may be infinite.

Measurability of $f={\sup }_n{f}_n$ follows from measurability of countable suprema. Since $f_n\le f$, monotonicity of the Lebesgue integral givesFor the reverse inequality, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. DefineThen $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\ge c,s{1}_{E_n}$,by continuity from below of a measure, applied to the finitely many level sets of $s$. Hence ${\lim }_n\int f_n\ge c\int s$. Letting $c\uparrow 1$ and taking the supremum over all simple $s\le f$, as in the definition of the Lebesgue integral, gives the reverse inequality and proves the theorem.

If $0<\theta <1$ and nonnegative measurable functions satisfy $h((1-\theta )x+\theta y)\ge f(x{)}^{1-\theta }g(y{)}^{\theta }$ for all $x,y\in {\mathbb{R}}^n$, then their Lebesgue integrals obeyThe quantile derivative identity and monotone substitution inequality prove the one-dimensional case; Tonelli theorem extends it by induction. It implies the Brunn–Minkowski inequality.

For $a<b$ and a real-valued function $f$ with a finite Riemann integral or Lebesgue integral on $[a,b]$, the signed area under a graph is its definite integral: geometric area above the horizontal axis contributes positively, and geometric area below it contributes negatively. If these two nonnegative areas are $A_{+}$ and $A_{-}$, thenThe second expression is the total geometric area under a graph. Taking the absolute value of the first expression does not usually give the second: cancellation has already removed information about how much area lies on each side.^[ref]