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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.

For a measurable function $T$, every f-divergence contracts under pushforward:This follows from conditional Jensen inequality applied to the jointly convex perspective $(a,b)\mapsto bf(a/b)$.

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]

On a finite-measure measurable set, a sequence of measurable functions converging pointwise almost everywhere converges uniformly outside a set of arbitrarily small measure. For tolerance $1/k$, the sets on which every index after $N$ has error at most $1/k$ increase to full measure as $N\to \infty$. Choose $N_k$ with exceptional measure at most $\epsilon 2^{-k-1}$ and intersect the good sets. Countable subadditivity bounds the discarded measure, and the resulting tail estimates prove uniform convergence. Finite measure is essential: $1_{[n,\infty )}$ on $[0,\infty )$ is a counterexample.

The essential infimum of a real measurable function is ${\mathrm{essinf}}_{\mu }f=-{\mathrm{esssup}}_{\mu }(-f)$. It is the largest lower bound that holds almost everywhere, with values allowed in the extended real number line. Unlike a pointwise infimum, it is unchanged by altering a representative on a null set.

For the function $g$ in essential infimum and essential supremum, the value $-3$ at one input changes the pointwise infimum, but the essential infimum remains zero under Lebesgue measure. If instead a function is $-3$ throughout an interval of any positive length and zero elsewhere, its essential infimum is $-3$. The distinction concerns the measure of the low-value set, rather than the magnitude of the low value.

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.

A measurable function is essentially bounded if some finite nonnegative real number $M$ bounds its absolute value almost everywhere. The exceptional null set may contain arbitrarily large values. With Lebesgue measure on $[0,1]$, a function equal to zero outside $\{1/n:n\in {\mathbb{N}}_{>0}\}$ and equal to $n$ at $1/n$ is essentially bounded despite being unbounded pointwise: its exceptional countable set is a null set, so $M=0$ works. Its L-infinity norm is zero because it represents the zero equivalence class in Lp space.

Contrast this with $f(x)=1/x$ on $(0,1]$. For every $M\ge 1$, the values exceeding $M$ occupy $(0,1/M)$, a set of positive Lebesgue measure. Hence no finite almost everywhere bound works, and this function is not essentially bounded. Being finite at every input does not imply essential boundedness.

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].

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 $\mathrm{essinf}f=-\mathrm{esssup}(-f)$.

Set $U(x)=U_0({\delta }_x)$. On a finite measurable space, points in the same atom of a sigma-algebra give the same Dirac measure, so $U$ is a measurable function and constant on atoms. Every probability measure is a finite mixture of representative Dirac measures with its atom masses as weights. Affineness then proves the expected-utility representation, even when the sigma-algebra is smaller than the power set.

For $0<p<1$, write $Q_p(f)=(\int |f{|}^{p}d\mu {)}^{1/p}$ to emphasize that this is a quasi-norm, as in the Lq quasi-norm branch. On two coordinates with counting measure, $u=(1,0)$ and $v=(0,1)$ give $Q_p(u)=Q_p(v)=1$, but $Q_p(u+v)=2^{1/p}>2$. At $p=1/2$, this is the explicit failure $4>1+1$ in the right panel of the diagram. The midpoint $(1/2,1/2)$ has $Q_p=2^{1/p-1}>1$, so the corresponding unit sublevel set is not a convex set.

A useful replacement survives: for nonnegative real numbers $a,b$, $(a+b{)}^p\le a^p+b^p$. Together with the scalar triangle inequality, this yieldsConsequently $d_p(f,g)=Q_p(f-g{)}^p$ is a metric on the equivalence classes of measurable functions with finite $Q_p$, even though $Q_p(f-g)$ itself need not be a metric. Do not replace the failed Minkowski inequality by a universal reversed bound for signed or complex functions: $g=-f\ne 0$ would give a zero sum but two positive separate sizes.