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Work on $([0,1],\mathcal{B},dt)$ as a probability space, with Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the filtration generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the dyadic slope martingale byThe endpoint $t=1$ may be assigned any value, since it is a null set for Lebesgue measure. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb{E}[G_{n+1}\mid {\mathcal{F}}_n]=G_n$ almost everywhere, so this is a martingale. The Lipschitz condition gives $|G_n|\le K$ everywhere except possibly at the freely chosen endpoint, where we take zero.

Apply the Lp martingale convergence theorem with $p=2$. Its limit $g$ has $|g|\le K$ almost everywhere and $G_n\to g$ in L1 norm as well, by Cauchy-Schwarz inequality. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded measurable function on the entire interval.

Set $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the linear interpolation of $f$ on the dyadic grid. The Lipschitz condition gives $\Vert f_n-f{\Vert }_{\infty }\le K{2}^{-n}\to 0$. AlsoThe two uniform limits coincide, giving the absolutely continuous function representationThe chosen bound holds for every $t$ after the null-set modification; the integral identity holds for every $x$ simultaneously.

The pointwise limit $f$ is a measurable function, since it can be expressed as ${\sup }_n{\inf }_{m\ge n}{f}_m$. Thus each set $E_n^{(k)}$ is a countable intersection of measurable sets within $A$. Deleting the condition at index $m=n$ weakens the requirements, whereas replacing $1/k$ by $1/(k+1)$ strengthens them. HenceThese are inclusions, not necessarily strict inclusions. The displayed sets describe eventual uniform error tolerances at an individual point.

Fix $k\ge 1$ and $x\in A$. By pointwise convergence, there is an $N=N(x,k)$ such that $|f_m(x)-f(x)|\le 1/k$ for every $m\ge N$. This is exactly $x\in E_N^{(k)}$, provingTogether with monotonicity in $n$ and finite measure of $A$, continuity from below of a measure gives $\lambda (E_n^{(k)})\uparrow \lambda (A)$. ConsequentlyThis quantitative form is what allows a single large-measure set to work for every tolerance, rather than choosing a different exceptional set for each point.

The mapping theorem for Poisson point processes applies provided $f$ is a measurable function and its pushforward measureis locally finite, meaningfor every bounded Borel set $B\subseteq {\mathbb{R}}^s$. If “Poisson process” means a simple point process, one also requires $f_{\ast }\Lambda$ to be atomless, equivalentlyUnder these conditions, $f(\Pi )$ is a Poisson point process with mean measure

A Bayesian inverse problem consists of a prior distribution ${\mu }_0$ on the unknown $u\in X$, a reference measure on the data space $Y$, and a jointly measurable likelihood $L(u;y)$ such that $L(u;\cdot )$ is a probability density function for ${\mu }_0$-almost every $u$. For observed data $y$, Bayes' theorem defines the posterior distribution byprovided $0<Z(y)<\infty$.

They are not independent. The Itô formula gives ${\beta }_t^2=t+2{\int }_0^t{\beta }_{s}d{\beta }_s$, and the bilinear Itô isometry therefore givesIf $B_t$ and ${\beta }_t$ were independent random variables, then $B_t$ would also be independent of the measurable function ${\beta }_t^2$, and centeredness would instead give $\mathbb{E}[B_t{\beta }_t^2]=\mathbb{E}[B_t]\mathbb{E}[{\beta }_t^2]=0$. This contradiction disproves independence.

A point process records random locations by a random measure that counts them. Examples include arrival times of a queue, positions of stars, and the finite eigenvalue configuration of a random matrix. The count in an observation window is a random variable; its expected value is supplied by the intensity measure of a point process, while its fluctuations and relationships with other windows require more information.

For a precise spatial convention, take $E={\mathbb{R}}^d$ with its Borel sigma-algebra and a probability space $(\Omega ,\mathcal{F},\mathbb{P})$. A locally finite point process $\Pi$ is a measurable function from $\Omega$ to the space ${\mathcal{N}}_{\mathrm{lf}}(E)$ of locally finite point measures, equipped with the sigma-algebra generated by evaluations $\nu \mapsto \nu (A)$ for Borel sets $A$. Each realization has the formwhere $J(\omega )$ is finite or countable and repeated locations are allowed; ${\delta }_x$ is a Dirac measure. Here a point measure is a counting measure with possible multiplicities, rather than the counting measure of a set of distinct locations.^[ref]

Local finiteness means ${\Pi }_{\omega }(B)<\infty$ for every bounded Borel set $B$, on a single almost sure event. In ${\mathbb{R}}^d$ it suffices to check the countable collection of balls of integer radius centered at the origin, and this is equivalent to finiteness on every compact set. On a general measurable space there is no notion of boundedness without additional structure: state a metric or an explicit family of observation windows. Another convention on locally compact spaces requires finiteness on compact sets; this need not coincide with finiteness on all bounded sets outside ${\mathbb{R}}^d$. Local finiteness does not require finitely many points in all of $E$.^[ref]

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt is a subset of the original domain of a function. The notation does not require an inverse function: the preimage of a singleton set can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The preimage of a singleton set is also its fiber of a function.

For subsets $U,V\subseteq Y$, preimages preserve both unions and intersections:For the intersection, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output subsets.

In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See Lebl, Proposition 0.3.15: preimages of unions, intersections and complements.

This makes preimages useful for transporting conditions. The solution set of $g(x)=0$ is $g^{-1}(\{0\})$, and a zero set is a preimage of zero. Requiring two output conditions takes the intersection of their preimages. In topology, a continuous function pulls every open set back to an open set; a measurable function pulls the specified measurable sets in its codomain back to measurable sets in its domain of a function. These definitions use preimages because their set operations remain consistent even when the function is noninjective.

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.

A probability density function represents probabilities as areas under a nonnegative measurable function. For a real random variable with an absolutely continuous probability distribution, it satisfiesfor every Borel set $A$. Its height is probability per unit of the outcome variable, not the probability of that exact outcome. Thus a probability density function may exceed one, while the probability of a singleton is zero. NIST describes the interval-probability interpretation through integration. ^[ref].

For the example $f_Y(y)=2y$ on $[0,1]$, $f_Y(0.8)=1.6$, but $\mathbb{P}(Y=0.8)=0$ and $\mathbb{P}(0.7<Y<0.9)=0.9^2-0.7^2=0.32$. The cumulative distribution function is $F_Y(y)=y^2$ within that interval, connecting density height to the local accumulation rate.

Changing the scale changes the density height. If $Z=100Y$, then $f_Z(z)=f_Y(z/100)/100=z/5000$ on $[0,100]$. At the corresponding value $z=80$, its height is $0.016$, while $\mathbb{P}(70<Z<90)=0.32$ is unchanged. The factor $1/100$ compensates for the wider integration variable. This is a change of variables, not a change in the underlying probabilities.

A probability density function is determined only almost everywhere: changing its values on a Lebesgue measure zero set changes no probabilities. The Cantor distribution has no such representation, and a mixed probability distribution can have both a point mass and a density component, so differentiating its cumulative distribution function does not recover all its mass.

A random element is a measurable function from a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ to a measurable space $(E,\mathcal{E})$. Its law is the pushforward measure ${\mathbb{P}}_X(A)=\mathbb{P}(X^{-1}(A))$ for $A\in \mathcal{E}$. Real-valued random variables and vector-valued random variables are examples; a stochastic sample path can also be a random element when its path-space sigma-algebra is specified. For a Polish space $E$, the usual choice is its Borel sigma-algebra. A large deviation principle concerns the laws of a family of such random elements.

A real-valued random variable is a measurable function $X$ from a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ to the real numbers with their Borel sigma-algebra. More general targets give random elements. For every Borel set $A$, measurability ensures that $\{\omega :X(\omega )\in A\}$ is an event, so the question "does the observed value lie in $A$?" has a defined probability.

The function $X$ is fixed; its input outcome is uncertain. This separates what we measure from how the uncertainty is modeled. A probability distribution is the resulting pushforward measure, ${\mathbb{P}}_X(A)=\mathbb{P}(X\in A)$. A cumulative distribution function, probability mass function, or probability density function describes that measure in an appropriate setting. The random variable and its probability distribution are different objects: the former also retains its relationship to other measurements on the same sample space.

For two independent uniform dice, let $U,V$ record their results and define $S=U+V$ and $M=\max (U,V)$. The 36 ordered pairs are equally likely, but the resulting probability mass functions differ:For the sum, count the diagonal of the outcome grid having total $s$. For the maximum, subtract the $(m-1{)}^2$ pairs with both results below $m$ from the $m^2$ pairs with both results at most $m$. These examples show how a deterministic function transforms a simple input probability distribution into a nonuniform output one.

The corresponding expected values are $\mathbb{E}S=7$ and $\mathbb{E}M=161/36$. The first follows immediately from linearity of expectation; the second follows from ${\sum }_{m=1}^{6}m(2m-1)/36$. Neither number must be an attainable result. An expected value describes an average under the probability measure, not a prediction that the next observation will equal it.

The distinction between a random variable and its probability distribution matters for dependence. $U$ and $V$ have the same discrete uniform distribution and are independent random variables. $U$ and a second copy of $U$ also have that same marginal probability distribution, but they always agree. Their joint probability distribution is concentrated on the diagonal; the marginals alone cannot say which relationship holds. This is why a model for several measurements must specify their joint law or an appropriate independence assumption.

Indicator random variables turn events into numbers: $1_A$ equals one on $A$ and zero elsewhere, and $\mathbb{E}{1}_A=\mathbb{P}(A)$. This connects event counting with expected value and variance, and lets tools for random variables answer questions about probabilities. More complicated functions of observations can describe totals, waiting times, or estimation errors using the same framework.

The reachable set ${\mathcal{R}}_T(x_0)$ collects all endpoints attainable from $x_0$ at exactly $T$ by admissible control inputs. If arrival any time up to $T$ is allowed, use ${\mathcal{R}}_{\le T}(x_0)={\bigcup }_{0\le t\le T}{\mathcal{R}}_t(x_0)$, with ${\mathcal{R}}_0(x_0)=\{x_0\}$. These definitions differ: a general dynamical system may drift away after early arrival, so waiting need not preserve the target. The reachable set is always relative to a declared input class and any constraints on the entire path.

For the double integrator starting from $(0,0)$, now require measurable functions as control inputs with $|u(t)|\le 1$ almost everywhere. Write $(p,v)$ for its endpoint at $T>0$. Two integrals determine the endpoint:Our mathematical example has the exact reachable setTo see the upper boundary, fix $v$: put $u=1$ first and $u=-1$ last, switching at $s=(T+v)/2$. This maximizes the weighted integral for position because early commands have the largest weight $T-t$; the required net velocity fixes $s$. Formally, $(T-t)-(T-s)=s-t$ has the same sign as this maximizing command, so changing to any other admissible command with the same integral cannot increase $p$. Reversing the order gives the lower boundary. Pointwise convex combinations of these two commands preserve $v$ and attain every intermediate $p$, so the formula describes the whole region, not merely a bound. The extreme commands are bang-bang control.

In particular, the bounded double integrator can move from rest to position $d$ and rest again exactly when $|d|\le T^2/4$. Its minimum duration is therefore $2\sqrt{|d|}$ in these normalized units. With freely chosen time every rest-to-rest distance is attainable; with a fixed deadline many are excluded. Changing the amplitude bound to $|u|\le a$ with $a>0$ scales this condition to $|d|\le a{T}^2/4$. This shows why specifying the controls matters even when the Kalman rank criterion gives full rank of a matrix.

In closed quantum control, a density operator reachable set lies inside its unitary orbit of a density operator, since coherent evolution preserves the spectrum. That orbit need not be fully reachable under the available control inputs.

A power law $q(t)=t^{\rho }$ responds to every fixed rescaling $t\mapsto ut$ by the same factor $u^{\rho }$. Regular variation keeps this scaling behaviour in the limit $t\to \infty$, while allowing departures from an exact power law. It is useful when the scaling exponent is stable but a smaller correction continues changing. The correction can even grow without bound.

An eventually positive measurable function $q$ has regular variation at infinity with index $\rho \in \mathbb{R}$ ifThe argument $ut$ is in the domain of a function for sufficiently large $t$. Each $u$ is held fixed as the limit is taken; allowing $u$ to change with $t$ is a different assertion. This is the scalar definition recalled by ^[ref]. The index is unique: taking $u=2$ determines $2^{\rho }$.

For example, on $t>1$ the function $q(t)=t^2\log t$ hasThus $q$ has regular variation of index two. Yet $q(t)/t^2=\log t\to \infty$: regular variation does not require asymptotic equivalence to a constant times a power law. Doubling $t$ gives $q(2t)/q(t)=4(1+\log 2/\log t)$. At $t=100,10^4,10^6$ this is approximately $4.60206,4.30103,4.20069$, respectively. The ratio approaches four slowly, even though the exponent is exactly two in the limiting scaling law.

The slowly varying function isolates this correction. Set $\ell (t)=q(t)/t^{\rho }$. ThenConversely, multiplying a slowly varying function by $t^{\rho }$ gives regular variation of index $\rho$. Hence the decomposition $q(t)=t^{\rho }\ell (t)$ is exact on its eventual domain of a function; it is the scaling relation, rather than the decomposition itself, that is asymptotic.

A Pareto distribution gives a decreasing example with a direct interpretation. For $x\ge 1$, take the tail probability $\mathbb{P}(X>x)=x^{-2}$. Its regular variation index is $-2$. Doubling a threshold divides its tail probability by four, andFor a tail probability with regular variation of index $-\alpha$, the same conditional probability tends to $2^{-\alpha }$. An exponential distribution behaves differently: its corresponding conditional probability is $e^{-\lambda t}\to 0$. This comparison explains why scaling limits are useful in extreme value theory; it does not establish that an observed dataset follows a particular tail model. The role of regular variation in heavy-tailed distributions is discussed in Lindskog, Resnick and Roy, introduction, p. 1.

An exponential function $q(t)=e^t$ is not regularly varying with any finite index: at $u=2$, $q(2t)/q(t)=e^t\to \infty$, rather than $2^{\rho }$. Boundedness alone is also insufficient, as the oscillating counterexample in slowly varying function shows. Checking a few rescaling ratios at finite arguments can illustrate a proposed law, but cannot prove its limiting definition.

Jovan Karamata developed this theory in his 1930 paper and returned to its fundamental results in 1933. The opening of his 1933 paper identifies the earlier work; pp. 57–58 connect the rescaling limit with a power law and a slowly varying factor. See J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bulletin de la Société Mathématique de France 61 (1933), 55–62. This gives a precise language for “almost a power” without discarding corrections that remain large.

The indicator function example reaches a squared L2 norm ratio of two. To approach four, concentrate more strongly near zero. For $0\le a<1/2$, use the measurable functionAll integrals use Lebesgue measure $dx$. Keeping both portions of the squared L2 norm givesConsequently,Every allowed member has ratio strictly below four. For any proposed constant $C<4$, a sufficiently large $a<1/2$ has ratio greater than $C$. Combined with Hardy's integral inequality, this proves that four is the smallest possible squared L2 norm constant.^[ref]

The endpoint is excluded: at $a=1/2$, ${\int }_0^1|f_a{|}^{2}dx={\int }_0^1{x}^{-1}dx=\infty$, and the inner portion of $\Vert H{f}_a{\Vert }_2^2$ also diverges. Its tail still has finite integral four. There is no finite L2 norm ratio at this endpoint, hence no equality example obtained by substituting it. The limit four is unattained by this family. Multiplying each $f_a$ by $\sqrt{1-2a}$ makes its L2 norm one without changing the ratio.

A simple function is a measurable function whose image is a finite set. Its value sets $A_j=\{x:s(x)=a_j\}$ form a finite measurable partition. For nonnegative $a_j$, its Lebesgue integral is ${\sum }_j{a}_j\mu (A_j)$, with the convention $0\cdot \infty =0$. The pieces need not be intervals, or even connected sets. See MIT lecture 10, Definitions 100 and 106.

For example, on $[0,1]$ let $s$ be $2$ on $[0,1/4)$, $1$ on $[1/4,3/4)$, and $0$ on $[3/4,1]$. Its Lebesgue integral is $2(1/4)+1(1/2)=1$. Its superlevel sets have Lebesgue measure $3/4$ for $0\le t<1$, $1/4$ for $1\le t<2$, and $0$ thereafter. The layer cake representation gives the same answer: $(3/4)(1)+(1/4)(1)=1$. These two calculations are the building blocks of the general Lebesgue integral.

A positive measurable function $\ell$ is slowly varying at infinity iffor every fixed $u>0$. It has regular variation of index zero. “Slowly” concerns relative change under multiplication of the argument; it does not mean that the function must be bounded, have a small derivative, or converge to a finite value.

Every positive constant is a slowly varying function. So are $\log t$, $1/\log t$ and $(\log t{)}^a$ for any fixed real number $a$, on $t>1$. Indeed,The first grows without bound, the second tends to zero, and the constant does neither. Another unbounded example is $\ell (t)=e^{\sqrt{\log t}}$:This function can grow much faster than $\log t$ while retaining regular variation of index zero.

Even convergence is unnecessary. The bounded functionoscillates between $e^{-1}$ and $e$. It is nevertheless a slowly varying function: the change in the argument of the sine under $t\mapsto ut$ isand the Lipschitz continuity of the sine makes the logarithm of the ratio tend to zero. The sequences $t_k=\exp (\exp (\pi /2+2\pi k))$ and $s_k=\exp (\exp (3\pi /2+2\pi k))$ give the two values $e$ and $e^{-1}$ for every nonnegative integer $k$, so there is no limit at infinity.

Contrast this with $h(t)=2+\sin (\log t)$, which is positive and bounded but is not a slowly varying function. Fix $u=e^{\pi /2}$. Along $t_k=e^{2\pi k}$ its rescaling ratio is $h(u{t}_k)/h(t_k)=3/2$; along $s_k=e^{\pi +2\pi k}$ the ratio is $1/2$. The persistent oscillations prevent even a limit for this one fixed rescaling. It is not regularly varying with any finite index either.

The useful distinction is whether a fixed proportional change of input eventually produces a negligible proportional change of output. A slowly varying function can therefore preserve the scaling exponent of a power law while changing its magnitude substantially. The decomposition in regular variation explains why a logarithmic factor can matter in numerical estimates without changing the limiting exponent.

For a real measurable function $m$ on a sigma-finite measure space, multiplication on $D(M_m)=\{f\in L^2:mf\in L^2\}$ is an unbounded self-adjoint operator. Testing its adjoint relation on $\{|m|\le n\}$ identifies the same maximal domain. Outside the essential range, multiplication by $(m-z{)}^{-1}$ is a bounded resolvent. Inside it, normalized indicators of finite-positive-measure sets where $m$ is close to $z$ are approximate eigenvectors, so the inverse cannot be bounded.

A measurable function is square-integrable when the integral of its squared absolute value is finite. Its equivalence class belongs to the L2 space; on a probability space, this means a finite second moment.

A function into a Banach space is strongly measurable if, outside a null set, it is a pointwise norm limit of measurable simple functions. A measurable function with values in a separable Banach space is strongly measurable: approximate values by a countable collection of balls of shrinking radii and then truncate the resulting countably valued approximants to simple functions. A strongly measurable function is Bochner integral integrable exactly when its norm is integrable. Finite second moments under a probability measure therefore imply existence of its Bochner integral expected value by the Cauchy-Schwarz inequality.