Connect distributional limits to integral transforms

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A deterministic value 1/n approaches zero, so its probability law approaches a point mass at zero. Yet the probability of being at most zero is always zero, while the limiting law assigns that event probability one. Distributional convergence therefore cannot mean convergence of every event probability. It tests bounded continuous functions and, equivalently on the real line, distribution functions at continuity points of the limit.

Predict

Predict whether weak convergence alone gives convergence of expectations. Let X_n equal n with probability 1/n and zero otherwise. Its laws converge weakly to zero, but E(X_n)=1. The unbounded function x is not an allowed bounded-continuous test. Uniform integrability supplies the extra control needed for expectation limits in suitable settings.

Experiment

Compare three tests for the deterministic sequence X_n=1/n.

Test	Values along sequence	Value at limit zero
h(x)=x/(1+abs (x))	1/(n+1)	0
Indicator of x≤0	0	1
Indicator of x≤1	Eventually 1	1

The first test is bounded and continuous. The second cuts through an atom of the limiting law. The third uses a continuity point of its distribution function. Identify the distinct hypotheses rather than treating all three rows as contradictory evidence.

Understand

Probability laws μ_n converge weakly to μ when integrals of every bounded continuous real function converge. On the real line this is equivalent to F_n(x)→F(x) at every continuity point of F. For the forward direction, sandwich the half-line indicator between continuous ramps with transitions near x. Their expectation gap tends to the limiting mass near x, which vanishes when x is not an atom. For the converse, choose continuity points a< b with arbitrarily small limiting tail mass. Distribution-function convergence makes the approximating laws’ tails small too. On [a, b], uniformly approximate a bounded continuous test function by a step function whose partition endpoints are continuity points. Interval probabilities converge; uniform approximation and the tail bound finish the proof.

The characteristic function is φ_X(t)=E exp (itX), with i²=−1. Its modulus is at most one and it is continuous by dominated convergence, with no moment requirement. Independence multiplies characteristic functions of sums by product integration.

First test the missing hypothesis

For X uniform on [−a,a], direct integration gives φ(t)=sin(at)/(at), extended by1 at zero. As a grows without bound this tends to zero at every fixed nonzero t, while remaining1 at zero. The limit is discontinuous and cannot be a probability characteristic function. All mass escapes each fixed bounded interval. This is why pointwise convergence alone is insufficient.

Can pointwise transforms hide escaping mass?

X is uniform on [−a,a]. Its characteristic function is sin(at)/(at), with value1 at t=0. Predict what happens to mass in the fixed interval[−4,4] when a grows.

Support half-width a0.25141664 Frequency window h0.0156250.06250.251Reset tightness comparison

Fixed axes t∈[−1,1], vertical range[−0.25,1.125]; the shaded window is[−h,h], and the red point stays at(0,1). The513 plotted samples are illustrative; the analytic formula, not visual resolution, determines the limit.

Mass in the fixed interval[−4,4]: 1.00000. At t=h, φ(h)=0.841471. The tail threshold from the proof is R=2/h=8.00000: actual P(|X|≥R)=0.00000, while the conservative analytic upper bound is 0.111111.

The upper bound min(1,a²h²/9) follows from the frequency-average inequality below and1−cos u≤u²/2. It is not estimated by sampling the curve. A zero actual tail at a very large R does not show that one fixed R works uniformly for an expanding family.

Repair the quantifiers

For every fixed nonzero t, sin(at)/(at) tends to0 as a tends to infinity, but the value at zero stays1. The pointwise limit is discontinuous there and is not a probability characteristic function. For every fixed finite R, mass in[−R,R] tends to0. Choosing a smaller h separately for each a only proves each individual law is tight; a tight family needs one R for all members at a given tolerance.

Twenty analytic settings, no simulation or saved state. Reset restores a=4,h=1/4. Use the written tasks to retain a prediction and proof repair; a finite plot does not prove a theorem.

Frequency information prevents escape

A family of real probability laws is tight if, for every ε>0, there is one finite R such that every law gives probability at most ε to |x|>R. Suppose characteristic functions φ_n converge pointwise to φ, which is continuous at zero. Necessarily φ(0)=1. For h>0, bounded Fubini gives

$$A_n(h):=\frac1{2h}\int_{-h}^{h}(1-\operatorname{Re}\varphi_n(t))\,dt=\mathbb E\left[1-\frac{\sin(hX_n)}{hX_n}\right].$$

At X_n=0 the ratio is1. The integrand on the right is nonnegative. If |hX_n|≥2, the absolute ratio is at most1/2, so the bracket is at least1/2. Therefore P(|X_n|≥2/h)≤2A_n(h).

For fixed h, dominated convergence on the finite frequency interval gives A_n(h)→A(h), the same average formed from φ. Continuity at zero gives A(h)→0 as h decreases to zero. Given ε, choose h first so A(h)<ε/4, then choose n large so A_n(h)<ε/2. Thus the tail bound is below ε for all sufficiently late laws at R=2/h. Each of the finitely many earlier probability laws has vanishing tails by continuity of measure; enlarge R to handle them simultaneously. This proves tightness of the whole sequence. Choosing h separately for every n would not prove this statement.

For the uniform workbench, 1−cos u≤u²/2 and E[X²]=a²/3 give A(h)≤a²h²/18. Hence its displayed conservative bound min(1,a²h²/9) is analytic, not a quadrature certificate. Its exact tail is max(0,1−2/(ah)).

Extract a probability limit on the real line

Here is the compactness step rather than an invocation of a general theorem. Enumerate the rationals. Successively extract subsequences on which the distribution functions F_n converge at the first rational, then the second, and so on; boundedness in[0,1] permits each extraction. Take the diagonal subsequence, with rational limits g(q), which are nondecreasing in q. Define F(x)=inf of g(q) over rational q>x.

This F is nondecreasing. It is right-continuous: if x_j decreases to x, then eventually x_j<q for each rational q>x, so lim F(x_j)≤g(q); take the infimum and use monotonicity for the opposite inequality. Tightness forces its endpoint limits to be0 and1: sufficiently negative rational endpoints have g at most ε, and sufficiently positive ones have g at least1−ε. Construct the law explicitly: for u in(0,1), let Q(u)=inf of all x with F(x)≥u. The endpoint limits make Q finite, and monotonicity plus right-continuity give Q(u)≤x if and only if u≤F(x). Thus Q is measurable and the pushforward of uniform measure on(0,1) has distribution function F. No mass is lost at infinity.

For rational p<x<q, monotonicity bounds the subsequential lower and upper limits of F_n(x) between g(p) and g(q). The supremum of g(p) for p<x equals F(x−), and the infimum for q>x equals F(x). To check the first equality, use g(p)≤F(p) and, for every y<x, choose rational p in(y,x) so F(y)≤g(p). Consequently F_n(x)→F(x) at every continuity point of F. The distribution-function criterion proved above now gives weak convergence of the extracted subsequence. The same construction applies to any subsequence of a tight sequence.

Characteristic functions determine the law

We need uniqueness to prevent different subsequences from approaching different probability laws. Prove it with Gaussian smoothing, without assuming a general Fourier inversion theorem. Let g_ε(z)=(ε√(2π))⁻¹exp(−z²/(2ε²)). The elementary Gaussian integral and integration by parts yield

$$g_\varepsilon(z)=\frac1{2\pi}\int_{\mathbb R}e^{-itz}e^{-\varepsilon^2t^2/2}\,dt.$$

For completeness, write J(z)=∫exp(−s²/2)exp(−izs)ds. Differentiation is dominated by |s|exp(−s²/2); integration by parts, with vanishing boundary terms, gives J′(z)=−zJ(z). Squaring J(0) and using polar coordinates gives J(0)=√(2π). Solve this ODE and rescale to obtain the displayed identity. These integrals are absolutely convergent.

Convolving a probability law μ with gε gives the density p(μ,ε)(x)=∫g_ε(x−y)dμ(y). Substitute the displayed identity. Fubini is justified by the integrable Gaussian frequency factor and total mass1, giving

$$p_{\mu,\varepsilon}(x)=\frac1{2\pi}\int_{\mathbb R}e^{-itx}\varphi_\mu(t)e^{-\varepsilon^2t^2/2}\,dt.$$

If μ and ν have equal characteristic functions, their smoothed densities are equal for every ε>0. For any bounded continuous f, realize smoothing on the product space as X+εZ, where Z is standard normal independent of X. Pointwise continuity gives f(X+εZ)→f(X), dominated by the fixed bound on f. Thus the smoothed integrals tend to the original integrals. μ and ν agree on all bounded continuous tests. Continuous ramps decreasing to each half-line indicator then show equal distribution functions, hence equal laws. This uniqueness proof requires no moments of μ or ν.

Finish Lévy’s continuity theorem and then the CLT

Extract a weak limit μ using tightness. Since cos(tx) and sin(tx) are bounded continuous tests, its characteristic function equals the given pointwise limit φ. Every other subsequential weak limit has the same transform and therefore the same law. If the whole sequence failed to converge weakly to μ, some bounded continuous test would have a subsequence of expectation differences bounded away from zero. Tightness extracts a further weakly convergent subsequence, whose unique limit is μ, a contradiction. This proves both existence of the limiting law and weak convergence. Conversely weak convergence implies characteristic-function convergence by those same sine/cosine tests; the transform of a probability law is continuous at zero.

For centered X with finite variance σ², expand exp (iu)=1+iu−u²/2+r(u), where r(u)/u² tends to zero at zero and is bounded on the entire real line. The global bound follows from the elementary exponential bound for large u and the Taylor remainder near zero. Consequently r(tX)/t² is dominated by a constant times X² and tends pointwise to zero. Dominated convergence gives φ_X(t)=1−σ²t²/2+o(t²).

For iid centered summands with σ²>0, the standardized sum has characteristic function [φ_X(t/(σsqrt (n)))]^n, tending to exp (−t²/2). The Gaussian calculation above identifies this as the standard normal characteristic function. It is continuous at zero, so the continuity theorem just proved gives weak convergence. Thus the finite-variance iid CLT follows from the expansion, product limit, tightness and uniqueness arguments; it does not require a finite third or fourth moment.

The primary reference is MIT18.175 Spring 2014 Lecture 15, which states the continuity theorem and its tightness/subsequence proof strategy. This lesson supplies the real-line extraction and Gaussian uniqueness details. The later section supplies Gaussian-damped interval inversion and finite-dimensional projection convergence; symmetric-cutoff inversion is proved with its boundary-atom weights and finite-dimensional rectangle extension.

Apply

Explain why the CLT does not say an individual observation becomes normal or that every finite sample has an exact normal distribution. Its conclusion concerns the limiting law of centered, scaled sums under the stated assumptions. Zero variance gives a deterministic centered sum and must be treated separately rather than divided by zero.

A bounded inversion theorem under an explicit L¹ hypothesis

Assume the characteristic function φ of a probability law μ is absolutely integrable on the real line. Define p(x)=(1/(2π))∫exp(−itx)φ(t)dt. The integral is absolutely convergent; dominated convergence gives continuity, and |p(x)|≤||φ||₁/(2π). This is a sufficient density criterion, not a claim that every characteristic function is integrable.

The Gaussian-smoothed density already proved above is pε(x)=(1/(2π))∫exp(−itx)φ(t)exp(−ε²t²/2)dt. Its difference from p is bounded uniformly in x by (1/(2π))∫|φ(t)|·|exp(−ε²t²/2)−1|dt, which tends to zero by dominated convergence. Since each pε is nonnegative, p is nonnegative. For every continuous compactly supported h, uniform convergence on its support gives ∫h pε→∫h p. The probabilistic smoothing argument also gives ∫h pε→∫h dμ. Therefore these two measures agree on such tests. Continuous approximations to interval indicators identify their masses, and an increasing sequence of compact intervals shows ∫p=1. Thus p is a bounded continuous density of μ.

Worked distinction. The normal transform exp(−t²/2) satisfies the hypothesis and yields its Gaussian density. A point mass at zero has φ(t)=1 and infinite L¹ norm. It has no Lebesgue density, despite having a continuous characteristic function. Smoothing it produces Gaussian densities whose peaks diverge as ε decreases, so the uniform-convergence step is exactly what fails.

Changed written task. A law has density on [−1,1] that is constant 1/2. Its transform is sin t/t. Explain why failure of absolute integrability would not prove absence of a density. Feedback: the theorem is sufficient, not necessary. This law already has an explicit density, while |sin t/t| has divergent integral by summing lower bounds on fixed subintervals of successive periods. The following section treats arbitrary probability laws through Gaussian damping, including endpoint atoms, and proves finite-dimensional projection convergence. The proof here extends the existing Gaussian uniqueness mechanism with one additional integrability hypothesis.

Predict, change the case, and repair the claim

Case Integrable transformAtom

φ(t)=exp(−t²/2) is absolutely integrable.

Which additional conclusion is justified?

The law has bounded continuous density given by inverse transform Every probability law has such a density All moments follow from integrability alone

Check this claim Reset practice

Before checking, write the hypothesis or calculation that decides your answer. Change cases and explain which assumption changed. These fixed cases are public guided practice, not a saved score or independent proof assessment.

Inversion without an integrable characteristic function

Every probability law admits a useful Gaussian-damped inversion, even when it has atoms. For finite a<b, integrate the smoothed density over (a,b), use Fubini (the interval is finite and the frequency Gaussian is integrable), then let ε decrease to zero:

$$\lim_{\varepsilon\downarrow0}\frac1{2\pi}\int_{\mathbb R}\varphi(t)e^{-\varepsilon^2t^2/2}\frac{e^{-ita}-e^{-itb}}{it}\,dt=\mu((a,b))+\frac12\mu(\{a\})+\frac12\mu(\{b\}).$$

At t=0 interpret the quotient as b−a, its continuous limit. To prove the limit, condition on X=x: P(a<x+εZ<b) tends to 1 for interior x,0 for exterior x, and 1/2 at either endpoint because Z is symmetric with no atom at zero. The conditional probability lies in[0,1], so dominated convergence integrates these limits. No moment or density assumption on X is required. This is a regularized inversion formula, not permission to drop the Gaussian factor inside a conditionally divergent integral.

For X=0 certainly and (a,b)=(0,1), the answer is 1/2, not μ((0,1))=0. For (−1,1) it is 1. The half-endpoint terms make the inversion compatible with atoms. Choose endpoints with zero mass to recover the actual interval probability; arbitrary intervals then follow by limits of such endpoints because a probability law has at most countably many atoms.

Symmetric-cutoff inversion without Gaussian damping

The damping can be removed with a different justification. First prove the Dirichlet integral. Integration by parts bounds |∫ᴿˢ sin u/u du|≤2/R for S≥R>0, so the improper integral exists and its partial integrals are uniformly bounded. For η>0 let J(η)=∫₀^∞exp(−ηu)sin u/u du. Differentiate under the exponentially integrable bound to get J′(η)=−1/(1+η²); J(η)→0 as η→∞, hence J(η)=arctan(1/η). The tail bound 2/R remains valid with exp(−ηu)/u, which is decreasing, uniformly for η≥0. Split at R, pass η→0 on the finite interval, then let R grow. It follows that ∫₀^∞sin u/u du=π/2. This proves both the limit and the uniform bound needed below.

At finite T, Fubini is legitimate because (exp(−ita)−exp(−itb))/(it)=∫ₐᵇexp(−ity)dy has modulus at most b−a. After inserting φ(t)=∫exp(itx)dμ(x), the symmetric frequency integral becomes an expectation of

$$K_T(x)=\frac1\pi\left[\int_0^{T(x-a)}\frac{\sin u}{u}\,du-\int_0^{T(x-b)}\frac{\sin u}{u}\,du\right].$$

The Dirichlet integral shows K_T(x) tends to 1 inside(a,b),0 outside[a,b], and 1/2 at either endpoint. Its uniform bound is independent of x,T, so dominated convergence with respect to the probability law μ proves

$$\lim_{T\to\infty}\frac1{2\pi}\int_{-T}^{T}\frac{e^{-ita}-e^{-itb}}{it}\varphi(t)\,dt=\mu((a,b))+\frac12\mu(\{a\})+\frac12\mu(\{b\}).$$

No absolute integrability of φ is assumed. The symmetric cutoff and endpoint weights are part of the theorem. A finite cutoff is an approximation with no universal error bound for arbitrary μ; pointwise density inversion is a different assertion.

For a box ∏ⱼ(aⱼ,bⱼ) in Rᵈ, replace the one-dimensional quotient by its product over coordinates and integrate over the frequency cube[−T,T]ᵈ. Finite Fubini gives E∏ⱼK_T,j(Xⱼ). Each factor is uniformly bounded and converges to 1 in its interval,0 outside its closed interval and 1/2 at an endpoint. Dominated convergence therefore gives the box’s interior probability plus the corresponding half-weight per boundary coordinate. If the boundary has zero mass the limit is exactly the box probability. This is an explicit multidimensional rectangle inversion, not an unqualified claim about every possible Fourier truncation shape.

The symmetric-cutoff theorem is compared with MIT18.175 Spring 2014 Lecture 15, checked 30 October 2026. The original derivation above fills in the sine-integral bound and the boundary-atom limit.

Random vectors: all projections determine the joint law

For X in Rᵈ define φ_X(t)=E exp(i t·X). Coordinatewise multiplication of the one-dimensional Gaussian identity gives the density formula

$$p_{X,\varepsilon}(x)=(2\pi)^{-d}\int_{\mathbb R^d}e^{-it\cdot x}\varphi_X(t)e^{-\varepsilon^2|t|^2/2}\,dt.$$

The Gaussian factor is integrable in every finite dimension, so Fubini applies. If every real projection a·X and a·Y has the same law, then φ_X(a)=φ_Y(a) for every a by testing each projection at frequency 1. Their smoothed densities coincide. For every bounded continuous h, dominated convergence on X+εZ gives E h(X+εZ)→E h(X); likewise for Y. Therefore their laws coincide. This proves the Cramér–Wold determination statement, not just a slogan about marginals. The coordinate projections alone are insufficient: (U,U) and (U,−U) for a symmetric nondegenerate U have the same individual marginals, but projection(1,1) gives 2U in the first and 0 in the second.

Prove the convergence version using smoothing

Suppose a·Xₙ converges in distribution to a·X for every a in Rᵈ. Then φₙ(t)→φ(t) pointwise. For fixed ε>0, dominated convergence in the Gaussian density formula gives pₙ,ε(x)→pε(x) at every x. Both are nonnegative densities integrating to 1. The identity ∫|pₙ,ε−pε|=2−2∫min(pₙ,ε,pε), and domination of the minimum by pε, prove convergence in L¹. This is the density form of Scheffé’s lemma, with its proof included.

For a bounded Lipschitz test h with Lipschitz constant L, coupling each law with an independent standard Gaussian Z gives

$$|\mathbb Eh(X_n)-\mathbb Eh(X)|\le2L\varepsilon\mathbb E|Z|+\|h\|_\infty\|p_{n,\varepsilon}-p_\varepsilon\|_1.$$

First let n grow at fixed ε, then let ε shrink. Thus all bounded Lipschitz tests converge. To reach all bounded continuous tests, note that convergence of each coordinate gives uniformly small coordinate tails; a union bound gives tightness in a common cube. On that cube a continuous h is uniformly continuous and can be uniformly approximated by a bounded Lipschitz function: interpolate its values on a fine rectangular grid, then extend using coordinatewise clipping to the cube. The approximation is bounded by the same bound on h; outside the cube its expectation error is controlled by the uniformly small tail probabilities. Therefore all bounded continuous tests converge. This proves Xₙ⇒X without invoking an unproved multidimensional subsequence compactness theorem. The forward direction follows immediately by composing a bounded continuous real test with x↦a·x.

Changed projection task and review criteria

Let Z₁,Z₂ be independent standard normals and Xₙ=(Z₁,Z₁+Z₂/n). For a=(s,t), a·Xₙ=(s+t)Z₁+(t/n)Z₂ converges to(s+t)Z₁, so Xₙ converges weakly to(Z₁,Z₁), a law concentrated on a line. Its covariance matrix tends to a singular matrix; a nonsingular density is not required for weak convergence. If you test only each coordinate, you cannot distinguish this limit from independent normals. Review criteria: retain arbitrary(s,t), compute the projected variance(s+t)²+t²/n², identify the line-supported joint limit, and explain why density inversion without damping cannot be assumed.

Primary theorem comparator: Simon Fraser University STAT450 Lecture 14, 1999, checked 30 October 2026. The proofs here use the previously established Gaussian identity and explicitly justify smoothing, L¹ density convergence and removal of smoothing. The symmetric-cutoff inversion above has its own proof; arbitrary infinite-dimensional laws are not claimed.

Continue

This course now proves the integrable iid strong law and real-line characteristic-function continuity theorem, including the finite-variance iid CLT. It does not claim a complete graduate probability course: pointwise density inversion without suitable regularity, infinite-dimensional theory and full martingale convergence require further study. The Gaussian-damped inversion and finite-dimensional Cramér–Wold results above have explicit proofs. Review which conclusions used measure construction, convergence, product laws, densities and conditional expectation.