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.

Suppose continuous functions $f_n$ converge by uniform convergence to $f$ on a domain of a function $X$, and $x_n\to x$ with all these inputs in $X$. The uniform limit theorem makes $f$ a continuous function, and the triangle inequality givesThe first term handles the changing approximation and the second handles the changing input. For $f_n(x)=x^n$ on $[0,1]$, the inputs $x_n=2^{-1/n}$ instead give $x_n\to 1$, $f_n(x_n)=1/2$, and $f(1)=1$. Pointwise convergence cannot supply the first uniform bound, and this pointwise limit also lacks continuity at one.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

On the half-line, $\sup f_n=1/(en)\to 0$, but ${\int }_0^{\infty }{f}_n=1$. The change of variable $x=ny$ shows why: the decreasing amplitude is spread over an increasing interval. Uniform convergence controls integrals on finite-measure domains; it does not alone control total mass on an infinite-measure domain.

For functions taking values in real numbers on a nonempty domain of a function $X$, the supremum error is the supremum of the absolute value of the approximation error over all inputs. It may be infinite and need not be attained. When finite, it is the supremum norm of $f_n-f$; uniform convergence is exactly $E_n\to 0$. Bounding this one quantity gives an accuracy guarantee for every input, including inputs that change with $n$.

Suppose $d$ is nonnegative, integrable and uniformly continuous on the positive half-line. If $d(t_n)\ge ϵ>0$ at times tending to infinity, uniform continuity supplies intervals of one fixed positive length around those times on which $d\ge ϵ/2$. An infinite disjoint subsequence of these intervals contradicts integrability. Thus the displayed implication holds. Mere smoothness at every finite time is insufficient: smooth peaks of shrinking widths can have finite integral and heights bounded away from zero.

A uniform algebra on $K$ is natural if every character of an algebra is evaluation at a point of $K$. The evaluation embedding then identifies its character space homeomorphically with $K$. A natural uniform algebra is characterized by the finite-generator criterion for a natural uniform algebra. In particular, $C(K)$ is natural; the disk algebra is natural when its underlying space is the closed disk.

The disk algebra consists of functions continuous on the closed unit disk and holomorphic in its interior, with the supremum norm. Polynomials are dense: dilations $f_r(z)=f(rz)$ converge uniformly to $f$ as $r\uparrow 1$, and each $f_r$ is uniformly approximated on the closed disk by its Taylor polynomials. A character of an algebra is therefore determined by its value $\lambda$ at the coordinate function; boundedness gives $|\lambda |\le 1$, and polynomial approximation yields $\phi (f)=f(\lambda )$. Consequently the disk algebra is a natural uniform algebra on the closed disk.

Let $C$ consist of continuous functions on the closed unit disk whose boundary values extend to a member of the disk algebra. The extension is unique by the maximum modulus principle; the extension map $T:C\to A(\overline{\mathbb{D}})$ is a surjective unital algebra homomorphism of norm at most one. Its kernel $I$ is the ideal of functions vanishing on the boundary. A character vanishing on $I$ factors through $T$, so it is $f\mapsto T(f)(\lambda )$ for some closed-disk point $\lambda$. For a character $\chi$ not vanishing on $I$, choose $h\in I$ with $\chi (h)\ne 0$. The rule $F\mapsto \chi (hF)/\chi (h)$ extends $\chi$ to a character of all continuous functions on the disk: multiplicativity follows from $\chi (hF)\chi (hG)=\chi (h)\chi (hFG)$. It is therefore evaluation at an interior point, since its value at $h$ is nonzero. These two closed-disk families agree exactly on their boundaries. Their glued compact space maps continuously and bijectively to the Hausdorff character space of an algebra, giving the stated homeomorphism. The algebra is closed because uniform convergence of boundary data gives uniform convergence of their analytic extensions.

For a positive Borel measure on ${\mathbb{R}}^d$, let $E_{\alpha }=\{M_u\nu >\alpha \}$, using open balls in the uncentered maximal function of a finite measure. Every ball witnessing the inequality lies entirely in $E_{\alpha }$. Cover a compact subset by finitely many witnessing balls and select disjoint balls greedily by decreasing radius. Their threefold dilations cover the compact set. Their total measure is at most $\nu (E_{\alpha })$, giving the bound by inner regularity of Lebesgue measure. For a signed or complex measure, apply this to its total variation measure.

An unbiased estimator $T$ is uniformly minimum-variance unbiased if its variance is no larger than that of every other unbiased estimator of the same parameter, for every parameter value. The Lehmann–Scheffé theorem gives such an estimator whenever an unbiased function of a complete sufficient statistic exists. Attaining the Cramér-Rao bound is sufficient for this property under its hypotheses, but is not necessary: for $T\sim \Gamma (n,\lambda )$ with $n>2$, $(n-1)/T$ is unbiased and has variance ${\lambda }^2/(n-2)$, while the information bound is ${\lambda }^2/n$. Completeness follows from uniqueness of the Laplace transform of $h(t)t^{n-1}$, so the estimator is still uniformly optimal among unbiased estimators.

For independent $X_i=\theta +Y_i$ with $Y_i$ exponentially distributed at known rate $\lambda$, the joint likelihood factors through the sample minimum $X_{(1)}$. Its excess over the endpoint is exponential with rate $n\lambda$, because $\mathbb{P}(\min Y_i>t)=e^{-n\lambda t}$. Thus the displayed estimator is an unbiased estimator, with variance $(n\lambda {)}^{-2}$. Even if $\theta \ge 0$, the estimator can be negative; truncation at zero introduces bias.

An ultralimit is the direct limit of a chain of structures formed by taking successive ultrapowers along their diagonal elementary embeddings. At limit stages use the elementary chain theorem. This is a different concept from a topological ultrafilter limit. Alternating relative ultrapower embeddings constructs isomorphic ultralimits of any two elementarily equivalent structures, without requiring the stronger single-ultrapower isomorphism theorem.

Let $S$ be a nonempty set and $\mathcal{P}\subseteq \mathcal{P}(S)$ be upward closed. An ultrafilter all of whose members lie in $\mathcal{P}$ exists exactly when every finite set partition of $S$ has a cell in $\mathcal{P}$. For sufficiency, the complements of sets outside $\mathcal{P}$ have the finite intersection property: a finite cover by such sets could be disjointified into a partition with no cell in $\mathcal{P}$. The ultrafilter lemma extends the generated proper filter. Necessity follows because an ultrafilter selects exactly one cell of every finite partition. This criterion requires partition richness of the whole ground set; it does not assert that the complement family is closed under unions.

The finite-trace independent family construction gives $2^{|X|}$ independent subsets of an infinite $X$. Each binary assignment generates a proper filter and, in ZFC, has an ultrafilter extension. Different assignments force incompatible membership decisions, giving $2^{2^{|X|}}$ distinct ultrafilters. The matching upper bound follows because every ultrafilter is a subset of $\mathcal{P}(X)$.

An ultrafilter with finite-sums members is an ultrafilter on the positive integers all of whose members are IP sets. Such an ultrafilter exists by extending the IP-star filter using the ultrafilter lemma. If a member were not IP, its complement would be an IP-star set and hence another member, contradicting properness. Conversely, every such ultrafilter contains the entire IP-star filter, since it cannot contain the non-IP complement of an IP-star set. This characterization does not assert that the ultrafilter is idempotent.

An ultrafilter on the positive integers with every member containing finite arithmetic progressions of arbitrarily large length. The Van der Waerden theorem and ultrafilter selection from a partition-rich family give existence. Such an ultrafilter must contain the complement of every set with bounded arithmetic-progression length, and it is a nonprincipal ultrafilter. No corresponding ultrafilter can have an infinite arithmetic progression in every member, by the dyadic block obstruction to infinite arithmetic progressions.

A uniform kappa-complete ultrafilter $U$ on an uncountable cardinal $\kappa$ is normal if every regressive function on a $U$-large set is constant on a $U$-large subset. Equivalently it is closed under diagonal intersections ${\Delta }_{\alpha <\kappa }{X}_{\alpha }=\{\beta <\kappa :(\forall \alpha <\beta )\beta \in X_{\alpha }\}$. Normality supplies homogeneous sets for pair-colourings by combining the chosen large tail at each earlier coordinate.

An ultrafilter on $P_{\kappa }(\lambda )$ is fine when it contains the set of small subsets containing $\alpha$, for each $\alpha <\lambda$. Fineness makes every coordinate of $\lambda$ represented in the corresponding ultrapower, though no one small subset is required to contain every coordinate.

In the classical risk model, condition on the first arrival of the Poisson process. If available capital just before it is $v$, ruin has probability ${\overline{F}}_X(v)+{\int }_0^v{f}_X(x)\psi (v-x)dx$. Integrate this expression at $v=u+ct$ against $\lambda e^{-\lambda t}$ and change variables from $t$ to $v$. Differentiating the lower limit gives the displayed equation. The claim-density convolution is continuous for bounded $\psi$, so the first-arrival representation also supplies the differentiability required by the equation.