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For standard Brownian motion started at zero and exit time $T$ from $(-b,a)$, $a,b>0$, stop the two exponential Brownian martingales with parameters $\lambda$ and $-\lambda$. Their stopped values are bounded by $e^{|\lambda |\max (a,b)}$. The optional stopping theorem and dominated convergence theorem give $e^{\lambda a}p+e^{-\lambda b}q=1$ and $e^{-\lambda a}p+e^{\lambda b}q=1$, where $p,q$ are the discounted upper and lower exit probabilities. Solving gives the displayed formula and $q=\sinh (\lambda a)/\sinh (\lambda (a+b))$. ThusAt $\lambda =0$, the first ratio has its removable value $b/(a+b)$, and the second equals one. Finiteness of $T$ follows from a uniformly positive one-unit-time chance to leave any fixed bounded interval.

Asymptotic regularization stops the gradient flow $x^{\prime }=-A^{\ast }(Ax-f)$ at $t=1/\alpha$, starting from zero. With the singular system of a compact operator convention $A{v}_j={\sigma }_j{u}_j$, its spectral filter givesThe scalar coefficient solves $c_j^{\prime }=-{\sigma }_j^2{c}_j+{\sigma }_j⟨f,u_j⟩$ with zero initial value. Since $1-e^{-s}\le \min (s,1)$ for $s\ge 0$, the operator norm is at most ${\alpha }^{-1/2}$. On the domain of the Moore–Penrose inverse of an operator, dominated convergence theorem of the squared spectral coefficients proves $R_{\alpha }f\to A^{†}f$. The noise-bias decomposition for linear regularization then gives noisy-data convergence when $\delta /\sqrt{\alpha }\to 0$.

On a finite measure space, if measurable functions are uniformly bounded and converge pointwise, then their integrals converge. This is the dominated convergence theorem with a constant dominating function.

For an integrable supermartingale with the displayed increment bound, $|X_{T\wedge n}|\le |X_0|+KT$. Thus the stopped values are dominated by one integrable random variable, including $X_T$. Apply optional stopping at the bounded times $T\wedge n$ and then the dominated convergence theorem. For a martingale the same proof gives equality.

For standard Brownian motion started at zero, $H_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely for $a\ge 0$ andThe Brownian reflection principle proves finiteness. For $u\ge 0$, the Exponential martingale for Brownian motion stopped at $H_a\wedge t$ is bounded by $e^{ua}$. The dominated convergence theorem gives $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$, and taking $u=\sqrt{2\lambda }$ gives the transform.

For bounded continuous boundary data $f$, the half-plane Poisson kernel is the law of $x+yC$. On one common probability space, $x+yC\to x_0$ almost surely as $(x,y)\to (x_0,0)$ with $y>0$. Boundedness of $f$ and dominated convergence theorem imply $\varphi (x+iy)\to f(x_0)$. This proves continuity at every finite boundary point even for tangential approach. No moment of the Cauchy distribution is needed.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

If $m\ge 1$ is an integer and $\mathbb{E}[|X{|}^m]<\infty$, the characteristic function has continuous derivatives through order $m$, withThus finite absolute moments justify recovering moments from derivatives.^[ref]

For a short justification of differentiation under the integral sign, the difference quotient for the $(k-1)$st derivative has integrand $(iX{)}^{k-1}{e}^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X{|}^k$, since $|e^{iu}-1|\le |u|$ for real $u$. Lower absolute moments are finite because $|X{|}^k\le 1+|X{|}^m$. Apply the dominated convergence theorem successively, then use the same integrable bound to obtain continuity of each derivative.

For the Bernoulli distribution with parameter $1/3$, direct differentiation gives ${\phi }_B^{\prime }(0)=i/3$ and ${\phi }_B^{\prime \prime }(0)=-1/3$, agreeing with $\mathbb{E}[B]=\mathbb{E}[B^2]=1/3$. Its variance is therefore $1/3-(1/3{)}^2=2/9$. Existence and continuity of a characteristic function alone do not authorize these moment calculations: the standard Cauchy distribution, whose expected value is undefined, still has a characteristic function. The stated finite absolute moment hypothesis is a sufficient condition for the derivative formula, not an assumption built into the definition of a characteristic function.

If $\mathbb{E}|X|<\infty$, the characteristic function $\varphi (u)=\mathbb{E}{e}^{iuX}$ satisfies ${\varphi }^{\prime }(0)=i\mathbb{E}X$. The bound $|(e^{iuX}-1)/u|\le |X|$ and the dominated convergence theorem justify differentiation at zero. For mean zero, $\psi (u)=|1-\varphi (u)|/|u|$ extends continuously by $\psi (0)=0$, and $|1-\varphi (u{)}^n|\le n|u|\psi (u)$ by a finite geometric series. This supplies a direct characteristic-function proof of the weak law of large numbers.

// Target: martingale.bigb

For a probability measure on the real line with characteristic function $\phi$, averaging $1-\cos (tx)$ over $[-h,h]$ gives $1-\sin (hx)/(hx)$. This is at least $1/2$ when $|x|\ge 2/h$. Integrating gives the displayed bound. Consequently pointwise convergence of characteristic functions to a function continuous at zero implies tightness of probability measures, by the dominated convergence theorem on $[-h,h]$ and a separate tail bound for the finitely many initial measures.

A bounded linear operator between Banach spaces is completely continuous when it maps weakly convergent sequences to norm-convergent sequences. Equivalently, it maps weakly compact sets to norm-compact sets: the Eberlein-Šmulian theorem gives the forward direction, while a weakly convergent sequence together with its limit is weakly compact and gives the reverse direction. The inclusion $C(\mathbb{T})\to L^2(\mathbb{T})$ for finite Lebesgue measure is completely continuous by pointwise convergence, the Uniform boundedness principle and the dominated convergence theorem, but is not a compact operator because of the orthogonal Fourier modes.

If $X_n\to X$ almost surely and $|X_n|\le H$ for integrable $H$, conditional expectations converge almost surely and in $L^1$. Conditional Fatou lemma applied to $H+X_n$ and $H-X_n$ bounds the conditional liminf and limsup by the same limit. For $L^1$ convergence, the contraction estimate $\mathbb{E}|\mathbb{E}(X_n-X\mid \mathcal{G})|\le \mathbb{E}|X_n-X|$ and dominated convergence theorem finish the proof. This also permits taking an unbounded measurable factor out of a conditional expectation by clipping it, provided the original product is integrable.

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

More explicitly, if $t_n\to t$, then $e^{i{t}_{n}X}\to e^{itX}$ pointwise and $|e^{i{t}_{n}X}|=1$. Apply the dominated convergence theorem to the real part and imaginary part separately. There is no finite moment hypothesis. For uniform continuity, the stronger estimatefollows by factoring out $e^{itX}$, whose absolute value is one. The integrand tends pointwise to zero and is bounded by two, so the dominated convergence theorem applies again. The bound is independent of $t$, which supplies uniform continuity on the entire real line.^[ref]

If the first absolute moment is finite, $|e^{iu}-1|\le |u|$ additionally gives $|{\phi }_X(t+h)-{\phi }_X(t)|\le |h|\mathbb{E}[|X|]$, a Lipschitz continuity estimate. The earlier argument matters because continuity survives even when this expected value is infinite. For convergence in distribution, the separate Lévy continuity theorem concerns limits of whole sequences of characteristic functions; the argument here establishes continuity of one characteristic function.

If $f$ is jointly continuous on a compact rectangle $J\times [a,b]$, then the displayed parameter integral is continuous on $J$. For $t_n\to t$, pointwise continuity gives $f(t_n,x)\to f(t,x)$. The extreme value theorem bounds $|f|$ by a constant $M$ on the rectangle, and $M$ is integrable on the finite interval $[a,b]$. The dominated convergence theorem gives $F(t_n)\to F(t)$.

The same conclusion holds on an unbounded parameter interval if $f$ is jointly continuous there: use a compact rectangle around each fixed parameter value. A global bound over the whole parameter range is unnecessary.

The integral $I(x)={\int }_0^{\infty }{e}^{xt-t^3/3}dt$ is a rescaled Scorer Hi function. It has the three useful limitsFor the negative-argument limit, scale $t=s/M$ and use the dominated convergence theorem. At zero, substitute $s=t^3/3$ and use the Gamma integral. For the positive-argument limit, the exponent has its maximum at $t=\sqrt{M}$, with second derivative $-2\sqrt{M}$; Laplace's method gives the displayed factor. These estimates describe a cubic endpoint-to-saddle transition.

For $1<p<\infty$ and $u,v$ in an Lp space, the derivative with respect to the real scalar $t$ has the displayed form; the integrand is zero where $u=0$. The pointwise chain rule gives the derivative, while the Holder inequality makes $(|u|+|v|{)}^{p-1}|v|$ integrable and permits the dominated convergence theorem. This variational derivative characterizes nearest points to closed convex sets and constructs the Lp duality density of a bounded linear functional.

Let $t_0$ be interior to a parameter interval. Suppose $h(t,\cdot )$ is a measurable function for each $t$, $h(t_0,\cdot )$ is a Lebesgue integrable function, and $h(\cdot ,x)$ is differentiable near $t_0$ for $x$ almost everywhere. If one integrable nonnegative function $D$ satisfies $|{\partial }_{t}h(t,x)|\le D(x)$ throughout that neighbourhood off one fixed null set, then $F(t)={\int }_{X}h(t,x)d\mu (x)$ is defined there and satisfies the displayed derivative formula at $t_0$.

The mean value theorem gives $|h(t,x)-h(t_0,x)|\le |t-t_0|D(x)$, proving local integrability. It also bounds the difference quotient by $D$. That quotient converges almost everywhere to ${\partial }_{t}h(t_0,x)$, so the dominated convergence theorem permits passing the limit through the Lebesgue integral.

In the dividend-free Black-Scholes model, let $\tau =T-t>0$, $K>0$, $S>0$, $\sigma >0$ andThen the displayed finite sum prices the European option with power put option payoff. Here $\Phi$ is the standard normal distribution function. Indeed, $\log S_T$ conditionally has a normal distribution with mean $\log S+(r-{\sigma }^2/2)\tau$ and variance ${\sigma }^2\tau$. Expand $(K-S_T{)}^n$ by the binomial theorem on $S_T<K$ and apply the lower truncated lognormal moment to each term. Equivalently,This expectation representation proves monotonicity and convexity in $S$. It also gives $P_n(0,t)=K^n{e}^{-r\tau }$ and $P_n(S,t)\to 0$ as $S\to \infty$ by the dominated convergence theorem. For positive $S$ the lognormal distribution smoothing makes the price smooth and strictly positive.

If random variables $X_i$ satisfy ${\sum }_{i}E|X_i|<\infty$, then ${\sum }_i{X}_i$ converges absolutely almost surely, is integrable, and its expected value equals ${\sum }_{i}E[X_i]$. Indeed, the monotone convergence theorem gives $E[{\sum }_i|X_i|]={\sum }_{i}E|X_i|<\infty$, so that nonnegative sum is finite almost surely. It dominates every finite partial sum in absolute value; the dominated convergence theorem then justifies passing the limit through the expected value. No independence is required.

For example, if $|X_i|\le M{q}^i$ for a fixed $M$ and $0<q<1$, the geometric series bound proves the hypothesis directly. This is useful when repeated observations are multiplied by geometrically shrinking distances or weights.

Let $S_n$ have integrable, mean-zero independent and identically distributed random variables as increments, taking integer values at most one, with $\mathbb{P}(X_1=1)>0$. For a positive integer $b$, put $T_b=\inf \{n:S_n=b\}$ and $M(\tau )=\mathbb{E}{e}^{\tau X_1}$, $\tau >0$. Then $T_b$ is finite almost surely. Indeed, $b-S_{n\wedge T_b}$ is a nonnegative martingale; finite convergence on $\{T_b=\infty \}$ would force its integer increments eventually to vanish, contrary to the Borel-Cantelli lemmas applied to $\{X_n=1\}$.

The exponential martingale of a random walk $e^{\tau S_n}/M(\tau {)}^n$ stopped at $T_b$ is bounded by $e^{\tau b}$ because $S_{n\wedge T_b}\le b$ and $M(\tau )\ge 1$. The dominated convergence theorem therefore preserves its mean one at the limit and proves the displayed first-passage time transform. This argument accommodates arbitrarily large negative jumps; the restriction concerns upward jumps. For a simple symmetric random walk, $M(\tau )=\cosh \tau$, giving $\mathbb{E}{e}^{-\alpha T_b}=(e^{\alpha }-\sqrt{e^{2\alpha }-1}{)}^b$.