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Suppose an unknown has prior distribution $\Pi$ and its conditional observation law $P_u$ has jointly measurable Radon-Nikodym derivative $L(u,m)$ relative to a fixed observation law $P_0$. If $L>0$ for $\Pi (du)P_0(dm)$-almost every pair, thendefines the posterior distribution for almost every observation. Indeed, $\int L(u,m)P_0(dm)=1$ for almost every $u$, so Tonelli theorem gives $\int Z(m)P_0(dm)=1$. Hence $Z$ is finite almost everywhere, and positivity of $L$ and Fubini's theorem give $Z>0$ almost everywhere. The data law is $Z{P}_0$, and integration of the proposed posterior against this data law recovers the joint law. This proves the conditional distribution property and also shows that the data law is an equivalent probability measure to $P_0$.

The zero set $\{t\ge 0:B_t=0\}$ of one-dimensional Brownian motion is almost surely closed, uncountable, and has zero Lebesgue measure. The last fact follows from Tonelli theorem because $\mathbb{P}(B_t=0)=0$ for every $t>0$.

If $K$ obeys the Hörmander integral kernel condition with constant $C$, and an integrable $b$ is supported in a cube $Q$ with $\int b=0$, then ${\int }_{(LQ{)}^c}|K\ast b|\le C\Vert b{\Vert }_1$ for $L=4\sqrt{d}$. Here $LQ$ is the concentric cube with multiplied side length. Subtract the kernel value at the cube center inside the convolution; outside $LQ$, every source displacement is smaller than half the observation displacement. Tonelli theorem and the integral kernel condition give the bound.

For $0<\alpha <2$, represent $u^{-1-\alpha }$ as $\Gamma (1+\alpha {)}^{-1}{\int }_0^{\infty }{v}^{\alpha }{e}^{-vu}dv$. The Tonelli theorem applies to the nonnegative integrand. The inner integral in $u$ is $1/[v(1+v^2)]$, leaving $\Gamma (1+\alpha {)}^{-1}{\int }_0^{\infty }{v}^{\alpha -1}/(1+v^2)dv$. Substitution $z=v^2$ and the Euler beta function together with the Gamma reflection formula give the result. The bounds at zero and infinity explain exactly the index range.

For independent integrable variables, $\mathbb{E}{\prod }_i{X}_i={\prod }_i\mathbb{E}{X}_i$. Tonelli theorem first gives $\mathbb{E}{\prod }_i|X_i|={\prod }_i\mathbb{E}|X_i|<\infty$; then Fubini's theorem factors the signed joint integral.

For $0<a<b$, the nonnegative function $e^{-xy}{1}_{\{x>0,a\le y\le b\}}$ is measurable and has integral ${\int }_a^{b}dy/y=\log (b/a)$. Tonelli theorem first proves integrability without assuming it; Fubini theorem then permits the order of integration to be exchanged. Integrating in $y$ gives the stated Frullani integral. The cancellation between the exponentials is essential at $x=0$.

Let $X_n$ be integrable independent and identically distributed random variables, and let $T\ge 1$ be a stopping time for their natural filtration, with $\mathbb{E}T<\infty$. Since $\{T\ge n\}$ is measurable before $X_n$ is observed, independence and the Tonelli theorem give $\mathbb{E}|X_T|\le {\sum }_n\mathbb{E}[|X_n|1_{\{T\ge n\}}]=\mathbb{E}|X_1|\mathbb{E}T$. The event $T=n$ need not be independent of $X_n$, so the selected increment can be biased.

The integrated mean squared error is $\mathbb{E}{\int }_D(\hat{m}(x)-m(x){)}^{2}dx$ over a specified domain $D$. The Tonelli theorem permits integration of the pointwise bias-variance decomposition of mean squared error.

For an $L^2$ kernel and an independent sample from a density $f$, the centred kernel density estimator hasIndependence removes cross terms and Tonelli theorem integrates the pointwise variance. Young's convolution inequality ensures $K_h\ast f\in L^2$ even without $f\in L^2$. The Cauchy-Schwarz inequality yields the expected-norm bound $\mathbb{E}\Vert {\hat{f}}_{n,h}-K_h\ast f{\Vert }_2\le \Vert K{\Vert }_2/\sqrt{nh}$.

For a nonnegative measurable function $g$,On a graph, applying this identity to $|g(x)-g(y)|$ expresses its total variation as the integral of the edge boundaries of the superlevel sets.

More generally, if $\phi :[0,\infty )\to [0,\infty )$ is differentiable, increasing, and satisfies $\phi (0)=0$, then Tonelli theorem and the fundamental theorem of calculus give

Let $F,G\ge 0$, $1<p<\infty$, and $G\in L^p$. Assume the displayed tail inequality and finite positive-threshold superlevel measures. First split $G$ at $a/2$ to obtain $a\mu \{F>2a\}\le {\int }_{\{G>a\}}G$. Layer cake representation and Tonelli theorem imply $F\in L^p$. Integrating the original inequality against $p{a}^{p-2}$ then gives $\Vert F{\Vert }_p^p\le q\int G{F}^{p-1}\le q\Vert G{\Vert }_p\Vert F{\Vert }_p^{p-1}$, with $q=p/(p-1)$. This is the analytic estimate behind the Doob Lp maximal inequality.

Let $X=\min (T,C)$, $V=1_{\{T\le C\}}$, and suppose $T$ and $C$ are independent, with continuous failure density $f=hS$. Then the Tonelli theorem givesThus $V-H(X)$ has mean zero. The identity also holds conditionally on covariates when censoring is independent conditionally on them. For an estimated hazard, the mean error is controlled by $\mathbb{E}|\hat{H}(X)-H(X)|$, not by pointwise fit alone.

// Target: survival-analysis.bigb

For geometric-attempt exponential backoff, assume each attempted transmission fails independently with a fixed probability $q<1$. The probability of reaching stage $k$ is $q^k$, and its mean waiting time is $2^k/p_0$. The Tonelli theorem gives $\mathbb{E}T=p_0^{-1}{\sum }_{k\ge 0}(2q{)}^k$, finite exactly when $q<1/2$. The packet succeeds almost surely for every $q<1$, so almost-sure success does not guarantee finite mean delay. A fixed independent failure probability is a simplifying assumption, not a theorem about collisions in an interacting network.

// Target: probability-and-statistics.bigb

A multiple integral accumulates a function over a region with several coordinates. The Fubini theorem and Tonelli theorem give conditions under which it equals successive one-dimensional integrals. A change of variables formula expresses it in new coordinates using the absolute Jacobian determinant.

For $d\ge 3$, $A_d=\Gamma (d/2-1)/(2{\pi }^{d/2})$. Substitute $u=|x{|}^2/(2s)$ to obtain the displayed identity for $x\ne 0$. The semigroup property and Tonelli theorem then give $P_t(|\cdot {|}^{2-d})(x)=A_d^{-1}{\int }_t^{\infty }{p}_s(x)ds$. The coefficient multiplying the time integral is the reciprocal of the coefficient in the full Green integral, a distinction important in Brownian motion hitting asymptotics.

For every nonprincipal Dirichlet character, $L(1,\chi )\ne 0$. A nonreal character is covered by Euler product positivity for L-function nonvanishing at height zero, since its squared character is nonprincipal.

For a real character, suppose $L(1,\chi )=0$. The simple zeta pole would cancel, making $F(s)=\zeta (s)L(s,\chi )$ an entire function. Its nonnegative zeta-times-real-L coefficients satisfy $a_n\ge 0$ and $a_{m^2}\ge 1$. Termwise derivatives at two give $(-1{)}^k{F}^{(k)}(2)={\sum }_n{a}_n(\log n{)}^k{n}^{-2}$. The Taylor series of this entire function at two converges at zero. Every term there is nonnegative, and Tonelli theorem identifies its sum asThe last series diverges because all square coefficients are at least one. This contradiction proves the claim. It is the positive-coefficient mechanism behind the more general Landau theorem for a Dirichlet series with nonnegative coefficients.

Since the kernel has integral one, The Tonelli theorem and the triangle inequality giveThis proves the stated convolution approximation estimate.

For the unlabelled compactness assertion let $M$ bound the norms of $S$, and fix $\eta >0$. Choose the kernel width $\delta$ smaller than the translation radius supplied for $\eta$, so that the same bound holds for all shifts of size at most this width. The preceding estimate gives $\Vert f\ast g-f{\Vert }_1<\eta$ uniformly. The smoothed family has uniform bound $M/\delta$ and uniform Lipschitz constant $M/{\delta }^2$. Choose $K$ with all original tails outside $[-K,K]$ below $\eta$, and put $L=K+\delta$. The Tonelli theorem givesOn $[-L,L]$ the smoothed family is relatively compact in the uniform norm by the Arzelà-Ascoli theorem. Take a finite uniform net with error below $\eta /(2L)$ and extend its centers by zero outside that interval. It is a finite $L^1$ net for the smoothed-and-truncated family with error $\eta$. Each original $f$ is within another $2\eta$ of its corresponding truncated smoothing. Thus $S$ is totally bounded in $L^1$. The displayed (a),(b) are the two hypotheses used together in this argument.

For the final family, Young convolution inequality gives $\Vert f_0\ast f{\Vert }_1\le \Vert f_0{\Vert }_1$. Its translations satisfywhich tends to zero uniformly over the family. Since $f$ vanishes outside $[-1,1]$,uniformly as well. Both compactness hypotheses apply, proving $S_0$ is totally bounded.

Let $\varphi (u)=\int e^{iuy}\mu (dy)$. Since $1-\cos (uy)\ge 0$, the Tonelli theorem giveswith the fraction defined to be one at zero. For $|x|\ge 1$, $\sin x/x\le \sin 1$. To verify the constant, the function is even; on $[1,\pi ]$ it decreases because $(x\cos x-\sin x{)}^{\prime }=-x\sin x<0$ on $(0,\pi )$ and $x\cos x-\sin x$ starts from zero. On $[\pi ,\infty )$, $\sin x/x\le 1/\pi <\sin 1$. Consequently the integrand is at least $1-\sin 1$ on $|y|\ge \lambda$, proving the characteristic-function tail boundFor pointwise convergence ${\varphi }_n\to \varphi$, where $\varphi$ is a characteristic function, the dominated convergence theorem applies on each finite integration interval, since the integrands lie between zero and two. HenceContinuity of $\varphi$ at zero and $\varphi (0)=1$ make the right side tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ so it is below $\epsilon /2$. The tail probability is then below $\epsilon$ for all sufficiently large $n$. Enlarge the compact interval to control the finitely many remaining measures, each of which has vanishing tails. This proves uniform tightness of the entire sequence, not just of its tail. It is the characteristic-function tightness bound principle with the sharper displayed constant.

For the integrable mean-zero summands, differentiation at zero is justified byThe dominated convergence theorem gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1=0$. DefineIt is continuous away from zero by continuity of the characteristic function, and continuous at zero by the zero derivative. Since $|\varphi (u)|\le 1$, the finite geometric series yields, for $n\ge 1$,and the case $n=0$ has both sides zero. Thus the requested uniform-in-$n$ estimate isThis is the characteristic-function derivative under an absolute first moment estimate.

Independence makes the characteristic function of $S_n/n=(X_1+\cdots +X_n)/n$ equal to $\varphi (u/n{)}^n$. Therefore $|1-\varphi (u/n{)}^n|\le |u|\psi (u/n)$. Applying the already proved characteristic-function tail bound to its distribution, for every $\epsilon >0$,The convergence follows because ${\sup }_{0\le u\le 1/\epsilon }\psi (u/n)\to 0$, and the remaining integral of $u$ is finite. Hence $\boxed{S_n/n\to 0\text{ in probability}}$, the mean-zero weak law of large numbers. This deduction uses the tail bound directly and does not presuppose the Lévy continuity theorem.

For the continuous adapted process $X$, the supremum on each compact time interval is attained. Let $D_t=(\mathbb{Q}\cap [0,t])\cup \{t\}$. Continuity gives ${\sup }_{s\le t}{X}_s={\sup }_{r\in D_t}{X}_r$, and henceEach event inside the union is in ${\mathcal{F}}_r\subseteq {\mathcal{F}}_t$, so this proves carefully that $T$ is a stopping time. Merely checking rational times for $X_r\ge \lambda$ would miss a path that touches the level at an isolated irrational time; the approximation from below avoids this issue. This is the closed-level hitting times of continuous adapted processes criterion. For $\lambda =0$, nonnegativity makes $T=0$ and the later inequality is immediate.

For $\lambda >0$, put $A=\{T\le t\}$ and $\tau =T\wedge t$. On $A$, continuity implies $X_{\tau }\ge \lambda$. To justify the optional sampling step without extra filtration assumptions, round $\tau$ upwards to a finite dyadic grid in $[0,t]$, obtaining stopping times ${\tau }_n\downarrow \tau$. The event $A$ belongs to ${\mathcal{F}}_{\tau }$, and hence to ${\mathcal{F}}_{{\tau }_n}$: directly, $A\cap \{\tau \le u\}=\{T\le u\}$ for $u<t$, while for $u\ge t$ it is $A$. Discrete optional sampling for the sampled submartingale givesAs $n\to \infty$, continuity gives $X_{{\tau }_n}\to X_{\tau }$. Nonnegativity and the Fatou lemma therefore yieldConsequently the Doob maximal inequality for a nonnegative submartingale takes the precise form

For the second-moment bound, suppose first that $\mathbb{E}{X}_t^2<\infty$; otherwise the inequality with an infinite right side is automatic. Set $M_R=X_t^{\ast }\wedge R$. The layer cake representation and the preceding tail estimate giveThe integrands are nonnegative, so the interchange is justified by the Tonelli theorem. Since $M_R$ is bounded, division is legitimate when its norm is nonzero, giving $\Vert M_R{\Vert }_2\le 2\Vert X_t{\Vert }_2$; the zero-norm case is immediate. Finally, monotone convergence as $R\to \infty$ givesThis truncated layer-cake proof of the L2 maximal inequality proves square integrability of the maximum rather than assuming it during the Cauchy-Schwarz inequality step.

For the population identity, write $X=\min (T,C)$, $V=1_{\{T\le C\}}$, and $G(t)=\mathbb{P}(C\ge t)$. Assume independent censoring, with $f_T(t)=h(t)S(t)$. Thenwhile the Tonelli theorem and the cumulative hazard function giveThis is mean accumulated hazard under independent censoring, proving $\mathbb{E}[V-H(X)]=0$. For the fitted martingale residual $Y=V-\hat{H}(X)$,Hence its expectation is approximately zero when the fitted hazard error at the observed time is small in mean. This spells out the required sense of a good estimate; pointwise consistency alone does not control a divergent tail. In particular, a Kaplan–Meier estimator with zero final survival gives infinite transformed times there and cannot supply finite residuals without a tail restriction. Neither exact finite-sample zero mean nor validity under informative censoring is being assumed.

To check an omitted explanatory variable $Z$, plot the martingale residuals against $Z$ or compare groupwise mean residuals, using a smooth trend or appropriate uncertainty intervals. Under a correct conditional survival model and conditionally independent censoring, $\mathbb{E}[Y\mid Z]$ should be approximately zero. A systematic positive trend indicates more observed failures than the fitted accumulated hazard predicts; a negative trend indicates fewer. Such patterns suggest adding $Z$, a nonlinear term or an interaction and reassessing the fit. The bounded-above, often long negative tail of martingale residuals means that the diagnostic is a mean-pattern check, not a requirement for symmetric Gaussian residuals.