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The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.

The bounded-time optional sampling theorem says that for an integrable discrete-time martingale $(U_n)$ and a bounded stopping time $S$, $\mathbb{E}[U_S]=\mathbb{E}[U_0]$. It requires no assumption about an eventual unbounded stopping time.

Apply this theorem with $S=n\wedge \tau$, or simply take the expectation of the stopped martingale in part (ii). For $N>0$ we obtainSince $P_{n\wedge \tau }\ge 0$,The monotone convergence theorem now yields $\mathbb{E}[\tau ]\le 3abc/N<\infty$. In particular $\tau <\infty$ almost surely, which has been established without assuming that optional stopping is valid at $\tau$.

For each sample path with finite $\tau$, $P_{n\wedge \tau }$ eventually equals $P_{\tau }=0$. Also $0\le P_{n\wedge \tau }\le N^3$. The dominated convergence theorem gives $\mathbb{E}[P_{n\wedge \tau }]\to 0$. Taking the limit of a sequence in the earlier identity therefore provesFor example, initial holdings $(1,1,1)$ give expectation one, as the first transfer must eliminate its giver. This computes the first elimination time, rather than the later time when one player owns all tokens.

Assume $X$ is square-integrable. Since $X$ and $Y$ have the same probability distribution, $Y$ is also square-integrable and $\mathbb{E}[Y^2]=\mathbb{E}[X^2]$. The product $XY$ is integrable by the Cauchy-Schwarz inequality.

We use the defining test-function identity for conditional expectation: for bounded $\mathcal{G}$-measurable $H$, $\mathbb{E}[XH]=\mathbb{E}[YH]$. Take $H=Y{1}_{\{|Y|\le m\}}$ and let $m\to \infty$. The dominated convergence theorem, with bounds $|XY|$ and $Y^2$, givesThis is the extension of the conditional expectation pull-out identity to this square-integrable multiplier. ConsequentlyA nonnegative random variable with zero expectation is zero almost surely. HenceGeometrically, this is the equality case in the L2 space orthogonal projection interpretation of conditional expectation.

A dominated differentiation under the integral sign suffices. Suppose $\varphi (\cdot ,t)$ is measurable, $\varphi (\cdot ,t_0)$ is integrable at one parameter value, and for almost every $x$ the map $t\mapsto \varphi (x,t)$ is continuously differentiable on an interval. If $|{\partial }_t\varphi (x,t)|\le H(x)$ throughout that interval for an integrable $H$, then the parameter integral is differentiable, andOne direct sufficient version assumes ${\partial }_t\varphi (x,t)$ is continuous in $t$ almost everywhere and has that common domination; dominated convergence theorem then proves the assertion.

Here $|f(x)e^{itx}|=|f(x)|$, so the complex-valued function $h$ is integrable for every real $t$. For a real increment $u\ne 0$, the elementary bound $|e^{iux}-1|\le |ux|$ givesThe right side is integrable by hypothesis and the integrand converges pointwise to $ixf(x)e^{itx}$. dominated convergence theorem, applied to real and imaginary parts, provesThis supplies the domination explicitly rather than merely differentiating a formal Fourier transform.

For fixed $x>0$, $f_n(x)\to e^{-x}\sin x$; for $x\le 0$, the limit is zero. On $0<x<n$, the supplied inequality gives $(1-x/n{)}^n\le e^{-x}$, soThis is the given integrable dominating function. By the dominated convergence theorem,

The dominated convergence theorem states: if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le g$ almost everywhere for an integrable $g\ge 0$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$.

To derive it from the monotone convergence theorem, first prove Fatou's lemma. For nonnegative measurable $h_n$, the functions $k_N={\inf }_{n\ge N}{h}_n$ increase to $\mathrm{liminf}{h}_n$ and satisfy $\int k_N\le {\inf }_{n\ge N}\int h_n$. Monotone convergence yields $\int \mathrm{liminf}{h}_n\le \mathrm{liminf}\int h_n$.

Now $|f|\le g$ almost everywhere, so $f$ is integrable. Apply Fatou's lemma to the nonnegative functions $2g-|f_n-f|$, which converge almost everywhere to $2g$:Hence $\int |f_n-f|\to 0$, and $|\int f_n-\int f|\le \int |f_n-f|$ proves the assertion. Common exceptional null sets may be removed at the beginning since the sequence is countable.

The monotone convergence theorem states that if nonnegative measurable functions $h_n$ increase almost everywhere to $h$, then $\int h={\lim }_n\int h_n$, allowing infinity. The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ for an integrable $g$, then $h$ is integrable and $\int h_n\to \int h$.

On the fixed parameter interval the integrand tends to $\cos \theta$ as $\kappa \to 0$ and is bounded by $\cos \theta$. The dominated convergence theorem, or uniform convergence here, givesThis equals Sam's time between departure and reunion. Physically, $dx/dt=-c\kappa \tan \theta$ tends uniformly to zero, so Alice's worldline approaches Sam's, and the time dilation disappears.

With $G$ a standard normal random variable,For $f\in C_b^1$, in the usual sense that $f$ and its continuous derivative are bounded, the mean value theorem bounds the difference quotient by $a(t)\Vert f^{\prime }{\Vert }_{\infty }$. Dominated convergence theorem therefore allows differentiation inside the integral:This also follows pathwise from $X(t,x+h)-X(t,x)=a(t)h$. If the textual description were read as assuming only bounded differentiability, some control of $f^{\prime }$ is needed for the displayed right side; the stated $C_b^1$ hypothesis supplies it. Gaussian smoothing itself still differentiates a bounded measurable $f$, using the derivative of the Normal distribution instead.

Using the Cauchy distribution from the preceding part gives one common probability space for all starting points:As $(x,y)\to (0,0)$ with $y>0$, $x+yC\to 0$ almost surely, since $C$ is a finite real random variable almost surely. Continuity of $f$ at zero gives $f(x+yC)\to f(0)$ almost surely, and boundedness supplies an integrable constant bound. Dominated convergence theorem therefore provesThe argument applies to every sequence approaching zero, including tangential approaches. Along the real boundary the extension is already $f$, which is continuous. Hence the extension is continuous at zero in the closed half-plane. No finite moment of the Cauchy distribution is required.

Let $p=\mathbb{P}(y_{\tau }=\alpha )$. For $k\ne 2$, the bounded stopped martingale from part (b) converges to $y_{\tau }^{2-k}$, so the dominated convergence theorem givesTherefore the boundary hitting probability from a diffusion scale function isFor $k=2$, applying Itô's formula to $\log Y_t$ gives a bounded stopped martingale instead, yieldingIf $k>2$ and $\beta >1$ is fixed, the first expression tends to zero as $\alpha \downarrow 0$. A continuous path that reaches zero before it reaches $\beta$ must first reach every such $\alpha$ before $\beta$, so that event has probability zero. Any path hitting zero in finite time has a finite maximum up to that time, and therefore belongs to one of these zero-probability events with an integer upper boundary $\beta >1$. Their countable union still has probability zero. Thus the process does not hit zero almost surely. This is the relevant part of the Hitting-zero classification for a Bessel process.

A measurable-function class is pointwise separable if it has a countable subclass such that every member is the pointwise limit of a sequence from that subclass. With an integrable common envelope, dominated convergence theorem makes both empirical averages and population expectations converge along each such sequence. Thus the supremum of the absolute empirical errors equals the supremum over the countable subclass and is measurable. A finite integrable bracketing cover provides an integrable envelope by taking the maximum absolute value of all its endpoints.

Three players with positive integer holdings exchange a token by choosing an unordered active pair uniformly, then choosing the transfer direction fairly. Let $N$ be the conserved total, and $\tau$ the stopping time of the first elimination. Given holdings $(x,y,z)$ before $\tau$, the conditional expectation of the change of their product is $-(x+y+z)/3=-N/3$: transfers in each pair decrease the mean product by the third holding. Thus the displayed stopped process is a martingale. Its nonnegative product term gives $\mathbb{E}[n\wedge \tau ]\le 3abc/N$, so the monotone convergence theorem first proves integrability of $\tau$. The bounded product then vanishes by the dominated convergence theorem, yielding $\mathbb{E}[\tau ]=3abc/N$. Continuing the deterministic time term beyond elimination instead produces a submartingale.

Scale the offered load and capacity together. Reversing the denominator of the Erlang loss formula givesFor each fixed $j$, the summand tends to $(C/\nu {)}^j$. If $\nu >C$, the dominated convergence theorem applies with the summable geometric series bound $(C/\nu {)}^j$. Otherwise each fixed finite partial sum has limiting value at least its number of terms, so the reciprocal diverges. Taking reciprocals proves the formula, including the critical case $\nu =C$.

For a European put option with deterministic discount factor $D$ and an integrable terminal stock price, $P(K)=D{\mathbb{E}}_{\mathbb{Q}}[(K-S_T{)}^{+}]$. If $\mathbb{Q}(S_T=K)=0$, bounded difference quotients and the dominated convergence theorem give the displayed derivative, the price of a digital put option. At an atom, the left derivative is $D\mathbb{Q}(S_T<K)$ and the right derivative is $D\mathbb{Q}(S_T\le K)$, so the ordinary derivative need not exist.

For $B(z)=\lambda z^m{\prod }_n{b}_{a_n}(z)$ with nonzero zeros $a_n$ counted with multiplicity, Jensen's formula gives the displayed identity. The negative logarithms of partial products increase, so the monotone convergence theorem permits termwise integration. The Blaschke condition implies ${\sum }_n-\log |a_n|<\infty$, which dominates the absolute summands. The dominated convergence theorem therefore gives a limit of zero as $r\uparrow 1$. Since $|B|\le 1$, the mean of $|\log |B||$ tends to zero as well.

A bounded continuous factor $k_1$ and an integrable factor $k_2$ give a bounded velocity-integral gain on bounded continuous functions, with bound $\Vert Kf{\Vert }_{\infty }\le \Vert k_1{\Vert }_{\infty }\Vert k_2{\Vert }_1\Vert f{\Vert }_{\infty }$. The dominated convergence theorem proves joint continuity of the velocity integral. Combined with damped transport, the factorial bound for a Volterra iterate proves existence on arbitrary finite time slabs.

In Feynman gauge, use the auxiliary photon propagator $-i{g}_{\mu \nu }/(k^2-{\lambda }^2+i0)$ and $r=\lambda /m>0$. The projected one-loop Feynman parameter integral isHere $x$ is the source's photon parameter $z$.^[ref]

To check the reduction, use $N_2=4z(1-z)m^2$ and ${\Delta }^{\prime }=(1-z{)}^2{m}^2+z{\lambda }^2$. The three-parameter expression has prefactor $e^2/(16{\pi }^2)$ multiplying $N_2/{\Delta }^{\prime }$. At fixed $z$, ${\int }_0^{1-z}du=1-z$, giving $e^2/(4{\pi }^2){\int }_0^{1}z(1-z{)}^2/[(1-z{)}^2+z{r}^2]dz$. Substituting $e^2=4\pi \alpha$ gives exactly the displayed integral. This starts from the projected QED numerator; it is not a derivation of the whole QED calculation from an arbitrarily chosen integral.

The regulator can then be removed by a short independent argument. Write its integrand as $f_r(x)$. For $0\le x<1$, $f_r(x)\to 2x$, andThe dominated convergence theorem therefore gives $I(r)\to 1$ and $F_2^{(1)}(0)=\alpha /(2\pi )$. At $x=1$ the regulated integrand is zero, whereas the limiting function can be assigned value $2$; one point has zero measure and changes no integral. For every $r>0$, $0<I(r)<1$, and decreasing $r$ increases the integral. The visible suppression near $x=1$ narrows as the regulator shrinks; it is not a uniform pointwise approximation at that endpoint.

This auxiliary mass regulates an infrared divergence, not an ultraviolet divergence; the finite-mass prescription alone is not fully gauge-invariant.^[ref]

The ultraviolet problem belongs to the full QED amplitude. Electron self-energy, wave-function renormalization and the charge vertex require consistent counterterms and renormalization conditions. The Ward identity for QED renormalization constants ties $Z_1$ to $Z_2$, rather than allowing the vertex and external electron normalization to be fitted independently. Minimal renormalizable QED has no independent Pauli term to adjust this finite one-loop $F_2(0)$: its dimension-five coefficient is predicted from the mass and electric charge inputs. Adding a free Pauli term would define a different theory with an extra parameter. Finiteness of the reduced integral alone proves neither all-order renormalizability nor finiteness of every QED observable.

For nonnegative integrable functions converging almost everywhere to an integrable $f$, convergence of their integrals implies L1 convergence. Indeed $\min (f_n,f)\to f$ almost everywhere and is dominated by $f$, so the dominated convergence theorem gives convergence of its integral. Use $|f_n-f|=f_n+f-2\min (f_n,f)$. This applies to probability densities, giving convergence in total variation distance.

If a negative differentiable concave utility function satisfies $\mathbb{E}U(\theta X)>-\infty$ on an open interval, outer secants at four points bound the sample derivative on a smaller compact interval. Negativity makes the endpoint values absolutely integrable. The dominated convergence theorem then gives $d\mathbb{E}U(\theta X)/d\theta =\mathbb{E}[X{U}^{\prime }(\theta X)]$ and continuous derivative when $U^{\prime }$ is continuous. This is a justification of differentiation, rather than an assumption of integrability of marginal utility.