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Put $h(t,x)=e^{-x^2-t^2/x^2}$. For every $t\ge 0$, $0<h(t,x)\le e^{-x^2}$, and the upper bound is a Lebesgue integrable function by part (b)(i). Since $h(t,x)\to e^{-x^2}$ for each $x>0$ as $t\downarrow 0$, the dominated convergence theorem gives right continuity at zero andTo apply dominated differentiation under the integral sign, fix $t_0>0$ and restrict $t$ to $[t_0/2,3t_0/2]$. ThenFor $x\ge 1$, $D(x)\le 3t_0{e}^{-x^2}$. For $0<x<1$, writing $a=t_0^2/4>0$, the substitution rule $u=1/x$ givesThus $D$ is integrable. The mean value theorem bounds nearby difference quotients by $D$, so the dominated convergence theorem justifies the derivative at $t_0$. As $t_0$ was arbitrary,Now make the change of variables $u=t/x$. The reversal of limits cancels the negative differential and givesConsequently $F^{\prime }(t)=-2F(t)$. This also follows in sign from the fact that the positive integrand decreases strictly as $t$ increases: the printed equation with a positive sign cannot hold. Solving the linear differential equation, or differentiating $e^{2t}F(t)$, and using right continuity at zero givesThis is a half-line instance of the Gaussian integral with an inverse-square term.

True. Replace the dominating representative by $|g|$ if necessary, and take the countable union of the exceptional Lebesgue null sets so that every bound holds off a single null set. Since the interval has Lebesgue measure one, Cauchy-Schwarz inequality givesThus $g^2$ is a Lebesgue integrable function. The squared functions satisfy $|f_n{|}^2\to 0$ pointwise and $|f_n{|}^2\le |g{|}^2$ almost everywhere. The dominated convergence theorem therefore yieldsThis uses the Lp inclusion on a finite measure space to turn the assumed fourth-power bound into the second-power domination actually needed.

The dominated convergence theorem states that if measurable functions converge almost everywhere and their absolute values are bounded by a single Lebesgue integrable function, their limit is integrable and their integrals converge to its integral.

Put $b_j=a_j/j$. The hypothesis makes $b_j$ summable. On the positive integers with counting measure define $f_n(j)=(j/n)b_j{1}_{j\le n}$. For each fixed $j$ it tends to zero and $0\le f_n(j)\le b_j$. Dominated convergence givesthe stated Kronecker lemma.

For a bounded interval union $B$, $S_j\sim N(0,j)$ has density bounded by $1/\sqrt{2\pi j}$. Hence $\mathbb{P}(S_j\in B)\le |B|/\sqrt{2\pi j}$, andBy Tonelli's theorem, the nonnegative series is finite almost surely. Kronecker lemma then makes the empirical occupation fraction of $B$ tend to zero. The family of finite rational-endpoint interval unions is countable, so intersecting their probability-one events proves simultaneous convergence for all of them. For the whole line, every indicator is one. ThusThere is no countably additive Borel measure extending these values: $\mathbb{R}={\bigcup }_{n\ge 1}(-n,n)$ would have measure zero by continuity from below, contrary to its assigned measure one. The escaped empirical mass cannot be captured by a probability measure on the real line.

This is the Gaussian random walk spends zero density of time in bounded sets phenomenon.

For the fixed observation interval, $N_3$ has Poisson distribution with mean and variance $3\lambda$. Hence $N_3/3$ is unbiased and has variance $\lambda /3$. Its score function is $N_3/\lambda -3$, and its Fisher information is $3/\lambda$. The Cramér-Rao bound for unbiased estimation of $\lambda$ therefore gives variance at least $\lambda /3$, attained by $N_3/3$.

For the stopping-time observation, $T_{10}$ has Gamma distribution with shape $10$ and rate $\lambda$. Direct integration of its density givesThus $k=9$, andThe score function is $10/\lambda -T_{10}$, so the Fisher information is $10/{\lambda }^2$. Its Cramér-Rao lower bound is ${\lambda }^2/10$, strictly below this estimator's variance.

Nevertheless, $9/T_{10}$ is uniformly minimum-variance unbiased. The observed $T_{10}$ is a sufficient statistic, and its family is complete: if ${\mathbb{E}}_{\lambda }h(T_{10})=0$ for all $\lambda >0$, then ${\int }_0^{\infty }h(t)t^9{e}^{-\lambda t}dt=0$ for all $\lambda >0$. Uniqueness of the Laplace transform, applied after an exponential tilt to make an Lebesgue integrable function, implies $h=0$ almost everywhere. The Lehmann–Scheffé theorem now proves the minimum-variance assertion. Failure to attain the Cramér–Rao bound does not disprove minimum variance within the class of unbiased estimators.

For the actual observations, the two likelihood functions areThey differ by a factor independent of $\lambda$. The likelihood principle consequently requires the same inferential conclusions, since the observed data carry proportional likelihoods. But the two minimum-variance unbiased estimates are $\boxed{10/3\text{ and }3}$ respectively. Minimum-variance unbiased estimation therefore violates the likelihood principle: its unbiasedness and variance refer to the sampling plan, including the stopping rule, rather than solely to the observed Likelihood function.

The precise conclusion is that finite one-sided limits must agree, so a jump discontinuity is impossible. We prove this using explicit kernels and derive every conjugate-series identity used. The distinction from a removable discontinuity is addressed at the end.

Write the Fourier coefficients asTranslation multiplies each coefficient by a unimodular factor and preserves the hypothesis, so place the proposed discontinuity at zero. Suppose the finite limits are $A=f(0^{-})$ and $B=f(0^{+})$. For $0<r<1$, geometric summation gives the Poisson kernel on the circle and the conjugate Poisson kernel on the circle:These series converge uniformly and absolutely for each $r<1$. Therefore they may be integrated term by term against the Lebesgue integrable function $f$. With normalized convolution, $P_r$ has multiplier $r^{|n|}$, while $Q_r$ has multiplier $-i\mathrm{sgn}(n)r^{|n|}$ for $n\ne 0$ and zero at $n=0$. More explicitly, at zero,The assumed vanishing of negative Fourier coefficients thus proves, rather than presupposes, the conjugate relation

We next show that the right-hand side is bounded as $r\uparrow 1$. The kernel $P_r$ is positive and even, and its normalized integral is one by its series. Its mass outside any fixed $(-\delta ,\delta )$ tends to zero; in fact its supremum there is $O_{\delta }(1-r)$. On $(-\delta ,0)$, $f$ is uniformly close to $A$ for sufficiently small $\delta$, and on $(0,\delta )$ it is uniformly close to $B$. Evenness places asymptotically half the mass on each side. The outside contribution tends to zero because $f$ is integrable and the outside kernel tends uniformly to zero. Bounding the inside errors by the one-sided approximation errors givesHence $(Q_r\ast f)(0)$ is bounded by the proved conjugate relation.

On the other hand, oddness of $Q_r$ givesFor $0<\delta <\pi$, differentiating $1-2r\cos t+r^2$ proves the exact identityAlso $Q_r\ge 0$ on $(0,\pi )$ and is uniformly bounded on $[\delta ,\pi ]$ as $r\uparrow 1$. Write $f(-t)-f(t)=A-B+e(t)$ with $e(t)\to 0$ as $t\downarrow 0$. The integral over $[\delta ,\pi ]$ is $O_{\delta }(1)$ by integrability. Given any $\eta >0$, choose $\delta$ so that $|e(t)|\le \eta$ on $(0,\delta )$. Positivity and the preceding identity show that the near-zero error, divided by $\log (1/(1-r))$, has limiting upper bound at most $\eta /\pi$. Letting $\eta \downarrow 0$ provesBoundedness of the numerator makes the left side tend to zero. Therefore $A=B$. This proves one-sided Fourier spectrum excludes jumps, and supplies the required conjugate-sum argument in full.

There is a point-value qualification to the phrase “first kind.” If it means finite unequal one-sided limits, the proof establishes exactly the stated result. If the definition also includes removable discontinuities, the literal statement is false for an arbitrary integrable representative: the function that equals zero everywhere except $f(0)=1$ has every Fourier coefficient zero but a removable discontinuity at zero. Integral hypotheses cannot detect a change at one point. Thus no jump is possible; equal one-sided limits need not equal an arbitrarily assigned point value.

A phase function is a real smooth functionthat is positively homogeneous of degree one in $\theta$ and has no critical point in all variables:The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of all $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and multi-indices $\alpha ,\beta$,

For a cutoff $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero, define the oscillatory integral byRepeated integration by parts makes the limit meaningful and independent of the cutoff.

The singular support is the complement of the largest open set on which the distribution is represented by a smooth function. Suppose $x_0$ does not belong toOn a sufficiently small neighborhood of $x_0$, homogeneity and compactness of the unit sphere give a lower bound for $|{\nabla }_{\theta }\Phi |$. The differential operatorsatisfies $L{e}^{i\Phi }=e^{i\Phi }$. Repeatedly transferring $L$ to the amplitude lowers its symbol order until the integral and all its $x$-derivatives converge absolutely. Hence $I_{\Phi }(a)$ is smooth near $x_0$, proving

For the stated distribution on ${\mathbb{R}}^3$, rotational symmetry lets us align the polar axis with $x$ and write $\rho =|x|$. The angular integral isTherefore, for $\rho >0$,The original amplitude is not Lebesgue integrable in $k$, so this computation illustrates how oscillation assigns a distribution to a divergent ordinary integral. The result is smooth away from the origin and has singular support $\{0\}$. It is a constant multiple of the three-dimensional Yukawa potential, satisfying $(-\Delta +1)u=4{\pi }^2{\delta }_0$ with the normalization used in the question.

For an integrable function $f:{\mathbb{R}}^n\to \mathbb{C}$, the Lebesgue differentiation theorem states thatfor almost everywhere $x\in {\mathbb{R}}^n$. In particular,at every Lebesgue point of $f$.

Almost every $x\in \mathbb{R}$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,For $t<0$, the same estimate over $[x+t,x]$ gives the identical limit. Thereforeat every Lebesgue point of $g$, so $G$ is differentiable $\lambda$-almost everywhere. This is the differentiation of an indefinite Lebesgue integral.

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

For $f\in L^1(\mathbb{R})$, its regular tempered distribution acts by integration, and the distributional transform agrees with the functionThe triangle inequality gives the uniform boundIf ${\lambda }_j\to \lambda$, then $f(x)e^{-i{\lambda }_{j}x}\to f(x)e^{-i\lambda x}$ pointwise and every integrand is bounded in absolute value by the Lebesgue integrable function $|f|$. The Dominated convergence theorem therefore proves that $\hat{f}$ is continuous. In fact the Riemann-Lebesgue lemma also gives $\hat{f}(\lambda )\to 0$ as $|\lambda |\to \infty$.

Because $x^{-2}$ is integrable on $[1,\infty )$, the Dominated convergence theorem applied to the defining integral proves that $v$ is continuous on $\mathbb{R}$.

For $\lambda >0$, the change of variables formula $t=\lambda x$ givesand for $\lambda <0$ the analogous formula is obtained with $e^{it}$. On either open half-line the lower endpoint stays away from zero locally, so repeated differentiation under the integral sign proves smoothness. Thus

The finite sigma-algebra ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$ is partitioned by the nonempty eventsThese are its atoms. Define the random variableand give it any finite value on the union of the null atoms. It is ${\mathcal{F}}_n$-measurable and integrable. Every $A\in {\mathcal{F}}_n$ is a union of atoms, soThis constructs the requested variable directly, without invoking the general existence theorem for conditional expectation.

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

Choose measurable versions of the two Lebesgue integrable functions and consider $F(x,y)=g(x-y)h(y)$. Translation and the corresponding linear change of variables preserve Lebesgue measurability; one can first use Borel versions and then disregard the resulting null-set changes. Tonelli's theorem and translation invariance giveFubini theorem therefore shows that the section in $y$ is integrable for almost every $x$. Define $(g\ast h)(x)$ by that section integral where it exists and by zero on the exceptional null set. Fubini theorem also makes this a measurable Lebesgue integrable function, withIts almost-everywhere defining formula is precisely the desired convolution. This construction explains why an everywhere-existing pointwise integral is not required.

Fubini's theorem for the product of two sigma-finite measure spaces states that an integrable measurable function on the product has integrable sections almost everywhere, the section integrals define Lebesgue integrable functions, and both iterated integrals equal the product integral. The integrability hypothesis is absolute integrability. For nonnegative measurable functions, Tonelli's theorem gives equality of the possibly infinite product and iterated integrals without first assuming finiteness.

The displayed function is not Lebesgue integrable. Its value at the origin can be assigned arbitrarily, since a point is a null set. On $0<x<1$ and $0<y<x/2$,Consequently Tonelli's theorem gives the divergent lower boundThe signed iterated integrals illustrate why the hypothesis matters. For $x>0$, $f(x,y)={\partial }_y[y/(x+y{)}^2]$, so integration first in $y$ and then in $x$ gives $1/2$. Integrating first in $x$ gives $-1/2$, using $f(x,y)=-{\partial }_x[x/(x+y{)}^2]$. Exceptional boundary sections have measure zero. Thus finite iterated values do not establish product integrability.

Choose $Z_{\infty }$ as the pointwise limit of a sequence of $(Z_n)$ where it exists, and as zero otherwise. The convergence event and every $Z_n$ are $\sigma (X)$-measurable, so this is an exactly $\sigma (X)$-measurable version, bounded by $L$. Part (a)(ii), the Doob-Dynkin lemma, supplies a Borel measurable function $g$ on $[0,1]$ such that $Z_{\infty }=g(X)$. Applying scalar clipping to $g$ with bounds $[-L,L]$ preserves that equality and ensures $|g(u)|\le L$ everywhere.

Part (c)(ii) and the conditional expectation formula from part (b)(iv) now giveSince the indices $n$ form a countable family, these identities may be taken on one common full-probability event.

For fixed $n,k$, both the slope and the displayed average are constant on the atom $A_{n,k}$, whose probability is $2^{-n}>0$. An equality holding almost surely of these constants forces the deterministic equalitySum over $0\le k<j$. The telescoping sum givesThus the desired identity holds at every dyadic endpoint, including $1$, without evaluating the undefined slope at $X=1$.

Finally, given $x\in [0,1]$, take dyadic endpoints $x_n\to x$. The Lipschitz condition implies $|f(x_n)-f(x)|\le L|x_n-x|$, and the bound on the Borel measurable function $g$ givesPassing to the limit of a sequence provesThis is a martingale proof of the integral representation of a Lipschitz function. It constructs a bounded Lebesgue integrable function representing the increments of $f$, and in particular proves absolute continuity of $f$. No prior assumption that $f$ has a derivative is needed.

The change of variables formula for a Lebesgue integrable function isThe absolute Jacobian determinant accounts for area regardless of orientation. The formula also holds for nonnegative measurable functions with possibly infinite integral.

Translating a Lebesgue measurable set by a real number preserves measurability and its Lebesgue measure. Translation invariance of Lebesgue outer measure and Carathéodory's criterion give the measurability assertion and equality. Applying the result to indicator functions, then simple functions, and finally increasing simple approximations gives ${\int }_{\mathbb{R}}u(x-t)dx={\int }_{\mathbb{R}}u(x)dx$ for every nonnegative Lebesgue measurable function. For a Lebesgue integrable function, apply the identity to its positive part and negative part.