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Suppose the continuous functions converged pointwise to the rationality indicator. Each $F_N$ is closed, sincePointwise convergence implies that every point belongs to some $F_N$: its eventual values lie near either zero or one. Thus $\mathbb{R}={\bigcup }_N{F}_N$. By the Baire category theorem, some $F_N$ contains a nonempty open interval $I$. Choose a rational $r\in I$ and an irrational $s\in I$. For sufficiently large $n\ge N$, convergence gives $f_n(r)>2/3$ and $f_n(s)<1/3$. The intermediate value theorem then forces $f_n$ to take the value $1/2$ between them, at a point in $I\subseteq F_N$. This contradicts the defining condition of $F_N$. Therefore the rationality indicator is not a pointwise limit of continuous functions, the rationality indicator is not Baire class one result.

The closed-set form of the Baire category theorem says: if a nonempty complete metric space is a countable union of closed sets, at least one of those sets has nonempty interior. Equivalently, a complete metric space cannot be a countable union of closed sets with empty interiors. In the application the space is $\mathbb{R}$ with its usual complete metric.

Writing $S=\sum X_i$, the likelihood function is ${\theta }^S(1-\theta {)}^{n-S}$ and its supremum occurs at $\hat{\theta }=S/n$, including the limiting boundary cases zero and one. The likelihood ratio compares $2^{-n}$ under the null with that supremum. ThusUse $0\log 0=0$ at boundary outcomes. Wilks theorem gives asymptotic ${\chi }_1^2$ under the regular interior null, because one parameter is constrained. Directly, setting $\hat{\theta }=1/2+Z$ and expanding the bracket gives $2Z^2+O(Z^4)$, henceUnder the null, $2\sqrt{n}{Z}_n\stackrel{d}{\to }N(0,1)$ and $Z_n=O_P(n^{-1/2})$. The remainder vanishes in probability, proving the same chi-squared distribution limit. If an observed $Z_n$ is not small, the quadratic numerical approximation is unreliable and the exact statistic should be used; exact binomial calibration is available. Persistent non-small deviations under a fixed alternative instead make the statistic grow linearly with $n$, rather than having the null chi-squared distribution law.

Assume an identifiable smooth parametric model, an interior true parameter, a consistent interior maximum-likelihood estimator, interchange of differentiation and integration, a uniform local law of large numbers for the Hessian, and nonsingular finite Fisher information. For one observation define the score function $s_{\theta }(X)={\nabla }_{\theta }\log p_{\theta }(X)$ andThe score function has expectation zero by differentiating the normalization of the density. Its sample sum satisfies $n^{-1/2}{S}_n(\theta )\stackrel{d}{\to }N(0,J)$ by the vector central limit theorem. The score function equation and Taylor's formula giveThe bracket divided by $n$ converges in probability to $-J$ by consistency and the assumed uniform law. Solving the equation and applying Slutsky theorem yieldsThe integral Hessian avoids incorrectly assuming a single intermediate point for a vector-valued Taylor formula.

The strong law of large numbers states that for independent identically distributed real variables with $\mathbb{E}|X_1|<\infty$, the sample average converges almost surely to $\mathbb{E}{X}_1$. The central limit theorem states that if additionally $0<\mathrm{Var}{X}_1={\sigma }^2<\infty$, thenThe vector form uses finite second moments and gives a centred multivariate Gaussian with the covariance matrix of one observation.

By Poisson thinning, the first arrival times of the coupon types are independent exponential distribution $T_j$ of rates $\lambda p_j$. Completion requires all those arrivals, so $T={\max }_j{T}_j$ andTo relate time to count rigorously, let $E_k$ be the successive independent exponential waiting times of mean $1/\lambda$. The completion count $L$ depends only on the coupon labels, which are independent of all $E_k$. ThereforeNonnegative integration justifies the sum, avoiding any unproved optional-stopping assertion. Thus the Poissonized coupon completion time and count identity is $\lambda \mathbb{E}T=\mathbb{E}L$. Changing variables $s=\lambda t$ in the tail integral giveswhich depends only on the type probabilities, not $\lambda$. These expectations are finite when all $p_j>0$; a missing type makes both infinite and completion impossible.

For an interval of length $t$, the joint probability generating function of the marked counts isThis factors into two Poisson generating functions. Independent parent increments on disjoint intervals and independent marks give independent increment vectors, so the factorization holds jointly over all intervals. Thus the two processes are independent Poisson processes of rates $\lambda p$ and $\lambda (1-p)$. For $m$ types the same calculation gives ${\prod }_j\exp [\lambda p_{j}t(s_j-1)]$, proving independent Poisson processes of rates $\lambda p_j$. A zero probability gives an identically zero process.

Fatou's lemma states that nonnegative measurable functions satisfyTo prove it, put $h_n={\inf }_{k\ge n}{f}_k$. These measurable nonnegative functions increase to $\mathrm{liminf}{f}_n$. By the monotone convergence theorem, $\int \mathrm{liminf}{f}_n={\lim }_n\int h_n$. Since $h_n\le f_k$ for every $k\ge n$, one has $\int h_n\le {\inf }_{k\ge n}\int f_k$. Taking the limit proves the inequality, including extended infinite integrals.

For a fixed $c>0$, the functions $|f_n|1_{\{|f_n|\le c\}}$ tend pointwise to zero and are dominated by the Lebesgue integrable constant $c$, since $\mu (\Omega )<\infty$. The dominated convergence theorem therefore gives convergence of their integrals to zero. SplitTake the upper limit in $n$ and then send $c\to \infty$. The specified uniform integrability condition makes the right side zero. It also ensures integrability of each $f_n$, by the same split for a sufficiently large $c$. Hence $\int |f_n|d\mu \to 0$, and in particular $\int f_{n}d\mu \to 0$.

For a counterexample without the tail condition, use Lebesgue measure on $(0,1)$ and $f_n(x)=n{1}_{(0,1/n)}(x)$. Every fixed positive $x$ eventually lies outside the shrinking interval, so $f_n(x)\to 0$, but ${\int }_0^1{f}_{n}dx=1$ for every $n$.

Take the truncated dyadic approximationIt has finitely many values and measurable level sets, so is simple. Downward rounding and truncation ensure $0\le g_n\le f$. For a fixed finite $f(x)$, eventually $n>f(x)$, and then $0\le f(x)-g_n(x)<2^{-n}$. Thus $g_n(x)\to f(x)$ everywhere. If extended value infinity is permitted, define $g_n=n$ there, also giving the required limit.

An $\mathcal{A}$-simple function is a measurable function taking finitely many values, written $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ with disjoint measurable $A_j$. For nonnegative simple functions define $\int sd\mu ={\sum }_j{a}_j\mu (A_j)$, with $0\cdot \infty =0$. Refining two disjoint descriptions shows this value is independent of the description. For nonnegative measurable $h$, defineFor a real measurable $f$, put $f^{+}=\max (f,0)$ and $f^{-}=\max (-f,0)$. It is Lebesgue integrable when $\int |f|d\mu <\infty$, equivalently both positive and negative parts have finite integrals. ThenThis includes Lebesgue integrable signed simple functions, whose integral is the same finite weighted sum.

If two simple closed geodesics were disjoint on a sphere-like surface, they would bound an annulus. The Gauss-Bonnet theorem applied to it gives ${\int }_{A}KdA+{\int }_{\partial A}{k}_{g}ds=2\pi \chi (A)=0$. Both boundary geodesic curvatures vanish, but positive Gaussian curvature makes the area integral strictly positive, a contradiction. Hence the two geodesics must intersect.

The assertion fails for nonzero constant signed geodesic curvature. On the unit round sphere a latitude of colatitude $\theta$, oriented in increasing longitude, has $k_g=\cot \theta$. The latitude of colatitude $\pi -\theta$ has the opposite value; reversing its orientation gives the same value as the first. Choosing $0<\theta <\pi /2$ produces two disjoint smooth simple closed curves with the same positive nonzero curvature. Reversing both handles negative values, and $\theta$ can be chosen for any specified nonzero $\lambda$. Thus constant nonzero geodesic curvature does not force intersection.

Let $J$ be rotation by ninety degrees in the oriented tangent plane, $Jv=N\times v$. In local coordinates seek a curve withThis smooth second-order ordinary differential equation, written as a first-order system for $(x,\dot{x})$, has a unique local solution with the specified initial position and velocity. Its speed is constant, sinceThe initial speed is one, so the parameter is arc-length and the signed geodesic curvature is $⟨{\nabla }_s\dot{\gamma },J\dot{\gamma }⟩=\lambda$. Conversely every requested curve satisfies this system, which proves the claimed uniqueness as well as existence on some $(-ϵ,ϵ)$.

For a flat matter-dominated universe, the Friedmann solution has $a\propto t^{2/3}$, $H=2/(3t)$ and $\bar{\rho }=1/(6\pi G{t}^2)$. Thus the density equation isTrying $t^s$ gives $s(s-1)+4s/3-2/3=0$, with roots $2/3$ and $-1$. ThereforeThe growing mode is proportional to the scale factor. When smooth dark energy or a cosmological constant dominates, the matter source density falls while expansion provides continued damping, suppressing growth. In the late constant-$\Lambda$ limit, $H\to H_{\Lambda }$ and $\bar{\rho }\to 0$, so $\ddot{\delta }+2H_{\Lambda }\dot{\delta }\simeq 0$ and $\delta$ tends to a constant plus a decaying $e^{-2H_{\Lambda }t}$ term. Matter perturbations therefore cease growing appreciably rather than continuing the matter-era law indefinitely.

A comoving material cell contains the same particles in its perturbed and background descriptions, so its mass is $\bar{\rho }{a}^3{d}^{3}q=\rho d^{3}r$. The Jacobian determinant is $a^3\det (I+{\nabla }_q\psi )$. Expanding the determinant givesSplit the potential into homogeneous and perturbed parts. There is a sign error in the printed hint: ${\nabla }^2{\Phi }_0=4\pi G\bar{\rho }$ requires $\nabla {\Phi }_0=+4\pi G\bar{\rho }r/3$, not the printed negative sign. Its acceleration $-\nabla {\Phi }_0$ agrees with $\ddot{a}/a=-4\pi G\bar{\rho }/3$.

For the perturbation, ${\nabla }_q^2\varphi =4\pi G\bar{\rho }{a}^2\delta =-4\pi G\bar{\rho }{a}^2{\nabla }_q\cdot \psi$. Under the usual homogeneous boundary convention the curl-free force therefore satisfieswhere ${\psi }_L$ is the longitudinal part of the displacement. Expanding the particle acceleration gives $\ddot{r}=\ddot{a}(q+\psi )+2\dot{a}\dot{\psi }+a\ddot{\psi }$. The background acceleration cancels the first term, leavingFor a longitudinal displacement this proves the displayed longitudinal displacement in linear dust perturbations equation with ${\psi }_L=\psi$. For unrestricted displacement, a transverse component has no density contrast and instead obeys ${\ddot{\psi }}_T+2H{\dot{\psi }}_T=0$; the full-vector equation printed in the question thus needs the longitudinal assumption. Taking the divergence removes this qualification and gives