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For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.

Let $T(x)=x+\alpha (\mathrm{mod}1)$ with irrational $\alpha$. This irrational rotation of the circle preserves Lebesgue measure and is ergodic. Apply the Birkhoff ergodic theorem to the indicator function $1_A$. Its limitis invariant, hence almost everywhere constant by ergodicity. Its integral must equal that of $1_A$, so the constant is $\mu (A)$. Thereforefor Lebesgue measure-almost every $x\in (0,1]$.

The function $\varphi (x)={\max }_{v\in C}⟨x,v⟩$ is the support function ${\sigma }_C$. For a nonempty compact convex set,so its convex conjugate is the indicator function ${\iota }_C$. Applying the Moreau decomposition,Multiplication of an indicator function by a positive scalar does not change it, and its proximal operator is the Euclidean projection onto a convex set. Therefore

Suppose first that the system is ergodic and $K_{F}g=g$. For each real $t$, the level setis invariant modulo a null set, so $\omega (E_t)\in \{0,1\}$. The distribution function of $\mathrm{Re}g$ can therefore jump only once, which makes $\mathrm{Re}g$ constant almost everywhere. The same argument applies to $\mathrm{Im}g$.

Conversely, if $E$ is invariant, then $K_F{1}_E=1_E$. If every invariant $L^2$ function is constant, the indicator function $1_E$ is almost everywhere zero or one, and hence $\omega (E)=0$ or $1$. This proves the invariant-function characterization of ergodicity.

Apply the preceding Fourier argument to $f=1_A$, retaining the frequencies needed to make the oscillation of $1_A\ast 1_A\ast 1_A$ strictly smaller than its mean ${\alpha }^3$. The standard optimized cutoff gives a set $\Gamma \subseteq \hat{G}$ withsuch that the nonnegative function $F=1_A\ast 1_A\ast 1_A$ cannot fall from a maximal value to zero under any shift in $B(\Gamma ,\alpha )$. If $x$ maximizes $F$, thenThe support of a convolution of indicator functions is the corresponding sumset, so

The pushforward measure of $\mu$ under $g$ is $g_{\ast }\mu$, defined for $B\in \mathcal{B}$ byThe Lebesgue decomposition theorem says that if $\nu$ and $\mu$ are sigma-finite measures on the same measurable space, then uniquely

Let $f:[0,\infty )\to \mathbb{R}\cup +\infty$ be convex with $f(1)=0$. If $\lambda$ dominates the probability distributions $P,Q$, with densities $p,q$, their f-divergence isusing the lower-semicontinuous perspective convention where $q=0$. This definition is independent of the dominating measure.

The data processing inequality for f-divergences statesTo prove it, take $\lambda =P+Q$ and let $\mathcal{G}=g^{-1}(\mathcal{B})$. If $p=dP/d\lambda$ and $q=dQ/d\lambda$, then the pullbacks of the densities of $g_{\ast }P$ and $g_{\ast }Q$ with respect to $g_{\ast }\lambda$ are respectively ${\mathbb{E}}_{\lambda }[p\mid \mathcal{G}]$ and ${\mathbb{E}}_{\lambda }[q\mid \mathcal{G}]$. The perspective $F(a,b)=bf(a/b)$ of a convex function is jointly convex. Conditional Jensen inequality therefore givesIntegration proves the claim.

The Squared Hellinger distance isIf $P,Q$ have densities $p,q$ with respect to a sigma-finite measure $\mu$, this becomes

Fix any probability distribution $Q$ and set $p_j=P_j(A_j)$, $q_j=Q(A_j)$, and $t_j=H^2(P_j,Q)$. Applying the data processing inequality to the indicator function of $A_j$ gives the Bernoulli Hellinger boundThe hinted inequality impliesThe function $t\mapsto \sqrt{t(1-t/4)}$ is concave on $[0,2]$. Since the $A_j$ form a set partition, ${\sum }_j{q}_j=1$, and Jensen inequality givesTaking the infimum over $Q$ proves the stated inequality.

The printed assertion needs an endpoint qualification: the enumeration includes zero, but $1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed discontinuities occur at every $r_j>0$.

Define the finite step function sumsThe assumed convergence of ${\sum }_{k}1/k^2$ and the bound on each indicator function giveThus the defining series converges everywhere and has uniform convergence. Part (b) proves that $f$ is Riemann integrable; it also gives, if desired,For $r_j>0$ and $0\le t<r_j$, every indicator function is nonincreasing, and the $j$th drops by one between $t$ and $r_j$. HenceTaking inputs from the left tending to $r_j$ proves a discontinuity there.

For completeness, continuity at zero follows directly from a finite-head and tail argument. Given $\epsilon >0$, choose $N$ with ${\sum }_{k>N}1/k^2<\epsilon$. Choose $\delta >0$ below every positive $r_k$ with $k\le N$. For $0\le t<\delta$, the first $N$ indicator functions have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\le {\sum }_{k>N}1/k^2<\epsilon$. This confirms that the endpoint exception is real. Replacing the printed enumeration of $[0,1)$ by one of $(0,1)$ would remove it.

Finally, the density of the rational numbers supplies a positive rational $r_j$ in every nonempty open subinterval of $[0,1]$. Thus a Riemann integrable function can have a discontinuity in every open subinterval: dense jumps do not prevent Riemann integrability.

For any chosen proper prior density $\rho (\mu ,\sigma )$ on $\mathbb{R}\times (0,\infty )$ with positive evidence, Bayes' theorem givesThe prior density comes from the first two random draws. The likelihood $L$ is the probability of satisfying all observed-label constraints after integrating the past duration noises. The factor $\Phi ((\mu -30)/\sigma {)}^5$ is the probability of the five-day outcome conditional on the shared parameters. The denominator is the probability of acceptance, which converts the numerator's joint probability into the required conditional probability.

For the independent uniform priors in part (a), the constant $\rho =1/(60\cdot 19)$ cancels and both integrals reduce to $0<\mu <60$, $1<\sigma <20$. A prior on the variance $v={\sigma }^2$ would require transforming the density: this chosen prior has density $1/(38\sqrt{v})$ for $1<v<400$ conditional on $\mu$. One must not silently replace it by a uniform prior on $v$.

There is no unique numerical probability until the prior distribution is chosen. As a check for the stated uniform priors, independent Gaussian quadrature calculations with 256 and 512 nodes per coordinate agree at the shown precision:The prior-predictive probability of all observed labels is approximately $0.00030422536$, explaining the rejection cost. These numbers are illustrations of the explicit model, not additional observed data.

A subtle point is that the five future days are independent only conditional on $(\mu ,\sigma )$. Integrating shared parameter uncertainty gives $\mathbb{E}[{\pi }^5\mid A]$, not $(\mathbb{E}[\pi \mid A]{)}^5$. The latter is about $0.08631110$ for these priors and understates the five-day probability. Jensen's inequality gives $(\mathbb{E}[\pi \mid A]{)}^5\le \mathbb{E}[{\pi }^5\mid A]$, while $0\le \pi \le 1$ gives $\mathbb{E}[{\pi }^5\mid A]\le \mathbb{E}[\pi \mid A]$. For two future late-arrival indicator functions, their posterior covariance is $\mathrm{Var}(\pi \mid A)\ge 0$. This is the posterior predictive repeated-success identity.

The precise conclusion isTo prove it, observe that the maximum rises exactly when the new height difference is positive:A block landing in a shorter column produces $D_m=0$ and no rise; every other case raises the maximum and leaves $D_m>0$. Starting from $H_0=0$, telescoping givesThe difference chain is irreducible and positive recurrent by part (i). Apply the countable-state Markov chain ergodic theorem to the bounded indicator function $g(k)=1_{\{k>0\}}$. Its stationary mean is ${\sum }_{k\ge 1}{\pi }_k=1-{\pi }_0=2/3$, giving the asserted almost sure convergence. No aperiodicity assumption is needed for this time average. This establishes the growth rate of two-column sticky deposition.

Put $A=\{|X|\ge Z\}$. Both $A$ and its indicator function are measurable, and $0\le Z{1}_A\le |X|$. Thus $Z{1}_A$ is an integrable random variable. Monotonicity of conditional expectation givesThe measurable-factor property of conditional expectation says that a bounded $\mathcal{G}$-measurable function $H$ satisfies $\mathbb{E}[HW\mid \mathcal{G}]=H\mathbb{E}[W\mid \mathcal{G}]$ for every integrable random variable $W$. Here the factor $Z$ need not be bounded. Apply that property first to $Z\wedge m$ and $W=1_A$, then use monotone convergence on the nonnegative products, including the conditional version, to obtainAll the left-hand conditional expectations are well defined; the final one is finite almost surely because $Z{1}_A$ is an integrable random variable. Since $Z>0$ almost surely, division now yieldsThis proves the conditional Markov inequality with a measurable threshold without assuming that $Z^{-1}|X|$ is an integrable random variable. The measurable-factor rule, with its product-integrability condition, is also stated in the standard conditional expectation theory.^[ref]

For every $n$, ${\mathcal{F}}_n$ is a sigma-algebra contained in $\mathcal{F}$, and including one more random variable among its generators can only enlarge it. Hence ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$; the trivial ${\mathcal{F}}_0$ is contained in every sigma-algebra. These are exactly the requirements for a filtration.

The constant $X_0=x$ is ${\mathcal{F}}_0$-measurable. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable. The pair $(X_n,U_{n+1})$ is then ${\mathcal{F}}_{n+1}$-measurable, and the event $\{U_{n+1}\le X_n\}$ is ${\mathcal{F}}_{n+1}$-measurable because comparison of two real coordinates defines a Borel set. Define its indicator function by $B_{n+1}$. The recursion becomesIt is a measurable function of that pair. Mathematical induction proves $X_n$ is ${\mathcal{F}}_n$-measurable for every $n$, so $(X_n)$ is an adapted process. The same recursion also proves $0<X_n<1$ at every finite time, a useful bound for the later integrability arguments.

Consider the Borel setIts indicator function is nonnegative and measurable, so the Tonelli theorem applies to the product measure, without needing any separate absolute-integrability argument. At a fixed $y\in (a,b]$, its section in the first coordinate is $(a,y]$; outside $(a,b]$ that section is empty. ThereforeThe last equality uses the definition of a cumulative distribution function on the half-closed interval. It retains all point atoms of a measure on the diagonal $x=y$, excludes a possible point atom of a measure at $x=a$, and includes $y=b$, exactly as required by the set $D$. If $a=b$, both sides vanish. This verifies the identity even when the probability measures are not non-atomic measures.^[ref]

Put $P_n=X_n{Y}_n{Z}_n$. Before $\tau$, all three players are active. Each unordered pair is selected with probability $1/3$, and each direction of transfer then has conditional probability $1/2$. Given the current holdings $(x,y,z)$, a transfer within the first pair has mean productFor transfers within the other two pairs the analogous values are $xyz-y$ and $xyz-x$. Averaging over pairs givesAt and after elimination the product is zero, since an eliminated player never re-enters. In particular the stopped process has incrementThe indicator function is ${\mathcal{F}}_n$-measurable, and the expression in parentheses has zero conditional expectation on its support. The stopped process is adapted and integrable: at fixed $n$, its absolute value is bounded by $N^3+nN/3$. HenceThis is the product martingale for three-player token exchange.

Let $\mathcal{G}=T^{-1}(\mathcal{F})=\{T^{-1}(A):A\in \mathcal{F}\}$. Inverse images commute with complements and countable unions, so this is a sub-sigma-algebra of $\mathcal{F}$. The random variable $X\circ T$ is $\mathcal{G}$-measurable.

The assumption $\mathbb{P}\circ T^{-1}=\mathbb{P}$ says that $T$ is a measure-preserving transformation. Consequently $X\circ T$ has the same probability distribution as $X$ and is integrable. It also implies $\mathbb{E}[W\circ T]=\mathbb{E}[W]$ for every integrable random variable $W$, first for indicator functions, then for simple functions, and then by positive and negative approximation.

To identify a conditional expectation, take any $A\in \mathcal{F}$. The assumed identity with the bounded random variable $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.

For each $n\ge 0$, choose an open arc $I_n$ about $e^{in}$ of arclength $\pi 2^{-n-2}$, and put $G^1={\bigcup }_{n\ge 0}{I}_n$. It contains every point used by every empirical measure, whereasThus ${\mu }_N^1(G^1)=1$ for all $N$. Let $K=\mathbb{T}\setminus G^1$. This is closed and ${\mu }_N^1(K)=0$ for all $N$. But $1/\pi$ is irrational, so the weak-star limit is normalized arclength andTherefore the proposed implication is false. Weak-star convergence tests continuous functions and does not assert convergence of the indicator function of an arbitrary closed set.

For the proposed joint probability density $\rho$, define the acceptance indicator functionOn the proposal's support the importance weight is $\pi (Y_i,V_i)/\rho (Y_i,V_i)=M{A}_i$. Therefore the requested augmented-space importance sampling estimator isIts expected value is $\theta$, provided ${\mathbb{E}}_p|\varphi (X)|<\infty$. This is not the ordinary estimator $n^{-1}{\sum }_{i}w(Y_i)\varphi (Y_i)$ from question 1. Taking the conditional expectation of this augmented estimator given all $Y_i$ produces that ordinary estimator, because $\mathbb{E}[A_i\mid Y_i]=w(Y_i)/M$. This is the Rao-Blackwell identity for a fixed-denominator rejection estimator.

Use the same acceptance indicator functions and set $K_n={\sum }_{i=1}^n{A}_i$, $S_n={\sum }_{i=1}^n{A}_i\varphi (Y_i)$. The rejection sampling estimator based on these $n$ proposals is the mean over the accepted observations:For $K_n=0$, any fixed convention, such as zero, is harmless asymptotically. The $A_i$ are independent and identically distributed random variables with Bernoulli distribution and success probability $1/M$, so the strong law of large numbers gives $K_n/n\to 1/M$ almost surely. In particular $K_n>0$ eventually almost surely. The exact identitythen holds whenever $K_n>0$. Wherever $S_n\ne 0$, it follows thatA qualification is missing from the printed claim: the quotient must be defined eventually. A convenient sufficient assumption is ${\mathbb{E}}_p|\varphi |<\infty$ and $\theta \ne 0$. Indeed, another application of the strong law of large numbers gives $S_n/n\to \theta /M$, ensuring that $S_n$ is eventually nonzero. This proves the stated quotient limit under that assumption.

As printed, the assertion is false as a universally defined quotient. For example, $\varphi \equiv 0$ makes both estimates zero for every $n$. Even excluding that example is insufficient: take $p=q$ to be the uniform distribution on $(0,1)$, $M=1$, and $\varphi (y)=1$ for $y<1/2$ and $-1$ otherwise. Then $S_n$ is a simple symmetric random walk, which returns to zero infinitely often almost surely, leaving the quotient undefined infinitely often.

With integrability but no assumption $\theta \ne 0$, the exact multiplicative identity remains valid, and both estimates converge to $\theta$ almost surely. In particular ${\hat{\theta }}_n^{\mathrm{IS}}-{\hat{\theta }}_n^{\mathrm{RS}}\to 0$ almost surely. One may alternatively define the quotient's removable $0/0$ cases as $M{K}_n/n$; that explicitly extended quantity always converges to one. These are valid repairs of the source's unrestricted statement.

Store one entry per occurrence in an immutable list. Use lower type bounds for arguments that might widen the element type, and remove only the first matching occurrence:A covariant type parameter cannot appear directly as an ordinary object method argument type. The fresh $U$ with $T<:U$ permits insertion, deletion queries and count queries without imposing such a negative occurrence of $T$. delete can retain result type CoMSet[T] because it introduces no new elements; widening its result to CoMSet[U] would also be valid. union must widen because the other multiset may contain the broader type. The class-private immutable elems is accessible on another instance of the same software class; it is not private[this].^[ref]

Let $c_M(y)$ be a multiset multiplicity. The implementations satisfyThe brackets denote an indicator function. Structural induction on the list proves removeOne removes precisely one equal entry if present, otherwise none. filterNot would incorrectly remove every occurrence. List concatenation adds multiplicities, implementing the paper's additive additive multiset union, rather than the maximum-multiplicity convention sometimes used for that name.

All methods preserve the original immutable object. insert takes constant time through structural sharing; count and delete take $O(n+1)$ time, and union takes $O(n+1)$ time to copy the receiver's list spine. Equality is ordinary stable value equality; finite representable counts are assumed.

The perimeter of a measurable set relative to an open domain is the total variation seminorm of its indicator function there. For a smooth region it equals the Hausdorff measure of the boundary lying inside the domain. Global perimeter uses the entire space as the domain; an internal interface shared by two regions appears in both relative perimeters and must be counted only once in a segmentation energy.

Define the indicator function of the rational numbers on the real line by $f(x)=1$ for rational $x$ and $f(x)=0$ otherwise. Every rational shift preserves rationality and irrationality, so every nonzero rational number is a period of a function. Conversely, an irrational shift sends $x=0$ from value one to value zero, so it is not a period. The period group of a function is exactly $\mathbb{Q}$, and the positive periods $1/n$ approach zero without a smallest one. The function is nonconstant and discontinuous everywhere; this is why continuity matters in continuous periodic functions have a least positive period.