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The new characteristics have invariant $x_1-x_2$. On the initial curve it is $D(s)=s$ for $s<0$ and $D(s)=s-s^2$ for $s\ge 0$. The latter branch folds at $s=1/2$, attaining its maximum $1/4$. Some characteristics meet the initial curve twice or more; generic values of $f$ at those intersections disagree, so no global solution constant along characteristics can exist.

At every initial point except $s=1/2$, the curve is non-characteristic, and the local existence and uniqueness theorem applies. At $s=1/2$, tangency requires $f^{\prime }(1/2)=0$, but this is not sufficient. The fold tangency obstruction for characteristic data requires locally $f(s)=F(D(s))$ with $F$ continuously differentiable: equal invariant values must have equal data, and $f^{\prime }(s)/(1-2s)$ must have a finite continuous limit. For example $f(s)=|s-1/2{|}^{3/2}$ satisfies $f^{\prime }(1/2)=0$ and agrees across the fold, but would force an infinite derivative of $F$ at $1/4$.

If the full compatibility condition holds, $F$ can be extended to invariant values above $1/4$ to give a local solution. Those values are not specified by the initial curve, so local uniqueness then fails.

The solution is constant on characteristics $x_1+x_2=\eta$. On the initial curve this invariant is $h(s)=s$ for $s<0$ and $h(s)=s+s^2$ for $s\ge 0$. This is a $C^1$ bijection of $\mathbb{R}$, with everywhere positive derivative. Its inverse isTherefore $\boxed{u(x_1,x_2)=f(h^{-1}(x_1+x_2)).}$ The inverse is $C^1$ even at zero, so for $C^1$ data this is a global $C^1$ solution. Every characteristic meets the initial curve exactly once.

Along the characteristics, $x_1+x_2$ is constant. Since $d{x}_2/dt=-1$, the amplitude equation gives $du/d{x}_2=-x_{2}u$. Matching the initial value at $x_2=0$ yieldsDifferentiation directly verifies both the partial differential equation and the initial data.

The characteristics solve ${\dot{x}}_j=a_j(x)$ and $\dot{u}=b(x,u)$, starting with $x$ on the initial hypersurface and $u=\phi (x)$. The non-characteristic hypersurface condition at a point is $a\cdot n\ne 0$, where $n$ is any nonzero normal there. It makes the map from initial-surface coordinates and characteristic time to $x$ locally invertible. Local uniqueness for the ordinary differential equations and the inverse function theorem then give a unique local $C^1$ solution near that point, for the stated $C^1$ coefficients and data.

Independence and the normal exponential moment give $\mathbb{E}[S_t\mid {\mathcal{F}}_s]=S_s\exp [(\mu +{\sigma }^2/2)(t-s)]$. ThusFor $\sigma \ne 0$, $B_t/t\to 0$ almost surely, so $\log S_t/t\to -{\sigma }^2/2$ and $S_t\to 0$ almost surely. Nevertheless $\mathbb{E}{S}_t=e^{x_0}$ for every finite $t$, whereas $\mathbb{E}{S}_{\infty }=0$. This Exponential martingale for Brownian motion is not uniformly integrable, so it does not converge in $L^1$. If $\sigma =0$, the process is instead constantly $e^{x_0}$.

A standard Brownian motion starts at zero, has continuous paths, and has independent increments $B_t-B_s\sim N(0,t-s)$. By the Brownian reflection principle, for $a>0$,Hence every positive integer level is reached with probability one. Intersecting these countably many probability-one events proves $\boxed{\underset{t\ge 0}{\sup }{B}_t=\infty \text{almost surely}.}$

Relative to a filtration $({\mathcal{F}}_t)$, a martingale is an adapted integrable process satisfying $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for $s\le t$. The martingale convergence theorem gives an almost-sure finite limit for a nonnegative martingale, with $\mathbb{E}[M_{\infty }]\le \mathbb{E}[M_0]$ by Fatou's lemma. In continuous time take the usual right-continuous version and filtration hypotheses. Convergence in $L^1$, and preservation of expectation at infinity, additionally require uniform integrability.

Assign each arriving car independently to a uniformly chosen lot. The Poisson thinning theorem gives three independent Poisson processes, each of rate $\lambda /3$. For an interval of length $t$, their joint probability generating function isThe factorization proves independence of the counts in that interval; independence over disjoint intervals proves independence of the processes.

For a renewal gap $T$ of mean $m$, the equilibrium first-delay density is $\mathbb{P}(T>t)/m$. Here $m=3/\lambda$ and the Erlang distribution survivor is $e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. The equilibrium delay of an Erlang renewal process is thereforeThis is an equal mixture of shapes one, two and three. For each equilibrium stream use this first delay, followed by independent shape-three gaps. A joint stationary version of the cyclic routing is obtained by choosing the initial position in the three-label cycle uniformly at random, independently of the Poisson arrivals. Its marginal streams have the stated equilibrium law, but remain dependent on one another.

Streams $a$ and $b$ are delayed renewal processes; $c$ is an ordinary renewal process. Their first delays have Erlang shapes one and two, while subsequent gaps have shape three. Stream $c$ has shape three for its first gap as well.

Within each stream every subsequent gap is the sum of three consecutive original exponential random variables, hence has an Erlang distribution of shape three and rate $\lambda$. Successive gaps within that stream use disjoint original gaps, so they are independent and identically distributed, and independent of its first delay. Gaps from different streams can overlap in their constituent original gaps, so the three streams are not independent.

Write the successive gaps of the rate-$\lambda$ Poisson process as independent exponential random variables $E_j$. The first arrivals in the three streams occur at $E_1$, $E_1+E_2$, and $E_1+E_2+E_3$. They have Erlang distributions of shapes one, two and three:

The uniqueness theorem for measures says that two measures agreeing on a generating pi-system agree on its generated sigma-algebra, provided they are finite with the same total mass, or there is a common increasing exhaustion by pi-system sets of finite measure.

For the finite case let $\mathcal{D}=\{A:\mu (A)=\nu (A)\}$. Equality of total masses gives closure under complements, and countable additivity gives closure under disjoint countable unions. Thus $\mathcal{D}$ is a Dynkin system containing the generating pi-system. The pi-lambda theorem puts the entire generated sigma-algebra inside $\mathcal{D}$.

For the exhaustion version, let $E_n\uparrow E$ with $E_n$ in the pi-system and $\mu (E_n)=\nu (E_n)<\infty$. The finite measures $A\mapsto \mu (A\cap E_n)$ and $A\mapsto \nu (A\cap E_n)$ agree on the generating pi-system, since it is intersection-closed. The finite proof applies, and continuity from below of a measure as $n\to \infty$ gives $\mu (A)=\nu (A)$ for every measurable $A$. A finiteness hypothesis is essential to this standard extension-uniqueness statement.

${\mathcal{E}}_1$ is a Dynkin system but not a pi-system. The whole ten-point set has even cardinality, complements preserve parity, and disjoint unions of even sets are even. Two even sets can intersect in a singleton, for example $\{1,2\}$ and $\{2,3\}$.

${\mathcal{E}}_2$ is neither. The infinite set $\mathbb{N}\setminus \{1\}$ belongs to it, but its singleton complement does not, so it is not a Dynkin system. The two infinite sets $\{1\}\cup \{2,4,6,\dots \}$ and $\{1,3,5,\dots \}$ intersect in the excluded singleton, so it is not a pi-system.

${\mathcal{E}}_3$ is a pi-system but not a Dynkin system. Intersections of bounded open intervals are bounded open intervals or empty. However, the family does not contain the whole underlying set $\mathbb{R}$. None of the three families is a sigma-algebra.

Every sigma-algebra is both a pi-system and a Dynkin system. Conversely, in a family having both properties, intersections and complements give finite unions and differences. For any sequence $(A_n)$ in it, replace $A_n$ by the disjoint sets $B_n=A_n\setminus {\bigcup }_{j<n}{A}_j$, which still belong to the family. Their disjoint union equals ${\bigcup }_n{A}_n$, proving closure under arbitrary countable unions. Thus the family is a sigma-algebra.

A pi-system is a family closed under finite intersections. A Dynkin system contains the whole underlying set, is closed under complements in it, and is closed under countable pairwise-disjoint unions.

A submersion from a nonempty compact manifold to ${\mathbb{R}}^{\ell }$, $\ell \ge 1$, would be surjective by the preceding result, since the target is connected. Its image would also be compact, contradicting noncompactness of ${\mathbb{R}}^{\ell }$. Thus no such submersion exists. The nonempty-domain qualification excludes the vacuous empty-manifold case.

Coordinate projections are open maps, so the local form of the submersion proves that $f$ is open. If $X$ is compact and nonempty, its continuous image is compact and hence closed in the Hausdorff manifold $Y$. Therefore $f(X)$ is a nonempty set both open and closed in connected $Y$, giving $f(X)=Y$.

At a point where $df$ is surjective, choose charts and reorder the domain coordinates so that an $\ell \times \ell$ minor of the derivative is invertible. The map $F=(f_1,\dots ,f_{\ell },x_{\ell +1},\dots ,x_k)$ has invertible derivative. By the inverse function theorem it gives local coordinates, in which $f$ is the projection $(u_1,\dots ,u_k)\mapsto (u_1,\dots ,u_{\ell })$. This is the submersion theorem.

The genus-two surface has fundamental group $G=⟨a_1,b_1,a_2,b_2\mid [a_1,b_1][a_2,b_2]=1⟩$. Connected double covering spaces correspond to index-two subgroups. Each is the kernel of a nonzero homomorphism $G\to C_2$; conjugacy creates no identifications because these kernels are normal. The commutator relation is automatic in $C_2$, so there are $2^4-1$ choices:To obtain a surjection onto the free group $F_2=⟨x,y⟩$, send $a_1\mapsto x$, $a_2\mapsto y$ and $b_1,b_2\mapsto 1$. The defining relation again vanishes, and the images generate $F_2$.