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For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

Take the indicator function $u(x_1,x_2)={\chi }_{\{x_1>1/2\}}$ inside $(0,1{)}^2$. For a compactly supported test vector field, integration in $x_1$ and then $x_2$ givesThe vertical derivative term integrates to zero. The absolute pairing is at most one, and test vector fields with ${\xi }_1=-1$ along all but arbitrarily short endpoint pieces approach one. Thus $|Du|(\Omega )=1$. More explicitly, the distributional derivative is $Du=e_1{\mathcal{H}}^1\upharpoonright \{(1/2,x_2):0<x_2<1\}$, where ${\mathcal{H}}^1$ is Hausdorff measure along the segment. Also $\Vert u{\Vert }_1=1/2$.

A Sobolev space function in $W^{1,1}(\Omega )$ has each weak derivative represented by an $L^1$ density with respect to planar Lebesgue measure. The nonzero segment measure here is singular with respect to that measure, so this requirement fails. Equivalently, the Sobolev fundamental theorem of calculus on lines would require an absolutely continuous representative on almost every horizontal slice, whereas each slice has a jump. ThereforeThis is the bounded-variation step outside W11 example.

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute integrability, bounded near zero by a constant times $t^{\mathrm{Re}s-2}$, permits termwise integration. The Gamma function integral then givesSpecify the Hankel contour to run from the negative real axis below the cut towards zero, circle zero counterclockwise with fixed radius $0<r<2\pi$, and return above the cut. Take $-\pi <\arg t<\pi$. The circle excludes every nonzero pole $2\pi i\mathbb{Z}$ of the denominator. For $\mathrm{Re}s>1$ its radius can shrink to zero, and the lower and upper rays give respectively $-e^{-i\pi s}\Gamma (s){\zeta }_R(s)$ and $e^{i\pi s}\Gamma (s){\zeta }_R(s)$. By the Gamma reflection formula,away from integer $s$ in that half-plane, with limiting equality at the removable points.

With the circle radius kept fixed, the contour integral is an entire function of $s$: the rays decay exponentially and converge locally uniformly, and the circular part stays away from zero. Multiplication by $\Gamma (1-s)$ is analytic for $\mathrm{Re}s<1$ and meromorphic globally. The apparent singularities at positive integers $s\ge 2$ are removable because the integral vanishes there and it already agrees with the original analytic zeta function nearby. Only $s=1$ remains a simple pole. Thus this constructs the meromorphic continuation to the whole plane and an analytic continuation on $\mathbb{C}\setminus \{1\}$. Keeping the circular segment is essential when the integral is not locally integrable at zero.

At $s=-1$ the two rays cancel, since $t^{-2}$ is single-valued. The Laurent expansionshows that the residue of $t^{-2}/(e^{-t}-1)$ is $-1/12$. Since $\Gamma (2)=1$, the residue theorem gives

For each $k$, choose $n_k$ with $\lambda (A\setminus E_{n_k}^{(k)})\le \epsilon 2^{-k-1}$ and set $B={\bigcap }_{k\ge 1}{E}_{n_k}^{(k)}$. Countable subadditivity gives $\lambda (A\setminus B)\le \epsilon /2$. For any tolerance $\delta >0$, choose $k$ with $1/k<\delta$. For all $m\ge n_k$ and every $x\in B$, $|f_m(x)-f(x)|\le 1/k$, proving uniform convergence on $B$.

If the functions are Borel measurable, $B$ itself is Borel and we can take $A_{\epsilon }=B$. If “measurable” is interpreted in the completed Lebesgue sense, use regularity of Lebesgue measure to choose a Borel subset $A_{\epsilon }\subseteq B$ differing from $B$ by a null set; the same measure bound and uniform convergence hold. Thus the requested Borel conclusion is valid under either convention. This proves the finite-measure Egorov theorem.

For an infinite-measure counterexample take $A=[0,\infty )$ and $f_n=1_{[n,\infty )}$. It converges pointwise to zero. Any subset of $A$ with complement of finite measure intersects every infinite tail $[n,\infty )$, so the supremum of $f_n$ there is one for every $n$. Hence finite measure is essential to the conclusion.

Let $q$ be conjugate to $p$, so $1/q=1-1/p$, including $q=1$ for $p=\infty$. Hölder's inequality implies, for every measurable finite-measure set $E$,This proves integrability on $A$ and gives uniform integrability there. Choose a set $B\subseteq A$ with $\lambda (A\setminus B)\le \delta$ on which convergence is uniform, using Egorov theorem. ThenFirst take $n\to \infty$ and then $\delta \downarrow 0$. Thus even the local $L^1$ difference tends to zero, and in particularFor $1<p<\infty$, simple functions supported on finite-measure Borel sets are dense in $L^q(\mathbb{R})$. The displayed convergence proves pairing convergence for such functions. Approximate any $h\in L^q$ by a simple $h_0$ and useBy the duality of Lp spaces, these are all continuous linear functionals on $L^p$, so $f_n$ converges weakly to $f$. It need not converge strongly: $f_n=1_{[n,n+1]}$ converges pointwise to zero but has $L^p$ norm one. For $p=\infty$ the pairing argument gives weak-star convergence against $L^1$, rather than asserting weak convergence against every functional in the much larger full dual.

A morphism of algebraic varieties is a map regular in local affine coordinates: for each point of the source, choose affine charts containing it and its image, and require each target coordinate to pull back locally to a quotient of polynomials with a denominator nonzero at that point. The map is continuous in the Zariski topology and pulls back regular functions on every target open set to regular functions on its inverse image.

Equivalently, one can take Zariski continuity and this local regular-function pullback property as the definition. In particular, a morphism between affine varieties is specified by globally regular coordinate functions, with values satisfying the defining equations of the target. Being an arbitrary set-theoretic function, or being merely Zariski continuous, is not sufficient.

Let $A=A(X)=k[x_1,\dots ,x_m]/I(X)$. Polynomial restrictions are regular functions, and a restriction is zero precisely when its class in the coordinate ring is zero. We prove surjectivity.

Using the usual irreducible-variety convention, view a regular function $f$ as an element of the fraction field of $A$. On a neighbourhood of every point it has the form $a/b$, where $a,b\in A$ and $b$ does not vanish at that point. Equality on a nonempty open subset of the irreducible variety is equality of rational functions. DefineThis is an ideal and contains a denominator nonzero at each point. Thus its zero locus is empty. The Hilbert Nullstellensatz implies $J=A$, so $1\in J$ and $f\in A$. ThereforeThe result also holds under a convention allowing reduced reducible affine varieties. One may cover by finitely many principal opens $D(b_i)$, represent $f$ in each localization, and increase powers until the local representatives glue after clearing denominators. Since these opens cover, the ideal generated by those powers is the unit ideal by the Nullstellensatz; a unit linear combination glues their numerators to one element of $A$. Irreducibility is used only for the shorter fraction-field proof, not for the assertion itself.

The Borel sigma-algebra $\mathcal{B}(\mathbb{R})$ is the smallest sigma-algebra containing all open subsets of $\mathbb{R}$. A real-valued Borel measurable function on $(E,\mathcal{E})$ is a map $f$ with $f^{-1}(B)\in \mathcal{E}$ for every $B\in \mathcal{B}(\mathbb{R})$. It suffices to test the sets $(a,\infty )$ for rational $a$, because these generate the Borel sigma-algebra.

Here “Borel function” describes measurability into the Borel structure of the target; the domain need not itself be a topological space.

Each function $(x,y)\mapsto |R_n(x)-R_n(y)|$ is Borel measurable on the product square, by composition with its coordinate projections and arithmetic of measurable real functions. Its value is one exactly when the $n$th digits disagree. Therefore the partial disagreement countsare nonnegative Borel functions and increase to the extended-real function $f$. For every finite $a$, $\{f>a\}={\bigcup }_N\{f_N>a\}$ is Borel. HenceInfinite disagreement is allowed, so restricting the target to finite real values would lose part of the stated function.

A time-homogeneous continuous-time Markov chain on a countable state space is a stochastic process with the Markov property: conditional on the current state, the law of its future depends on neither its past nor the current time. In the usual conservative, nonexplosive setting its sample paths are right-continuous step functions. Its Q-matrix $Q=(q_{ij})$ has $q_{ij}\ge 0$ for $j\ne i$, $q_i=-q_{ii}={\sum }_{j\ne i}{q}_{ij}<\infty$, and infinitesimal transition probabilities ${\mathbb{P}}_i(X(h)=j)=q_{ij}h+o(h)$ for $j\ne i$.

On entering $i$, the holding time has distribution $\mathrm{Exp}(q_i)$; independently, the next state is $j$ with probability $q_{ij}/q_i$. The successive states form the jump chain, whose transition probabilities areIf $q_i=0$, the state is absorbing, its holding time is infinite, and the jump-chain convention is $p_{ii}=1$. Nonexplosion ensures this construction defines the process for all finite times rather than accumulating infinitely many jumps in finite time.

Inductively, $X_0,\dots ,X_n$ are functions of ${\Delta }_0,\dots ,{\Delta }_n$, so ${\mathcal{F}}_n\subseteq \sigma ({\Delta }_0,\dots ,{\Delta }_n)$. The average $f_n(X_0,\dots ,X_n)$ is therefore ${\mathcal{F}}_n$-measurable and independent of ${\Delta }_{n+1}$. If $X_0,\dots ,X_n$ are integrable, thenStarting with $X_0={\Delta }_0$, induction proves integrability of every $X_n$. Adaptation holds by definition of the natural filtration. Independence and zero mean now legitimately giveHence the process is a martingale. Only first moments are used; no second-moment or boundedness assumption on the increments is needed.

A Lie point symmetry is a local one-parameter family of diffeomorphisms of $(t,\mathbf{x})$ that takes solution graphs of the differential equation to solution graphs wherever the transformed graph remains a graph over time. Its infinitesimal generator is $Y=\tau (t,\mathbf{x}){\partial }_t+{\sum }_i{\eta }_i(t,\mathbf{x}){\partial }_{x_i}$. Its first prolongation isFor a regular first-order equation the infinitesimal criterion is ${\mathrm{pr}}^{(1)}Y(\Delta )=0$ on $\Delta =0$; it says that the prolonged vector field is tangent to the equation submanifold.

With the constant nondegenerate canonical matrix $J$, define the Poisson bracket by $\{f,g\}=(\nabla f{)}^{T}J\nabla g$ and the Hamiltonian vector field by $V_f=J\nabla f$. Our convention gives $V_f[h]=\{h,f\}$. Applying the Jacobi identity to a test function $h$,soA first integral $F$ obeys $dF/dt=\{F,H\}=0$. Consequently $[V_F,V_H]=0$, and the local flows commute. The time-independent transformation generated by $V_F$, with $t$ unchanged, therefore maps integral curves of a vector field for $V_H$ to integral curves of a vector field: it is a Lie point symmetry.

Conversely, such a time-independent symmetry gives $[V_F,V_H]=0$, hence $V_{\{F,H\}}=0$. Nondegeneracy of $J$ makes $\{F,H\}$ constant on each connected component. At a fixed point ${\mathbf{x}}_0$, $J\nabla H({\mathbf{x}}_0)=0$ implies $\nabla H({\mathbf{x}}_0)=0$, so that constant is zero on its component. Thus the converse holds on a connected phase space containing a fixed point, or on each component that contains one.

The printed global converse needs this connectedness convention: a fixed point on one component does not control the constants on other components. For example take one component with $H=(q^2+p^2)/2$, $F=0$, and another with $H=p$, $F=q$. The Hamiltonian vector fields commute everywhere and there is a fixed point on the first component, but $F$ is not a first integral on the second. Even on a connected space the fixed-point condition matters: $H=p$, $F=q$ commute but $\dot{F}=1$.

Let a deterministic finite automaton recognizing the regular language have states $Q$, start state $q_0$, accepting states $F$, and transition function $\delta$. Construct a context-free grammar with one nonterminal $A_q$ for every state, start symbol $A_{q_0}$, and productionsInduction on the length of a word shows that $A_q$ derives $w{A}_r$ exactly when the automaton reaches $r$ from $q$ after reading $w$. The terminal production can then end the derivation precisely at an accepting state. Thus the grammar generates exactly the recognized language, proving every regular language is a context-free language. This is actually a right-linear grammar, a special case of a context-free grammar.