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If the sum of Lebesgue outer measures on the right is infinite, the inequality is automatic. Otherwise, fix $\epsilon >0$ and cover each $A_n$ by a countable family of intervals whose total length is less than $m^{\ast }(A_n)+\epsilon 2^{-n}$. These covers exist by the definition of infimum. The combined countable family covers ${\bigcup }_n{A}_n$ and has total length at most ${\sum }_n{m}^{\ast }(A_n)+\epsilon$. Take the infimum over covers of the union, then let $\epsilon$ tend to zero. No measurability assumption on the $A_n$ is needed.

If $A\subseteq B$, every countable interval cover of $B$ is also a cover of $A$. Taking the infimum of the lengths over the larger family of covers can only decrease the result. Thus Lebesgue outer measure is monotone on all subsets, with no measurability assumption.

Translating every interval of a countable cover of $A\subseteq \mathbb{R}$ by $a$ produces a cover of $A+a$ with the same lengths. Taking the infimum of cover lengths gives $m^{\ast }(A+a)\le m^{\ast }(A)$. Translating back by $-a$ gives the reverse inequality. This proves translation invariance for Lebesgue outer measure on every subset, without first assuming the set is Lebesgue measurable.

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.

The Lebesgue inner measure approximates a subset of the real line from within by compact sets, while Lebesgue outer measure approximates it from outside by countable interval covers. For a finite closed interval $I$ containing $E$,Indeed, every compact $K\subseteq E$ gives $m^{\ast }(I\setminus E)\le m(I)-m(K)$. Conversely, an open cover of $I\setminus E$ with total length arbitrarily close to its outer measure has compact complement in $I$ lying inside $E$, giving the reverse inequality in the limit.

For $m^{\ast }(E)<\infty$, equality of inner and outer measures holds precisely when $E$ is Lebesgue measurable. For an unbounded set, measurability is equivalently checked on all bounded intersections $E\cap [-n,n]$. Equality $m_{\ast }(E)=m^{\ast }(E)=\infty$ alone is insufficient: adjoining a bounded nonmeasurable subset to a disjoint half-line makes both quantities infinite without making the union measurable.

A bound $\mathbb{E}[B_{j+1}\mid B_j]\le f(B_j)$ implies $\mathbb{E}{B}_{j+1}\le \mathbb{E}f(B_j)$. It does not by itself imply $\mathbb{E}{B}_{j+1}\le f(\mathbb{E}{B}_j)$. For $f(b)=b^2/n$, Jensen's inequality goes in the opposite direction. The missing contribution is the variance: $\mathbb{E}[B_j^2]=(\mathbb{E}{B}_j{)}^2+\mathrm{Var}(B_j)$.

A null weight contributes zero to a Laplace integral on its domain of convergence. It can be added to a weight without changing the represented function, so an integration-by-parts differential equation for the weight is not an unconditional uniqueness theorem.

For example, let $\Gamma$ enter along angle $\pi /3$ from infinity to zero and leave along angle $2\pi /3$ to infinity. The entire weight $h(\zeta )=e^{{\zeta }^2}$ is null for every complex $z$. Throughout the wedge between those rays, $\mathrm{Re}({\zeta }^2)\le -r^2/2$; the factor $e^{z\zeta }$ is dominated by this quadratic decay. Closing the contour by a large circular arc and applying the Cauchy integral theorem therefore proves the zero integral. All derivatives with respect to $z$ converge locally uniformly as well.

On the same contour, the quartic weight $e^{{\zeta }^4/4}$ gives a nontrivial entire solution. Its rays can be deformed within the quartic decay sectors for a contour integral to angles $\pi /4$ and $3\pi /4$, where the value at $z=0$ is $-\sqrt{2}{4}^{-3/4}\Gamma (1/4)\ne 0$. Adding the null weight changes neither this integral nor its vanishing endpoint terms, but the sum does not satisfy $g^{\prime }={\zeta }^{3}g$. Thus the usual Laplace method selects a convenient weight by cancelling the interior integrand separately; a necessity claim needs additional injectivity assumptions.

Retaining integer assignment equations while dualizing extra capacity inequalities gives an easily evaluated upper bound for a maximization problem. With multipliers $\lambda \ge 0$ and capacities $b$, it is $\theta (\lambda )={\max }_{x\text{ assignment}}[c^{T}x+{\lambda }^T(b-Ax)]$. It is a convex maximum of affine functions. Minimizing it exploits the convex hull of assignments, so it can have a finite optimum when only a fractional mixture satisfies the side constraints and no integer assignment does. This separates a useful relaxation bound from a certificate of integer feasibility.

For the real-coefficient differential operator $Mv=(p{v}^{\prime }{)}^{\prime }+qv$, integration by parts givesIf $p$ tends to zero at both endpoints and $u,v,u^{\prime },v^{\prime }$ remain bounded there, the boundary term vanishes without prescribing endpoint values. Thus regular endpoint behaviour can supply the domain condition that makes the expression symmetric. Endpoint degeneracy alone is insufficient for arbitrary singular functions: for $p=x(1-x)$ and $v=\log x$, one has $p{v}^{\prime }=1-x\to 1$ at zero. A self-adjoint differential operator requires a specified admissible domain, rather than only a formal differential expression.

Continuous square-integrable martingales on a fixed finite horizon form a complete space under the displayed terminal norm, modulo indistinguishability of stochastic processes. A terminal-norm Cauchy sequence admits a subsequence whose successive differences have norms below $2^{-n}$. The Doob L2 maximal inequality and Markov's inequality make the probabilities that these differences have suprema above $2^{-n/2}$ summable. The Borel-Cantelli first lemma then gives an almost surely uniformly convergent telescoping series, so the limit has continuous paths.

Conditional-expectation contractivity gives convergence in $L^2$ at every fixed time, and passage to the limit in the martingale identities proves that the limit is a martingale. Adaptedness can be retained by taking the pointwise finite subsequence limit when it exists, and zero otherwise. The original Cauchy sequence converges in terminal norm to this limit. This completeness allows the Itô isometry to extend stochastic integration from simple predictable integrands to the full square-integrable integrand space while retaining a continuous martingale version.

Suppose measurable $a_n$ on a Lebesgue measurable set $E$ satisfy ${\sum }_n{\int }_E|a_n|<\infty$. The nonnegative partial sums $S_N={\sum }_{n\le N}|a_n|$ increase to a measurable extended-valued $S$. The monotone convergence theorem gives ${\int }_{E}S={\sum }_n{\int }_E|a_n|<\infty$, forcing $S<\infty$ almost everywhere. Each $a_n$ is finite almost everywhere too; discard the union of these countably many null sets when defining signed partial sums. The series ${\sum }_n{a}_n$ then converges absolutely outside a null set.

Suppose $p(T)q(T)=0$ for a linear operator $T$, and Bézout's identity for polynomials gives $ap+bq=1$. The kernel decomposition for coprime polynomials is $V=\ker p(T)\oplus \ker q(T)$. The displayed operators are the corresponding projection operators: $P$ is the identity on $\ker p(T)$ and zero on $\ker q(T)$, while $Q$ has the opposite action. Thus $P+Q=I$, $P^2=P$, $Q^2=Q$ and $PQ=QP=0$. These actions also prove independence from the choice of Bézout coefficients. Every summand is an invariant subspace of $T$, since the projections are polynomials in $T$ and commute with it.

For $T=\lambda I+N$ with $N^2=0\ne N$ and a polynomial $p$, the binomial formula gives $p(T)=p(\lambda )I+p^{\prime }(\lambda )N$. Hence its module on the two-dimensional space is indecomposable exactly when $p^{\prime }(\lambda )\ne 0$: in that case the nonzero nilpotent part is one Jordan block; otherwise the operator is scalar and every line is a submodule. For $p(X)=X^2$ over a field of characteristic different from two, the criterion is $\lambda \ne 0$.

For continuous $P\ge 0,Q$, a scalar second variation $q=\int (P{\phi }^{\prime 2}+Q{\phi }^2)$ is strictly positive on nonzero fixed-endpoint Lipschitz variations if and only if every regularized form $q_{\eta }$ for $\eta >0$ has no conjugate point in $(a,b]$, and the original weak Jacobi boundary problem has no nonzero Lipschitz solution. The Jacobi positivity criterion applies to each regularized coefficient $P+\eta >0$. Letting $\eta$ decrease to zero gives $q\ge 0$. For a nonnegative quadratic form, $q(\phi )=0$ forces its mixed bilinear form with every test function to vanish, by considering $q(\phi +t\psi )$ for both signs of $t$. This is the weak Jacobi equation, so the kernel condition gives strict positivity. Necessity follows by the same argument in reverse. It distinguishes strict positivity from a uniform coercive estimate when the derivative coefficient degenerates.

For a plate occupying $x>0$, the boundary part of the first variation of its isotropic thin-plate bending energy is, up to an overall nonzero sign,At a free edge the variation and its normal derivative can be prescribed independently. Their coefficients must therefore vanish, giving the two displayed natural boundary conditions: zero bending moment and zero effective transverse shear. The mixed-derivative coefficient includes the derivative along the edge of the twisting moment; keeping only the normal derivative of $w_{xx}+\nu w_{yy}$ would miss that contribution.

Suppose the rational canonical form of $T$ is $C(f_1)\oplus \cdots \oplus C(f_r)$, with monic nonconstant invariant factors of a linear operator satisfying $f_1\mid \cdots \mid f_r$. Each block has minimal polynomial and characteristic polynomial $f_j$, because a cyclic operator has equal minimal and characteristic polynomials. The determinant of a block diagonal matrix is the product of block determinants, giving $c_T={\prod }_j{f}_j$. A polynomial kills the direct sum exactly when it kills each block, equivalently when every $f_j$ divides it. Thus $m_T$ is their least common multiple, which the divisibility chain reduces to $f_r$. On the zero vector space both polynomials are conventionally one.

A Newton-Cotes quadrature is an interpolatory quadrature rule whose quadrature nodes are equally spaced. Integrating the Lagrange interpolation polynomial gives quadrature weights $W_j={\int }_a^b{\ell }_j(x)dx$. A Newton-Cotes closed quadrature includes both endpoints; an open rule uses interior quadrature nodes. Simpson's rule is the closed rule with three quadrature nodes. Its quadrature weights are $(b-a)(1,4,1)/6$.

A compatible transfinite family of bijections can enumerate the square of an ordinal by adjoining a new row and then the remaining new column. Starting with $g_0=\varnothing$, extend at a successor by $g_{\alpha +1}(\alpha ,\beta )={\gamma }_{\alpha }+\beta$ for $\beta \le \alpha$ and $g_{\alpha +1}(\beta ,\alpha )={\gamma }_{\alpha }+(\alpha +1)+\beta$ for $\beta <\alpha$. Thus ${\gamma }_{\alpha +1}={\gamma }_{\alpha }+(\alpha +1)+\alpha$. At limits take compatible unions. The two new domain pieces are disjoint; the column is empty at the initial step $\alpha =0$. Each step enumerates consecutive ordinal intervals, proving bijectivity by transfinite induction. Finite stages have ${\gamma }_n=n^2$. Successor and limit cardinality arguments then give the square equality for well-orderable infinite cardinals without using that equality to justify the construction.