Write $x\in Z(A{\otimes }_{k}B)$ uniquely as $x={\sum }_j{a}_j\otimes b_j$ using part (i). For each $a\in A$, the equality $(a\otimes 1)x=x(a\otimes 1)$ becomesThe coefficient uniqueness from part (i) gives $a{a}_j=a_{j}a$ for every $a$, so each $a_j$ belongs to the center of an associative algebra $Z(A)$. Consequently $x\in Z(A){\otimes }_{k}B$.

Choose a basis $z_1,\dots ,z_t$ of the vector space $Z(A)$ and rewrite $x={\sum }_{i=1}^t{z}_i\otimes d_i$. For $b\in B$, commuting with $1\otimes b$ givesExtend the linearly independent family $z_i$ to a basis of $A$ and apply its dual basis to the first factor. It follows that $b{d}_i=d_{i}b$ for every $b$, hence $d_i\in Z(B)$. This proves $Z(A{\otimes }_{k}B)\subseteq Z(A){\otimes }_{k}Z(B)$.

Conversely, if $z\in Z(A)$ and $w\in Z(B)$, thenThe pure tensors $a\otimes b$ span the tensor product, so $z\otimes w$ is central, as is every sum of such pure tensors. Therefore the center of a tensor product of associative algebras isThese tensor products are identified with their natural vector subspaces of $A{\otimes }_{k}B$; extending bases justifies these embeddings over the field $k$.

Use the left $A$-module structure $c\cdot (a\otimes b)=ca\otimes b$ on the tensor product of associative algebras. Expanding each $b\in B$ in the basis $b_1,\dots ,b_m$ shows that every tensor can be writtenThus the displayed family generates the module.

For uniqueness, let ${\beta }_j:B\to k$ be the corresponding dual basis, with ${\beta }_j(b_i)={\delta }_{ij}$. Applying the linear map ${\mathrm{id}}_A\otimes {\beta }_j$ to a linear relation ${\sum }_i{a}_i\otimes b_i=0$ gives $a_j=0$. Hence the family is linearly independent in a module, andIn particular $1\otimes b_1,\dots ,1\otimes b_m$ is a basis of a module, with coefficients uniquely in $A$.

Construct the sum successively, retaining a simple module only when it adds a new direct summand. Suppose $S=V_1+\cdots +V_{j-1}$ is already expressed as a direct sum of modules selected from these simple modules. The intersection $S\cap V_j$ is a submodule of the simple module $V_j$, so it is either zero or all of $V_j$. In the latter case $S+V_j=S$. In the former caseStarting with $S=0$ and proceeding by mathematical induction gives a direct sum of modules equal to $V_1+\cdots +V_n$. Thus the sum is a semisimple module. The argument proves the useful finite sum of simple submodules lemma without claiming that the originally listed submodules are themselves independent.

An $A$-module is completely reducible if it is a direct sum of modules each of which is a simple module:Here a simple module is nonzero and has no submodules other than zero and itself. This is precisely the definition of a semisimple module. The indexing set need not be finite unless $V$ has finite dimension; the zero module is included as the empty direct sum of modules.

The Bell state $|\Omega ⟩$ is an eigenvector of $J_r$ with eigenvalue $(1+3r)/2$. Its three-dimensional orthogonal complement has eigenvalue $(1-r)/2$. The Choi criterion for complete positivity therefore givesFor sufficiency one can also write a Kraus representation explicitly:In this range the four coefficients are nonnegative and sum to one, giving a physical mixture of Pauli conjugations. This is the complete positivity range of the depolarizing channel in the Bloch-scaling parameterization.

Use the unnormalized Choi matrix convention $J_r={\sum }_{j,k=0}^1|j⟩⟨k|\otimes {\mathcal{E}}_r(|j⟩⟨k|)$, which has trace two. In the ordered basis $|00⟩,|01⟩,|10⟩,|11⟩$,Equivalently $J_r=2r|\Omega ⟩⟨\Omega |+(1-r)I_4/2$. The convention obtained by applying the channel to one half of the normalized Bell state gives $J_r/2$; its positivity criterion is the same.

The output of a state of Bloch radius $u$ has eigenvalues $(1\pm ru)/2$. These are nonnegative for every $0\le u\le 1$ exactly whenIf $|r|>1$, a pure input already gives a negative eigenvalue. Conversely, $|r|\le 1$ preserves the Bloch ball, hence positivity for states and, by positive scaling, for all positive operators. This is positivity of the map alone; physical implementation on part of an entangled system needs complete positivity.

By linearity, the isotropic qubit map acts asIt contracts or expands every direction by the same factor and reverses the vector when $r<0$. More generally ${\mathcal{E}}_r(A)=rA+(1-r)\mathrm{tr}(A)I/2$, so the map preserves trace and Hermiticity for every real $r$.

True. Replace the dominating representative by $|g|$ if necessary, and take the countable union of the exceptional Lebesgue null sets so that every bound holds off a single null set. Since the interval has Lebesgue measure one, Cauchy-Schwarz inequality givesThus $g^2$ is a Lebesgue integrable function. The squared functions satisfy $|f_n{|}^2\to 0$ pointwise and $|f_n{|}^2\le |g{|}^2$ almost everywhere. The dominated convergence theorem therefore yieldsThis uses the Lp inclusion on a finite measure space to turn the assumed fourth-power bound into the second-power domination actually needed.

False. A typewriter sequence sweeps successively through finer dyadic partitions. For $k\ge 0$ and $j=0,\dots ,2^k-1$, define the indicator functionIts L2 norm satisfiesAs $n$ tends to infinity so does its block index $k$, proving norm convergence to zero. However, every $x\in (0,1)$ belongs to exactly one of the half-open intervals in each block, so $f_n(x)=1$ infinitely often. For $k\ge 1$ there is also an interval in each block that does not contain $x$, so $f_n(x)=0$ infinitely often. Thus there is no pointwise convergence at any $x\in (0,1)$, let alone convergence almost everywhere.

Let $(f_n)$ be a Cauchy sequence in the L2 space $L^2(\mathbb{R})$, choosing measurable representatives. There is a subsequence $(f_{n_k})$ withWrite $d_k=f_{n_{k+1}}-f_{n_k}$ and $S_N={\sum }_{k=1}^N|d_k|$. Repeated Minkowski inequality gives $\Vert S_N{\Vert }_2\le {\sum }_{k=1}^N{2}^{-k}\le 1$. As $S_N^2$ increases pointwise, the monotone convergence theorem givesHence $S={\sum }_k|d_k|$ is finite almost everywhere and represents an L2 function. Definewhere the series converges, and set $f=0$ on the exceptional Lebesgue null set. It is a measurable function, and $|f|\le |f_{n_1}|+S$ almost everywhere, so Minkowski inequality gives $f\in L^2(\mathbb{R})$.

For fixed $k$, Fatou's lemma applied to squares of finite tails and then Minkowski inequality giveThus the subsequence converges to $f$ in norm. Given $\epsilon >0$, choose a sufficiently late $n_k$ with $\Vert f-f_{n_k}{\Vert }_2<\epsilon /2$ and with $\Vert f_n-f_{n_k}{\Vert }_2<\epsilon /2$ for every sufficiently large $n$, using the original Cauchy sequence property. The triangle inequality then gives $\Vert f_n-f{\Vert }_2<\epsilon$. ThereforeThis proves the needed instance of completeness of Lp spaces without assuming the result being proved.

Use Lebesgue measure restricted to the Lebesgue measurable set $E$. For $1\le p<\infty$, start with the measurable real-valued functions $f$ on $E$ satisfying ${\int }_E|f{|}^{p}dm<\infty$. Define an equivalence relation by $f\sim g$ when $f=g$ almost everywhere, and let $L^p(E)$ consist of their equivalence classes. Addition and scalar multiplication are defined using representatives. The Lp norm isFor $p=\infty$, use measurable functions bounded almost everywhere, apply the same identification, and define the L-infinity norm byThese expressions are independent of the representative. For finite $p$, zero norm is equivalent to $f=0$ almost everywhere: each set $\{|f|>1/n\}$ must have measure zero, and their union is $\{f\ne 0\}$. The same implication follows from the essential supremum for $p=\infty$. Absolute homogeneity is immediate; Minkowski inequality proves the triangle inequality and closure under addition. Thus these are normed vector spaces. Using complex-valued representatives gives their complex versions by the same construction.

Minkowski inequality states that on a measure space $(X,\Sigma ,\mu )$, for $1\le p\le \infty$ and measurable real or complex $f,g$ of finite Lp norm,For $p=1$, integrate the scalar triangle inequality $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the same result by taking the essential supremum.

For $1<p<\infty$, first ensure that the left side is finite. The convex function $s\mapsto s^p$ on $[0,\infty )$ givesso $f+g$ has finite Lp norm. Set $q=p/(p-1)$ and $H=\Vert f+g{\Vert }_p$. Since $(p-1)q=p$, Holder inequality givesIf $H=0$ the result is immediate; otherwise divide by $H^{p-1}$. This proves the triangle inequality for every claimed exponent.

Put $h(t,x)=e^{-x^2-t^2/x^2}$. For every $t\ge 0$, $0<h(t,x)\le e^{-x^2}$, and the upper bound is a Lebesgue integrable function by part (b)(i). Since $h(t,x)\to e^{-x^2}$ for each $x>0$ as $t\downarrow 0$, the dominated convergence theorem gives right continuity at zero andTo apply dominated differentiation under the integral sign, fix $t_0>0$ and restrict $t$ to $[t_0/2,3t_0/2]$. ThenFor $x\ge 1$, $D(x)\le 3t_0{e}^{-x^2}$. For $0<x<1$, writing $a=t_0^2/4>0$, the substitution rule $u=1/x$ givesThus $D$ is integrable. The mean value theorem bounds nearby difference quotients by $D$, so the dominated convergence theorem justifies the derivative at $t_0$. As $t_0$ was arbitrary,Now make the change of variables $u=t/x$. The reversal of limits cancels the negative differential and givesConsequently $F^{\prime }(t)=-2F(t)$. This also follows in sign from the fact that the positive integrand decreases strictly as $t$ increases: the printed equation with a positive sign cannot hold. Solving the linear differential equation, or differentiating $e^{2t}F(t)$, and using right continuity at zero givesThis is a half-line instance of the Gaussian integral with an inverse-square term.

Let $J={\int }_0^{\infty }{e}^{-x^2}dx$. It is positive and finite: on $(0,1)$ the integrand is at most one, and on $[1,\infty )$ it is at most $e^{-x}$. Since the planar integrand is nonnegative and measurable, Tonelli's theorem givesThe change of variables formula in polar coordinates, with $x=r\cos \theta$, $y=r\sin \theta$ and absolute Jacobian determinant $r$, yieldsOne can apply the formula first on $\delta <r<R$, $0<\theta <\pi /2$, where the coordinate map is a diffeomorphism, and then exhaust the quadrant using the monotone convergence theorem. Taking the positive square root proves the half-line Gaussian integral:Here $dx$ is understood in the one-dimensional integral; it is missing from the printed question.

Again the function is continuous on its domain and therefore a measurable function. For $0<x<1$, $|(\log x)e^{-x}|\le -\log x$, andThis standard improper integral follows from the antiderivative $x-x\log x$ and $x\log x\to 0$ at zero. For $x\ge 1$, $0\le \log x\le x$, and integration by parts gives ${\int }_1^{\infty }x{e}^{-x}dx=2/e$. ThusThe logarithmic singularity is integrable and the exponential factor controls the tail, so this too is a Lebesgue integrable function.

The function is continuous on $(0,\infty )$, hence a measurable function. On $(0,1)$ its absolute value is at most one, while on $[1,\infty )$ the elementary estimate $|\sin u|\le |u|$ gives $|\sin (1/x^3)|\le x^{-3}$. By comparison for the Lebesgue integral,It is therefore a Lebesgue integrable function. The oscillations near zero do not obstruct absolute integrability because their amplitude stays bounded on an interval of finite measure.

True. Suppose instead that $B=A_1\times A_2$ were a Lebesgue measurable set in the plane. The section consequence of Fubini's theorem says that there is a Lebesgue null set $N\subseteq \mathbb{R}$ such thatis a Lebesgue measurable set for every $x\notin N$. This version applies to completed Lebesgue measure; it can be obtained on bounded rectangles and then extended by a countable exhaustion, so no finite-measure assumption on $B$ is needed.

For every $x\in A_1$ the section is exactly $A_2$, which is not measurable. Thus $A_1\subseteq N$. By completeness of Lebesgue measure, every subset of $N$ is measurable, contradicting the hypothesis on $A_1$. Therefore

False. Take $E=\{0\}$ and any non-measurable set $A\subseteq \mathbb{R}$. The vertical line $\{0\}\times \mathbb{R}$ is a planar Lebesgue null set: each segment $\{0\}\times [-N,N]$ can be enclosed in rectangles of arbitrarily small area, and the line is a countable union of these segments. Consequently its subset $E\times A$ is a Lebesgue measurable set of planar measure zero, by completeness of Lebesgue measure.

A nonempty factor can therefore have too little Lebesgue measure to transmit the other factor's failure of measurability. Its exceptional vertical section at $x=0$ is consistent with Lebesgue measurable sections being guaranteed only almost everywhere.

The exact criterion isNecessity follows already from one approximation with error less than one: a compact set $K\subseteq \mathbb{R}$ is bounded, hence has finite Lebesgue measure, so $m(E)=m(K)+m(E\setminus K)<\infty$.

Conversely, suppose $m(E)<\infty$. By continuity from above of a measure, the decreasing sets $E\setminus [-N,N]$ have measures tending to zero. Choose $N\ge 1$ so that $m(E\setminus [-N,N])<\epsilon /2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an open set $V\supseteq B$ with $m(V\setminus B)<\epsilon /2$. DefineIt is closed and bounded, hence a compact set by the Heine-Borel theorem, and $K\subseteq E$. The discarded portion satisfiesThe second term is at most $m(V\setminus B)$. This proves the claimed inner regularity of Lebesgue measure with an arbitrarily small absolute error.