The hook length of a cell is one plus the number of cells to its right in the same row and below it in the same column. The hook-length formula isFor $\lambda =(n-k,k)$, with $0\le k\le n/2$, the second-row hook product is $k!$. In the first row, the hooks are $n-k-j+2$ for $1\le j\le k$ and $n-k-j+1$ for $k<j\le n-k$, so their product is $(n-k+1)!/(n-2k+1)$. ConsequentlyAt $k=0$, interpret the second binomial coefficient as zero.

A tabloid records the sets of entries in each row of a tableau. Let $C_t$ be the subgroup permuting entries independently within each column. The polytabloid isand the Specht module $S^{\lambda }$ is spanned by these vectors. A standard polytabloid is one whose tableau increases along rows and down columns.

Order tabloids by first sorting each row increasingly, then comparing the resulting row-reading words lexicographically. For a standard tableau $t$, every nonidentity column permutation produces a strictly larger word. In the first row affected by that permutation, each replacement comes from farther down the same column and is larger than the entry it replaces; all earlier rows remain unchanged. Sorting the affected row therefore increases its word. Also no nontrivial column permutation fixes its row sets, since a cell is the unique intersection of its row and column. Hence the least tabloid in $e_t$ is $\{t\}$ with coefficient one.

Distinct standard tableaux have distinct row tabloids. In a nonzero linear relation among their polytabloids, choose the least leading tabloid with a nonzero coefficient. None of the vectors with larger leading tabloids can contribute to it, so its coefficient cannot cancel. This contradiction proves linear independence of the standard polytabloids.

Use induction on $\dim V$. By part (b), there is a one-dimensional invariant linear subspace $U=Kv$ on which $L$ acts as zero. The induced operators on $V/U$ remain nilpotent, so induction gives a simultaneous strictly upper triangular basis there. Part (c) lifts it together with $v$ to such a basis of $V$.

Thus all elements strictly lower one common complete flag. A nested Lie bracket of sufficiently many flag-lowering operators lowers it past zero. Consequently the lower central series of $L$ terminates, proving that the Lie algebra is nilpotent. In fact its nilpotency class is at most $n-1$ for $n>1$; the zero algebra covers the one-dimensional case.

Start with the given basis of $U$, and follow it by arbitrary lifts of the given basis of $V/U$. Since $U$ is an invariant linear subspace, every acting endomorphism has block formwhere $A$ and $B$ are its matrices on $U$ and $V/U$. Both are strictly upper triangular in the chosen bases. Therefore the whole matrix has zero entries on and below its diagonal. The same lifted basis works simultaneously for all elements of $L$.

We prove the common-zero-vector form of Engel's theorem by strong induction on the dimension of the acting Lie algebra of operators, allowing its action on a nonzero finite-dimensional space to have a kernel. The zero-dimensional case is immediate.

First, every proper Lie subalgebra $H<L$ has a strictly larger normalizer of a Lie subalgebra. The action of $H$ on $L/H$ by $h\cdot (x+H)=[h,x]+H$ consists of nilpotent operators by part (a). Its image has dimension at most $\dim H<\dim L$. By induction it has a nonzero common-zero vector $x+H$, giving $[H,x]\subseteq H$ with $x\notin H$. Hence $N_L(H)>H$.

Choose a maximal proper Lie subalgebra $H$. The preceding paragraph forces $N_L(H)=L$, so $H$ is an ideal of a Lie algebra. Moreover $L/H$ has dimension one: otherwise any nonzero one-dimensional subspace of $L/H$, which is a Lie subalgebra, would lift to an intermediate proper subalgebra. By induction the common kernelis nonzero. It is $L$-invariant, since $h(xv)=x(hv)+[h,x]v=0$ for $v\in W$, $h\in H$, $x\in L$. Take $x\notin H$. Its restriction to $W$ is nilpotent and so has a nonzero vector $v$ in its kernel. As $L=H+Kx$, this vector is killed by every element of $L$:This completes the induction. The underlying space is understood to be nonzero, as required for the asserted vector to exist.

On the space of endomorphisms of $V$, left and right multiplication by $x$ commute, and $\mathrm{ad}x$ is their difference. The binomial theorem therefore givesFor $m=2n-1$, every term has either $j\ge n$ or $m-j\ge n$, so every term vanishes when $x^n=0$. Restriction to the invariant Lie subalgebra $L$ provesThe argument works over every field; no division by a binomial coefficient is used.

Because $f$ is continuous, $g(x,y)=e^{-x^2}f(e^{-x^2/2}y)$ is continuous on ${\mathbb{R}}^2$, so $|g|$ is a nonnegative measurable function. Apply Tonelli's theorem before assuming integrability. For each fixed $x$, set $z=e^{-x^2/2}y$, so $dy=e^{x^2/2}dz$. Lebesgue substitution on an interval givesTherefore the Gaussian integral yieldsThis proves that $g$ is Lebesgue integrable on ${\mathbb{R}}^2$. Fubini's theorem now permits changing the order for $g$ itself. The same substitution without absolute values gives the stronger exact identityTaking absolute values and using the triangle inequality for an integral provesThe constant is sharp: any nonzero nonnegative continuous integrable $f$ gives equality. This is a weighted dilation kernel with $a(x)=e^{-x^2}$ and $b(x)=e^{-x^2/2}$. The inner $x$ integral in the requested order is guaranteed to be integrable for almost every $y$, which is all the iterated integral requires.

On sigma-finite measure spaces, including the real line with Lebesgue measure, Tonelli's theorem says that a nonnegative measurable function $H$ has equal product and iterated integrals:with infinity allowed. Finiteness is not a prerequisite.

Fubini's theorem says that if a measurable signed or complex $H$ is absolutely integrable on the product, then its sections are integrable for almost every value of the other variable, the iterated integral functions are integrable, and the same equalities hold with finite values. For completed Lebesgue measure, exceptional sections need not themselves be measurable; the integrals use the almost-everywhere defined sections. Thus Tonelli can first establish absolute integrability using $|H|$, after which Fubini permits changing the order for $H$.

Fix any real $t_0$ and restrict $t$ to $|t-t_0|<1$. The parameter derivative isThis bound is integrable on $(0,1)$ and is uniform throughout the parameter neighborhood. Part (ii) supplied integrability at $t_0$, so dominated differentiation under the integral sign proves differentiability there. As $t_0$ was arbitrary,For $t>0$, $e^{tx}\ge 1$; for $t<0$, $e^{tx}\le 1$. Since ${\int }_0^1(1-x{)}^{-1/2}dx=2$, these comparisons giveThe first two inequalities are actually strict, because the exponential differs strictly from one throughout $(0,1)$. Differentiating once more under the same type of bound gives $F^{\prime \prime }(t)={\int }_0^{1}x{e}^{tx}/\sqrt{1-x}dx>0$. Hence $F$ is strictly convex and lies above its tangent $2t$ at zero. These properties are collected in exponential integral with a square-root endpoint singularity.

Fix $t\in \mathbb{R}$. The mean value theorem for the exponential givesConsequentlyThe function is continuous on $(0,1)$ and hence measurable, so $u(\cdot ,t)$ is Lebesgue integrable for every real $t$. In particular, the numerator cancels the apparent $1/x$ singularity at zero: $u(x,t)\to t$ as $x\downarrow 0$. Only the integrable square-root singularity at one remains. At $t=0$ the function is identically zero.

A sufficient theorem is dominated differentiation under the integral sign. Let $t_0$ be interior to a parameter interval, suppose $u(\cdot ,t)$ is measurable for each $t$, and suppose $u(\cdot ,t_0)$ is integrable. Assume $u(x,\cdot )$ is differentiable near $t_0$ for $x$ outside one fixed null set, and that throughout this neighborhoodfor one nonnegative Lebesgue integrable function $H$. Then $F(t)={\int }_{E}u(x,t)dx$ is defined near $t_0$, is differentiable at $t_0$, andIndeed the mean value theorem bounds both $|u(x,t)-u(x,t_0)|$ by $|t-t_0|H(x)$ and the difference quotient by $H(x)$. The first bound proves local integrability; the second lets the dominated convergence theorem pass the limit of difference quotients through the integral.

Use the power series of the exponential function:Each term is a Lebesgue integrable function, andThe sum of these nonnegative integrals is at most $2{\sum }_{n\ge 0}1/n!=2e<\infty$. Applying integration of a series with summable L1 norms, just proved in part (ii), therefore justifies exchanging the series and the integral:The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.

The monotone convergence theorem states that, for nonnegative measurable $f_n$ on a Lebesgue measurable set $E$ with $f_n\uparrow f$ almost everywhere,allowing infinity. Fatou's lemma states that, for nonnegative measurable $f_n$,Now write $C={\sup }_n{\int }_E|g_n|<\infty$. A pointwise limit superior of measurable functions is measurable, and it agrees with $g$ off the exceptional null set. By completeness of Lebesgue measure, assigning any values on that null set preserves measurability. Therefore $g$ is measurable. Since $|g_n|\to |g|$ almost everywhere, Fatou's lemma givesThus $g$ is finite almost everywhere and is a Lebesgue integrable function. This is an almost-everywhere limit under a uniform L1 bound. The argument proves integrability of the limit, not convergence in L1 space.

Both functions are continuous on the positive domains in question, so they are measurable. For the first function, the decreasing substitution $t=1/x$ and $|dx/dt|=t^{-2}$ giveFor every integer $k\ge 1$,Summing this lower bound gives a divergent harmonic series. Thus $f$ is not Lebesgue integrable on $(0,1)$.

For the second function, $|\sin x|\le x$ on $(0,1)$ and $|\sin x|\le 1$ everywhere giveTherefore $g$ is Lebesgue integrable on $(0,\infty )$. The sine power kernel integrability threshold is $1<p<2$: one endpoint requires $p<2$, and the other requires $p>1$. The signed improper integral for $f$ does converge by the Dirichlet test for improper integrals, but conditional improper convergence does not give a finite integral of $|f|$.

Let $\varphi :I\to J$ be a $C^1$ diffeomorphism between intervals, which may be unbounded. For a nonnegative Lebesgue measurable function $f$ on $J$, Lebesgue substitution on an interval givesallowing the value infinity. The same identity applied to $|f|$ shows that $f$ is Lebesgue integrable exactly when the transformed integrand is integrable; in that case the displayed formula holds for signed or complex $f$ as well. The absolute value on the derivative makes the formula valid for both increasing and decreasing substitutions. Finite endpoints can be omitted without changing a Lebesgue integral, because individual points are null sets.

The dominated convergence theorem says that if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le H$ almost everywhere for a single Lebesgue integrable function $H\ge 0$, then $f$ is integrable andHere, for $x>0$, put $H_0(x)=x^{-1/2}{e}^{-x}$ and $f_n(x)=H_0(x)(1-(-1{)}^n{e}^{-nx})$. The exponential function gives $e^{-nx}\to 0$ at every such $x$, so $f_n\to H_0$. Also $|f_n(x)|\le 2H_0(x)$. This dominator is integrable: on $(0,1)$, $H_0(x)\le x^{-1/2}$, whose integral is two; on $[1,\infty )$, $H_0(x)\le e^{-x}$, whose integral is finite. DCT therefore proves the required limit.

The limiting value can also be evaluated. Substitute $x=t^2$ and use the Gaussian integral:In fact, scaling the second exponential term gives the exact integral $\sqrt{\pi }[1-(-1{)}^n/\sqrt{n+1}]$. The integral error decays as $n^{-1/2}$ despite exponential convergence at each fixed positive $x$; the slow part lies near the endpoint zero.

For a Lebesgue measurable function $f:\mathbb{R}\to \mathbb{R}$, the preimage $f^{-1}(U)$ of every open set $U$ is a Lebesgue measurable set. For a Borel measurable function, each such preimage must instead be a Borel set. Equivalently, both definitions can test all Borel target sets: the collectionis a sigma-algebra, since preimages preserve complements and countable unions. Containing all open sets therefore makes it contain the Borel sigma-algebra.

Now let $U$ be open. Since $h$ is a Borel measurable function, $h^{-1}(U)$ is Borel. Since $g$ is a Lebesgue measurable function, $g^{-1}(h^{-1}(U))$ is a Lebesgue measurable set. ButThus $h\circ g$ is a Lebesgue measurable function. The intermediate sigma-algebra is Borel; this is why Borel postcomposition preserves Lebesgue measurability.