A useful differentiation under the integral sign theorem is this: if $f(\cdot ,y)$ is integrable, $f$ is differentiable in $y$ almost everywhere, and on every compact parameter interval $|{\partial }_{y}f(x,y)|$ is bounded by an integrable function of $x$, then $F^{\prime }(y)={\int }_I{\partial }_{y}f(x,y)dx$, under the usual pointwise continuity hypotheses. It follows by applying dominated convergence to difference quotients. The theorem can be iterated when a corresponding bound holds for the second derivative.

On a compact interval $0<a\le y\le b$, the original integrand is bounded by $b/(1+x^4)$, using $|\sin (x^{2}y)|\le b{x}^2$. Its first and second parameter derivatives arewith integrable bounds $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$. Thus $F$ is twice differentiable. Subtracting the integrals givesThe final integral is finite: its integrand tends to one at zero and is absolutely bounded by $t^{-2}$ at infinity. Consequentlywhich establishes the required parameter-independent constant without needing an explicit Fresnel integral evaluation.

The Lebesgue outer measure of $E$ is the infimum of ${\sum }_j|I_j|$ over countable coverings by open intervals. A set is a null set when this infimum is zero, equivalently when it has such a cover of total length less than every prescribed positive number.

The Cantor set is obtained from $[0,1]$ by deleting the open middle third of every remaining interval at each stage. After $n$ stages it lies inside $2^n$ closed intervals of length $3^{-n}$. Slightly enlarging these finitely many intervals to open intervals shows its outer measure is at most $(2/3{)}^n$ for every $n$, hence zero.

For the modified construction, let $L_n$ be the total length of the $2^n$ intervals in $K_n$. The removed fraction at stage $n$ is exactly $1/(2^n+2)$, soThe product telescopes because $2^n+2=2(2^{n-1}+1)$, givingSince $K\subseteq K_n$, the same finite interval-cover argument gives $m^{\ast }(K)\le L_n$ for every $n$. Taking the infimum provesOnly interval lengths and the definition of outer measure were used; the three numbered clauses in the source describe the construction rather than separate requests to prove additional statements.

A finite soluble group is one whose derived series $G^{(0)}=G$, $G^{(j+1)}=[G^{(j)},G^{(j)}]$ reaches the trivial group. Equivalently it has a finite normal series with abelian factors. If $H\le G$, induction using commutator containment gives $H^{(j)}\le G^{(j)}$ for every $j$. Thus termination for $G$ implies termination for $H$, proving every subgroup of a soluble group is soluble.

For $S_4$, the chainhas abelian factors $V_4$, $C_3$ and $C_2$. Here $V_4$ consists of the identity and the three double transpositions. It is normal because conjugation permutes those double transpositions. Hence $S_4$ is soluble.

An example neither soluble nor simple is $S_5$. Its subgroup $A_5$ is proper, nontrivial and normal, so $S_5$ is not simple. The group $A_5$ is nontrivial and perfect: its generating three-cycles are commutators of double transpositions, using two additional symbols, so $[A_5,A_5]=A_5$. For example the commutator of $(12)(45)$ and $(23)(45)$ is a three-cycle, and conjugation and inversion give all three-cycles. Thus $A_5$ is not soluble, and subgroup inheritance then proves $S_5$ is not soluble either.