For $x\ge 1$, the bound $|\sin u|\le |u|$ gives $|g(x)|\le x^{-3/2}$, whose tail integral is finite. Thus $g{1}_{[1,\infty )}$ is integrable.

Near one, $h(x)=[(x-1)(x+1){]}^{-1/2}$ has an integrable inverse-square-root singularity. At infinity, however, $h(x)\ge 1/x$ for $x>1$, and the harmonic tail diverges. Thus $h{1}_{[1,\infty )}$ is not integrable. Assigning $h(1)=0$ changes neither conclusion, since one point has Lebesgue measure zero.

On $(0,1)$, $|g(x)|\le 1$, so ${\int }_0^1|g|\le 1$. For the second function,Both functions are measurable, and therefore both restrictions are Lebesgue integrable. The oscillation of $g$ near zero does not affect absolute integrability on this finite interval.

Since $|f|=f^{+}+f^{-}$, the previous part and closure under addition prove measurability of the absolute value.

The converse fails. Choose a nonmeasurable subset $N$ of $[0,1]$ and define $f=1$ on $N$, $f=-1$ on its complement. Then $|f|\equiv 1$ is measurable, whereas $f^{-1}([0,\infty ))=N$ is not. Thus measurability of $|f|$ does not imply measurability of $f$.

If $f$ is measurable, then for $a\le 0$, $\{f^{+}\ge a\}=E$, and for $a>0$, $\{f^{+}\ge a\}=\{f\ge a\}$. Thus the given superlevel criterion proves $f^{+}$ measurable. Part 2 makes $-f$ measurable, and the same argument applied to $(-f{)}^{+}$ proves $f^{-}$ measurable.

Conversely $f=f^{+}-f^{-}$. If both parts are measurable, part 2 and the permitted closure under addition prove $f$ measurable. Therefore $f$ is measurable exactly when its positive and negative parts are measurable.

For each real $a$,The final strict-superlevel set is measurable by part 1 and the given criterion for measurable functions. A complement relative to the measurable domain is measurable, so $-g$ is measurable.

Use the countable identityIf $f(x)>b$, some $n$ has $1/n\le f(x)-b$, proving the nontrivial inclusion. Each preimage on the right is measurable by statistical hypothesis, so closure of a sigma-algebra under countable unions proves the strict-superlevel preimage is measurable.

Again $s=(a/p)\in \{1,-1\}$. By Euler's criterion, $(s/p)=s^{(p-1)/2}$. If $p\equiv 1(\mathrm{mod}4)$, the exponent is even and the answer is one. If $p\equiv 3(\mathrm{mod}4)$, it is odd and the answer is $s$. Hence

Write $s=(a/p)$. If $p\mid a$, both sides of the proposed identity are zero. Otherwise $s=\pm 1$. The multiplicativity of the Legendre symbol and Euler's criterion giveFor the middle equality, Euler's congruence determines the sign exactly, since both the symbol and $s^{(p-1)/2}$ are in $\{1,-1\}$ and $p$ is odd.

Euler's criterion statesFor $p\nmid a$, the sign is $+1$ precisely for a quadratic residue and $-1$ for a quadratic nonresidue. For $p\mid a$, both sides are zero.

The Legendre symbol isIt depends only on the residue class of $a$ modulo the odd prime $p$.

Reduce the monic integer polynomial modulo $p$. Every $u\in {\mathbb{F}}_p$ satisfies $u^p=u$, so $u^p-u-1=-1\ne 0$. Thus the reduction has no root in ${\mathbb{F}}_p$. If it had a proper factor, part 2 would force such a root, a contradiction. Hence it is irreducible over ${\mathbb{F}}_p$.

The criterion of irreducibility by degree-preserving reduction modulo a prime, which follows from Gauss's lemma, now applies because the polynomial is monic and its degree is preserved on reduction. Consequently $X^p-X-1$ is irreducible in $\mathbb{Q}[X]$.

The intended hypothesis must be a proper monic factor, with $1\le r\le p-1$. Without that restriction the printed deduction is false: every polynomial has the displayed trivial factorization with $r=p$, even when it has no root in $F$.

By part 1, the roots in a splitting field are distinct and have the form $\alpha +j$. The degree-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using Vieta's formulas givesBoth $a_1$ and the sum of prime-field elements lie in $F$. Since $1\le r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha =(-a_1-{\sum }_{j\in J}j)/r\in F$, and part 1 proves the polynomial splits into distinct linear factors over $F$. In particular an Artin–Schreier polynomial of prime degree is either irreducible or splits completely.

In field characteristic $p$, the intermediate binomial coefficients in the $p$th power vanish. HenceRepeated translation gives all $p$ roots $\alpha +j$, $j\in {\mathbb{F}}_p$. They are distinct because the prime field embeds in $F$. They exhaust the roots by degree; also the formal derivative of $X^p-X-a$ is $-1$, so this is a separable polynomial. If one root belongs to $F$, they all do, andThis is the root-translation property of an Artin–Schreier polynomial.

Suppose the unique optimum $x^{\ast }$ were $(1-t)u+tv$ for distinct feasible points $u,v$ and $0<t<1$. Feasibility is preserved by convex combinations. Optimality gives $c^{T}u\le c^T{x}^{\ast }$ and $c^{T}v\le c^T{x}^{\ast }$, while linearity gives their strictly weighted average equal to $c^T{x}^{\ast }$. Both inequalities must therefore be equalities. Then $u$ and $v$ are also optimal, contradicting uniqueness. Hence the unique optimum is an extreme point of the feasible set.