Euler's criterion states, for an odd prime $p$,Apply it to $a=-1$. The integer $(-1{)}^{(p-1)/2}$ is $1$ when $p\equiv 1(\mathrm{mod}4)$ and $-1$ when $p\equiv 3(\mathrm{mod}4)$. Since those signs are distinct modulo an odd prime, the modular congruence determines the actual Legendre symbol:This is the first supplementary law for quadratic reciprocity.

For an odd prime $p$, the Legendre symbol isIt depends only on the congruence class of the integer $a$. The values $1$ and $-1$ distinguish a nonzero quadratic residue from a quadratic nonresidue; zero is a separate case.

Under the unital ring convention used elsewhere in this wiki, no such pair exists. In a two-element ring the elements are $0$ and $1$. Additive cancellation forces $1+1=0$, and the identity axiom forces $1\cdot 1=1$. The distributive property forces products involving zero to vanish. Every operation is therefore determined, giving the prime field ${\mathbb{F}}_2$ as the unique possibility up to ring isomorphism.

If the word ring allows a nonunital ring, there are instead two non-isomorphic examples: ${\mathbb{F}}_2$ and the zero-multiplication rng on the additive group $\mathbb{Z}/2\mathbb{Z}$. In the second example the unique nonzero element $u$ satisfies $u+u=0$ and $u^2=0$; every product is zero, so associativity and both distributive laws hold. An isomorphism would have to send $1$ to $u$ because it preserves zero and is bijective, but then it would send $1^2=1$ to both $u^2=0$ and $u$, which is impossible. These exhaust the rngs with two elements, since $u^2$ can only be $0$ or $u$. Specifying the identity convention is essential for this part.

No field has six elements. A finite field has positive characteristic: the sequence $0,1,2\cdot 1,\dots$ must repeat. By part (a), this characteristic is a prime number $p$. Translation by $1$ partitions the field into cycles of length $p$, since returning to a starting point after $k$ steps is equivalent to $k1=0$. Thus $p$ divides six and $p=2$ or $p=3$.

The multiples of $1$ form a copy of the prime field ${\mathbb{F}}_p$. For completeness, the inverse of a nonzero residue $a$ lies among these multiples: the integer Bezout identity $ab+cp=1$ gives $(a1)(b1)=1$. The entire field is a vector space over this subfield. A maximal linearly independent set is finite and spans, so it is a basis of some positive size $d$. Unique basis coordinates give exactly $p^d$ elements. This proves finite field cardinality is a prime power here. But six is neither a power of two nor a power of three, a contradiction.

A suitable field with four elements isWrite $\alpha$ for the class of $t$. Monic polynomial division over a ring gives a unique representative $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: existence follows by division, and uniqueness follows because a nonzero multiple of the degree-two modulus cannot have degree less than two. The four elements are therefore $0,1,\alpha ,\alpha +1$.

The defining relation gives ${\alpha }^2=\alpha +1$, and hence $\alpha (\alpha +1)=1$. Thus the nonzero elements have inversesThe quotient ring is commutative and every nonzero element is a unit, so it is a field. This verifies the field property directly, without quoting a theorem about irreducible-polynomial quotients.

No such ring exists: every finite integral domain is a field. Here is the proof. For $a\ne 0$, the map $L_a:R\to R$ defined by $L_a(x)=ax$ is injective, since $ax=ay$ gives $a(x-y)=0$ and therefore $x=y$. An injective self-map of a finite set is surjective, so some $x$ satisfies $ax=1$. Since an integral domain is commutative, $xa=1$ as well. Thus every nonzero element has a multiplicative inverse, making $R$ a field.

The conclusion still holds if an identity was not initially required. Surjectivity first gives an element $e$ with $ae=a$. Then $a(ex-x)=0$ implies $ex=x$ for every $x$, and commutativity makes $e$ a two-sided identity. Applying the preceding argument with $e$ supplies all inverses.

No such ring exists. Characteristic two gives $r+r=0$ for every $r$. Choose a nonzero element $a$, which exists in a five-element set. Translation $r\mapsto r+a$ is an involution without fixed points: applying it twice gives $r$, while $r+a=r$ would force $a=0$ by additive cancellation. Its disjoint pairs would partition all five elements, which is impossible because five is odd.

This argument applies even to a nonunital ring. With the unital convention, it is also the case $n=2$ of finite ring characteristic divides cardinality.

Take $R=\mathbb{Z}/4\mathbb{Z}$, a quotient ring with four residue classes. The class $\overline{2}$ is nonzero but $\overline{2}\overline{2}=\overline{0}$. Thus $R$ has a zero divisor and is not an integral domain.

The first-crossing maximal inequality for refining partitions follows from countable disjointness and the strict average bound:Thus this partition maximal operator has weak-type constant one.

The conclusion of differentiation along refining metric partitions isThe crucial geometric property is shrinking diameter, not merely refinement: since $x\in M_k(x)$ and $\mathrm{diam}{M}_k(x)<1/k$, continuity makes the corresponding oscillation tend to zero for every continuous test function. Continuous $L^1$ approximation and the maximal inequality then transfer this to $f$, exactly as before: the exceptional-set bound is $2\Vert f-g{\Vert }_1/\epsilon$ for an oscillation threshold $2\epsilon$. Countability makes the maximal function measurable, and positive cell measure makes every average defined. Refinement supplied the disjoint first-crossing argument; shrinking diameter supplies differentiation.

Every $x$ in the maximal superlevel set has a first index $k(x)$ at which the average on its unique cell $M_k(x)\in {\mathcal{P}}_k$ exceeds $t$. Refinement implies that all earlier cells containing $x$ contain this cell, and their averages are at most $t$. Thus $M_{k(x)}(x)\in {\mathcal{S}}_t$ and the selected cells cover the superlevel set.

Two cells at different levels are either disjoint or nested, because of refinement and the partition property. If selected cells of levels $k<l$ intersect, the level-$l$ one lies in the earlier cell whose average is above $t$, contradicting its first-crossing property. Cells at the same level are disjoint unless identical. Hence the selected cells form a countable disjoint partition of the superlevel set. A cell repeated unchanged at successive levels is counted once, at its first crossing; equal earlier ancestors are included in the comparison condition.

For a set $M$ of finite measure apply the preceding result to $f=1_M$. Since its local average minus its point value is bounded in absolute value by its mean absolute oscillation, it givesFor a general measurable $M$, apply the same argument to $1_{M\cap [-j,j]}\in L^1(\mathbb{R})$. At every interior point of $(-j,j)$, sufficiently short averages agree with those of $1_M$. The countable union of these interiors covers the line, and a countable union of null exceptional sets remains null. This proves the Lebesgue density theorem without a finite-measure assumption on $M$. The source's $\chi$ is the indicator of $M$.

Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow 0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the triangle inequality yieldsTherefore the maximal inequality and Chebyshev's inequality giveDensity makes the right side arbitrarily small. For each positive rational $\epsilon$ the exceptional set is null, proving $Df=0$ almost everywhere, the Lebesgue differentiation theorem. Values where the chosen representative is not finite already form a null set.

For the fundamental theorem define $F(x)={\int }_a^{x}f(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its measure is small by absolute continuity of the integral. At a Lebesgue point of $f$, for $h\ne 0$,Thus $F^{\prime }=f$ almost everywhere and $F(b)-F(a)={\int }_a^{b}f$. Differentiability at every point is not claimed for an arbitrary integrable function.

The centered Hardy-Littlewood maximal function isThe Hardy-Littlewood maximal inequality gives, for example,This valid constant follows from the disjoint-interval covering argument: a compact subset of the superlevel set has a finite cover by intervals of average above $t$; select disjoint intervals whose triples cover it, bound their total length by $\Vert f{\Vert }_1/t$, and use inner regularity. An optimal constant is not required.

For the finite nonnegative measures there exist a unique nonnegative integrable density $h\in L^1(\kappa )$, up to $\kappa$-almost-everywhere equality, and a unique nonnegative measure ${\mu }^s\perp \kappa$, such thatThis combines the Lebesgue decomposition theorem with the Radon-Nikodym theorem. In particular ${\mu }^a=h\kappa$ and $h=d{\mu }^a/d\kappa$; if $\mu \ll \kappa$ the singular term is zero.

Suppose $\mu =\alpha +\sigma =\beta +\tau$ are two such decompositions. Choose measurable supports $S,T$ of the singular parts with $\kappa (S)=\kappa (T)=0$, and put $N=S\cup T$. Absolute continuity makes both $\alpha$ and $\beta$ vanish on $N$. On $X\setminus N$ the two singular parts vanish, so for every measurable $A$,Subtracting from the finite measure $\mu$ gives $\sigma =\tau$. This proves uniqueness of the Lebesgue decomposition without subtracting infinite masses.

For the specified $M$, the restriction to $M$ is singular because $\kappa (M)=0$, and the complementary restriction is absolutely continuous by assumption. Their sum is $\mu$. Uniqueness therefore gives

The restriction of a measure to a measurable subset is $(\mu {|}_M)(A)=\mu (A\cap M)$ for every measurable $A$. Absolute continuity of measures, written $\mu \ll \kappa$, means $\kappa (A)=0⟹\mu (A)=0$. The measures are mutually singular when a measurable $S$ satisfies $\kappa (S)=0$ and $\mu (X\setminus S)=0$. Thus singularity means concentration on a null set of the other measure; it does not assert that every measurable set is null for one of them.

If $S\subseteq M_{-}$, then ${\mu }^{+}(S)=0$, so $\mu (S)=-|\mu |(S)\le 0$. Suppose instead that $S\cap M_{-}=\varnothing$ but $\mu (S)<0$. The preceding construction gives $R\subseteq S$ with ${\mu }^{+}(R)=0$ and $|\mu |(R)=-\mu (R)>0$. Then $M_{-}\cup R$ is ${\mu }^{+}$-null with variation strictly larger than that of $M_{-}$, a contradiction. Hence $\mu (S)\ge 0$ on the complement of $M_{-}$. This is a Hahn decomposition.

On $M_{+}=X\setminus M_{-}$ every partition has nonnegative signed measures, so $|\mu |(E)=\mu (E)={\mu }^{+}(E)$ for $E\subseteq M_{+}$. On $M_{-}$, ${\mu }^{+}$ vanishes. Splitting an arbitrary measurable set between these two sets givesIn particular ${\mu }^{+}\ll |\mu |$, and the exhibited Radon-Nikodym derivative verifies the Radon-Nikodym theorem for this pair directly.

Start with $R_0=A$. Having chosen disjoint $A_1,\dots ,A_i$, put $R_i=A\setminus {\bigcup }_{j\le i}{A}_j$ andThe empty set shows $s_i\ge 0$, and finite total variation bounds it above. If $s_i>0$, choose $A_{i+1}\subseteq R_i$ with $\mu (A_{i+1})\ge s_i/2>0$; if $s_i=0$, choose $A_{i+1}=\varnothing$. This is the required recursion, including the initial step $i=0$.

Disjointness and nonnegativity give ${\sum }_i\mu (A_i)\le |\mu |(A)<\infty$, hence $\mu (A_i)\to 0$. Write $R=A\setminus {\bigcup }_i{A}_i$. For every measurable $B\subseteq R$,Thus every measurable subset of $R$ has nonpositive measure. Each finite partition of $R$ consequently has variation sum exactly $-\mu (R)$, and therefore $|\mu |(R)=-\mu (R)$. It follows that ${\mu }^{+}(R)=0$. Finally countable additivity gives

From now on $|\mu |(X)<\infty$, so all signed quantities below are finite. Since $|\mu |(E)\ge |\mu (E)|$, the set functionis nonnegative. Countable additivity follows from that of the two terms, so ${\mu }^{+}$ is a finite positive measure.

Set $s=\sup \{|\mu |(M):{\mu }^{+}(M)=0\}$. This is finite. Choose measurable $M_j$ with ${\mu }^{+}(M_j)=0$ and $|\mu |(M_j)\to s$. Thenis ${\mu }^{+}$-null by countable subadditivity, and monotonicity gives $|\mu |(M_{-})\ge s$. The definition of $s$ gives the reverse inequality. Hence $M_{-}$ attains the maximum, rather than merely approximating it.

For a signed measure define its total variation measure byThe value at the empty set is zero. For disjoint measurable sets $A_j$ with union $A$, any finite partition $E_r$ of $A$ givesTaking the supremum proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus {\bigcup }_{j\le k}{A}_j$ givesIf one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\epsilon \downarrow 0$ and $k\to \infty$ proves countable additivity, including infinite values. Thus $|\mu |$ is a positive measure. This uses only countable additivity of $\mu$, without assuming a Hahn decomposition.