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The sequence $\lfloor n\alpha \rfloor$ for $n=1,2,\dots$, with a fixed positive irrational number $\alpha$. Complementary Beatty sequences for $\alpha ,\beta >1$ partition the positive integers when $1/\alpha +1/\beta =1$.

If $\alpha ,\beta >1$ are irrational numbers with $1/\alpha +1/\beta =1$, the Beatty sequences $\lfloor n\alpha \rfloor$ and $\lfloor n\beta \rfloor$ partition the positive integers.

In exact arithmetic, after $k$ completed halvings the retained root bracket has width $W_k=b_k-a_k$ and midpoint $m_k$, whereThe absolute value bound holds for every zero of a function $x_{\ast }$ in the current root bracket, including the one approached by continued bisection method updates. A zero of a function discarded with an earlier half need not satisfy later bounds. Before update $k\ge 1$, the evaluated candidate is $m_{k-1}$, whose corresponding guarantee is $(b-a)/2^k$. After update $k$, the new midpoint $m_k$ has the smaller guarantee $(b-a)/2^{k+1}$, even without evaluating $f(m_k)$.

The guaranteed error halves at every step; the actual absolute value error need not decrease at every step or have successive ratio $1/2$. In the square root example it increases from $|m_6-\sqrt{2}|\approx 0.000151062$ to $|m_7-\sqrt{2}|\approx 0.003755188$, while both errors remain below their guarantees. An exact midpoint zero of a function instead gives zero error and immediate termination. For this example $\sqrt{2}$ is an irrational number, so no finite exact halving can produce it as a midpoint.

For a brief proof outline, each retained half has half the previous width and contains a zero of a function by the intermediate value theorem. Every point in the retained root bracket is at most half its width from its midpoint. If the process does not stop at a zero of a function, the nested interval theorem gives one common point as the widths tend to zero; continuity and the opposite endpoint signs force its function value to be zero. Uniqueness in the initial root bracket is unnecessary.

The rational numbers $q_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$ lie in $[1,2]$ and obey $0\le \sqrt{2}-q_n<10^{-n}$. They form a convergent sequence in the real numbers, with limit of a sequence $\sqrt{2}$. Every subsequence has that same real limit. Because $\sqrt{2}$ is an irrational number, no subsequence converges to a point in the rational numbers with their usual metric. Boundedness alone therefore does not give the corresponding theorem in every metric space; completeness of the real numbers is doing real work here. The floor function simply truncates the decimal expansion in this example.

Cauchy's functional equation asks for functions $f:\mathbb{R}\to \mathbb{R}$ satisfyingThe identity must hold for every pair of real numbers, and a solution is an additive function. It says that combining two inputs before applying $f$ gives the same result as combining their outputs.^[ref]

For example, $f(x)=3x$ satisfies the identity because $3(x+y)=3x+3y$. At $x=2/3$ and $y=5/6$ the two sides giveFor positive widths, a rectangle of height three makes this visible: cutting its width into $x$ and $y$ partitions its area into $3x$ and $3y$. The same identity also holds for negative inputs, where $3x$ is a signed output rather than an ordinary area.

Every function $f(x)=cx$ with a fixed real number $c$ solves Cauchy's functional equation. The surprising question is whether these are all the solutions. Rational homogeneity of an additive function forces $f(q)=qf(1)$ for every rational number $q$, but reaching an irrational number requires an additional argument. Continuity at just one point or being a bounded function on an interval of positive length supplies that argument; without such a condition, discontinuous additive functions exist.^[ref]

This makes Cauchy's functional equation a useful test of the boundary between algebra and analysis: a rule about addition fixes all rational multiples, while a tiny amount of continuity fixes the entire real line. For example, the continuous exponential functional equation reduces to an additive function after taking a logarithm, so this result explains why its nonzero solutions are exponential functions.

A Dedekind cut can be represented by a nonempty proper subset $A$ of the rational numbers satisfying two conditions: if $q\in A$ and $r<q$ is rational, then $r\in A$; and $A$ has no maximum. The complementary set contains all the rational values on the other side of the boundary. Using the convention that the lower set never includes its boundary avoids two representations of a rational boundary.

The set $\{q\in \mathbb{Q}:q<0\}$ represents zero. The set $\{q\in \mathbb{Q}:q<0\text{ or }{q}^2<2\}$ represents $\sqrt{2}$, even though no rational number occupies that boundary by the irrationality of the square root of two. Thus a Dedekind cut describes an irrational number entirely using rational comparisons. Ordering these cuts by the subset relation and defining their arithmetic gives a construction of the real numbers. ^[ref].

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

Every nonempty open real interval contains an irrational number. Choose a rational number $q$ strictly inside the interval using density of the rational numbers; then $q+\sqrt{2}/N$ remains inside for sufficiently large positive integers $N$ and is irrational. Thus the irrational numbers form a dense subset of the real line. Together with rational density, this gives approaching sequences of both types at every real point, useful for testing continuity of functions defined differently on the two sets.

For every irrational number $\alpha$ and positive integer $N$, there are integers $p,q$ with $1\le q\le N$ such thatThe proof applies the pigeonhole principle to the $N+1$ fractional parts of $0,\alpha ,\dots ,N\alpha$.

Assuming the axiom of choice, Cauchy's functional equation has solutions that are not linear maps over the real numbers. Regard $\mathbb{R}$ as a vector space over the rational numbers. Because $\sqrt{2}$ is an irrational number, $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$. Zorn's lemma extends them to a Hamel basis $B$; assigning arbitrary real outputs to its basis vectors and extending by finite rational linear combinations gives an additive function.^[ref]

For a particular choice, assign $g(1)=3$ and $g(b)=0$ for every $b\in B\setminus \{1\}$. Every real number has a unique finite linear combination $x={\sum }_{b\in F}{q}_{b}b$ with $F\subset B$ finite and $q_b\in \mathbb{Q}$. Definewhere the coefficient $q_1$ is zero if $1\notin F$. The uniqueness of the Hamel basis representation makes this definition unambiguous, and adding the coefficients proves $g(x+y)=g(x)+g(y)$. On the rational two-dimensional vector subspace spanned by $1$ and $\sqrt{2}$, the rule is particularly simple:For instance, $u=2/3+\sqrt{2}$ and $v=5/6-2\sqrt{2}$ give $g(u)=2$, $g(v)=5/2$, and $g(u+v)=9/2$. Thus the map still preserves addition, even though $g(\sqrt{2})=0\ne 3\sqrt{2}$ rules out being a linear map over the real numbers.

The rational approximants $q_n$ from additive function continuous at one point is linear expose the failure of continuity: $q_n\to \sqrt{2}$ but $g(q_n)=3q_n\to 3\sqrt{2}\ne g(\sqrt{2})$. Moreover $h_n=\sqrt{2}-q_n\to 0$ while $g(h_n)=-3q_n\to -3\sqrt{2}$. The map is continuous nowhere, since continuity at any point would imply additive function continuous at one point is linear. It is unbounded on every interval of positive length by additive function bounded on an interval is linear.

The axiom of choice is used to obtain the Hamel basis; the coefficient rule does not provide an explicit way to find the representation of an arbitrary real number. A plot sampled only at rational numbers would show exactly the same values as $f(x)=3x$, so that plot cannot reveal this discontinuous additive function.

The essential infimum and essential supremum describe the lower and upper bounds of a measurable function after disregarding null sets for a specified measure. A pointwise infimum or supremum sees every input; an essential bound sees whether values below or above a proposed threshold occur on a set of positive measure. Thus "essential" means insensitive to changes on a null set, rather than insensitive to unusual values. This distinction makes essential bounds compatible with Lebesgue integrals and the equivalence classes used in Lp spaces. ^[ref].

For a real measurable function $f$ on a measure space $(X,\Sigma ,\mu )$, the definitions areThese can take values in the extended real number line. On a space of positive total measure, finite essential bounds obey ${\mathrm{essinf}}_{\mu }f\le f\le {\mathrm{esssup}}_{\mu }f$ almost everywhere. If $s={\mathrm{esssup}}_{\mu }f$ is finite, every threshold $b<s$ has $\mu (\{f>b\})>0$; the corresponding lower-edge statement is $\mu (\{f<a\})>0$ for every $a>{\mathrm{essinf}}_{\mu }f$. These are the measure-based counterparts of the approximation property of a supremum and approximation property of an infimum.

Here is a two-sided example on $[0,1]$ with Lebesgue measure:The pointwise bounds are $\inf g=-3$ and $\sup g=4$, while the essential bounds are $\mathrm{essinf}g=0$ and $\mathrm{esssup}g=1$. Both exceptional singletons are null sets. But for $0<a<1$, the inputs where $g<a$ still have measure $a$, and those where $g>a$ have measure $1-a$: deleting or adding two individual inputs cannot change those lengths. Hence neither essential bound can be moved inward.

The choice of measure is part of the answer. The same $g$ has essential bounds $-3$ and $4$ under counting measure, because every nonempty set has positive measure. Under the Dirac measure at $1/4$, both essential bounds equal $-3$. A null set for one measure need not be a null set for another.

Nor must the exceptional inputs be isolated. The Dirichlet function, equal to one at rational numbers and zero at irrational numbers, has pointwise supremum one on $[0,1]$, but essential supremum zero for Lebesgue measure: the rational numbers form a countable set that is a null set, even though they are dense. Conversely, a value ten on an interval of length $0.001$ affects the essential supremum just as decisively as it would on half the domain. Small positive measure is not zero measure.

For fixed $m$, the inner limit is one precisely when $m!x$ is an integer, and zero otherwise, because $|\cos (\pi y)|=1$ exactly at integers and powers of a number with modulus below one tend to zero. If $x=p/q$ is a rational number, $q$ divides $m!$ for every $m\ge q$, making the inner limit eventually one. If $x$ is an irrational number, multiplication by a nonzero integer cannot make it integral, making every inner limit zero. The order of the two limits is essential to this argument.

A factorial series with integral digits $0\le a_n<n$ is an irrational number exactly when positive digits occur infinitely often and digits below their maximum occur infinitely often. Under these two conditions every scaled tail $N!{\sum }_{n>N}{a}_n/n!$ lies strictly between zero and one. If the sum were a rational number $A/B$, choose $N\ge B$; then $N!A/B$ and the scaled finite head are integers, contradicting the strict tail bound. Conversely, an eventually zero digit sequence gives a finite rational number sum, and an eventually maximal sequence gives a finite head plus $1/N!$, again a rational number.

If Euler's number were $a/b$, choose $n\ge \max (b,2)$ and multiply its factorial-series tail by $n!$. This produces an integer, but the tail is positive and smaller than ${\sum }_{j\ge 1}(n+1{)}^{-j}=1/n<1$. The contradiction proves that the number is an irrational number. The argument rules out a hypothetical rational remainder by placing it in an impossible integer interval.

For integers $a,b>1$, rationality of ${\log }_{a}b$ is equivalent to equality of some positive integer powers $a^p=b^q$. Exponentiation proves one direction and taking logarithms proves the other. Thus multiplicatively independent integers give an irrational number as their logarithm ratio. For distinct prime bases and arguments, unique factorization supplies that independence.

For a nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

Checking only midpoints is not enough for an arbitrary subset of a real vector space. The rational numbers contain the midpoint of every pair of rational numbers, but the line segment from $0$ to $1$ also contains irrational numbers. For instance $\sqrt{2}/2$ lies on that line segment and is not rational. This shows why the definition quantifies over every real $t\in [0,1]$.

There is a useful additional hypothesis: a closed subset $C\subseteq {\mathbb{R}}^d$ containing the midpoint of every pair of its points is a convex set. Repeated bisection includes all points $(1-t)x+ty$ with dyadic rational $t\in [0,1]$. These coefficients approach every real $t\in [0,1]$; being a closed subset then includes each limit. This is a short proof outline, and the rational numbers example identifies precisely why a closed subset was required.

Inclusion gives $|\mathbb{R}\setminus \mathbb{Q}|\le \mathfrak{c}$. If $\kappa =|\mathbb{R}\setminus \mathbb{Q}|$, the disjoint decomposition of $\mathbb{R}$ givesThe irrational numbers are infinite, for instance $n+\sqrt{2}$ are pairwise distinct irrationals. Standard infinite cardinal addition gives ${\aleph }_0+\kappa =\max ({\aleph }_0,\kappa )$. Since Cantor's theorem gives $\mathfrak{c}>{\aleph }_0$, the equality forces $\kappa =\mathfrak{c}$. ThereforeThis is the permitted cardinal-arithmetic formal proof; it does not subtract infinite cardinals as though they were finite numbers.

To prove the Archimedean property, suppose the positive integers were bounded above in $\mathbb{R}$. By the least upper bound axiom they would have a finite supremum $s$. Since $s-1$ is not an upper bound, some positive integer $k$ satisfies $k>s-1$. Then the positive integer $k+1>s$, contradicting the definition of $s$. Thus the positive integers are unbounded. Equivalently, for any positive real $a,b$, some positive integer $n$ satisfies $na>b$, by applying unboundedness to $b/a$.

For fixed $m$, put $r_m={\cos }^2(m!\pi x)$. The standard properties needed are $0\le {\cos }^{2}t\le 1$, and ${\cos }^{2}t=1$ exactly when $t\in \pi \mathbb{Z}$. If $0\le r<1$, then $r^n\to 0$; for $0<r<1$ this also follows from the Bernoulli inequality, since $r^{-n}\ge 1+n(r^{-1}-1)\to \infty$. Therefore the inner limit isIf $x=p/q$ is a rational number, with $q\ge 1$, then $q$ divides the factorial $m!$ whenever $m\ge q$. Hence $m!x$ is an integer for all sufficiently large $m$, and the outer limit is one. This includes the endpoints $x=0,1$. If $x$ is an irrational number, $m!x$ cannot be an integer for any $m$, since division by $m!$ would make $x$ rational. Then every inner limit is zero. The factorial cosine iterated limit detects rationality result is consequentlyEach inner limit was taken with $m$ fixed; the argument does not interchange the two limits or assert a joint limit as both indices grow.

For a bounded function on an interval $I$, write ${\omega }_I(f)={\sup }_{I}f-{\inf }_{I}f$ for its oscillation of a function on an interval. For a partition $P$, the difference between its upper Darboux sum and lower Darboux sum is ${\sum }_{I\in P}|I|{\omega }_I(f)$. The Riemann integrability criterion says that this can be made arbitrarily small.

Since ${\omega }_I(f_1+f_2)\le {\omega }_I(f_1)+{\omega }_I(f_2)$, choose good partitions for the two functions and take their common refinement. Darboux sum refinement monotonicity makes both oscillation sums small on this refinement, so their sum is Riemann integrable. The same argument handles differences and scalar multiples.

Both $t\mapsto \max (t,0)$ and $t\mapsto |t|$ are Lipschitz continuous with constant one. Thus ${\omega }_I(f^{+})\le {\omega }_I(f)$ and ${\omega }_I(|f|)\le {\omega }_I(f)$. This calculation shows that Lipschitz composition preserves Riemann integrability, so the positive part and the absolute value are integrable.

The converse for the absolute value is false. Define $f=1$ on rational numbers and $f=-1$ on irrational numbers. Then $|f|=1$ is integrable, but every interval has supremum one and infimum minus one, so every Darboux difference for $f$ is $2(b-a)$. Hence $f$ is not Riemann integrable.

Finally,is integrable by the closure properties just proved. This establishes that Riemann integrability is closed under addition and maximum.

A countable supremum can fail, even for a uniformly bounded family. Enumerate the rational points of $[a,b]$ as $q_1,q_2,\dots$ and set $f_n=1_{\{q_n\}}$. Each singleton indicator function is Riemann integrable with integral zero: isolate its single nonzero point in intervals of arbitrarily small total length. Butis the nonintegrable Dirichlet function. This proves that a pointwise supremum need not preserve Riemann integrability.