Iid strong law fails at infinite absolute mean

Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series: For every $x\in \mathbb{R}$,

$\frac{d}{dx}\arctan x=\frac{1}{1+x^2},\arctan x={\int }_0^x\frac{dt}{1+t^2}\mathrm{.}$

For $\mathrm{\mid }x\mathrm{\mid }<1$,

$\arctan x={\sum }_{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n+1}}{2n+1}\mathrm{.}$

At the endpoint, the ordinarily convergent alternating series satisfies

$\frac{\pi }{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \mathrm{.}$

Probability laws correspond to distribution functions: Assume the Axiom of Countable Choice.

1. Let $X$ be a real random variable, let ${\mathbb{P}}_X$ be its law, and let $F_X(x)=\mathbb{P}(X\le x)$. Then $F_X$ is nondecreasing and right-continuous, satisfies ${\lim }_{x\to -\mathrm{\infty }}{F}_X(x)=0,{\lim }_{x\to +\mathrm{\infty }}{F}_X(x)=1,$ and obeys ${\mathbb{P}}_X((a,b])=F_X(b)-F_X(a)(a<b)\mathrm{.}$
2. Conversely, if $F:\mathbb{R}\to \mathbb{R}$ is nondecreasing and right-continuous with ${\lim }_{x\to -\mathrm{\infty }}F(x)=0,{\lim }_{x\to +\mathrm{\infty }}F(x)=1,$ then there is a unique Borel probability measure $\mu$ on $\mathbb{R}$ such that $\mu ((a,b])=F(b)-F(a)(a<b),$ equivalently $F(x)=\mu ((-\mathrm{\infty },x])(x\in \mathbb{R})\mathrm{.}$

The recursion theorem: Let $(N,0,\sigma )$ be a Peano system (def-peano-system), in particular the natural numbers $\mathbb{N}$ (def-natural-numbers). For any set $A$, any element $a\in A$, and any function $f:A\to A$, there is a unique function $g:N\to A$ such that $g(0)=a$ and $g(\sigma (n))=f(g(n))$ for all $n\in N$.

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure $\nu$ on $(S,\mathrm{\Sigma })$ is the common law of a countable independent family of $S$-valued random elements.

A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: Let $a<b$ be reals and let $f:[a,b]\to \mathbb{R}$ be continuous on $[a,b]$ (def-continuity-real). Then $f$ is bounded (def-bounded-set) and Riemann integrable on $[a,b]$ (def-darboux-integral).

The proof gives more than integrability: it gives a partition that works. For every real $\epsilon >0$ the uniform partition into $N$ parts already satisfies $U(f,P)-L(f,P)<\epsilon$, as soon as $N$ is large enough that $(b-a)\mathrm{/}N$ is below the $\delta$ that uniform continuity supplies for $\epsilon \mathrm{/}(2(b-a))$. Uniform continuity is exactly what makes one $\delta$ serve all $N$ subintervals at once, and it is the only place where the compactness of $[a,b]$ is used.

The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$: Let $a<b$ be reals, let $G:[a,b]\to \mathbb{R}$ be differentiable at every point of $[a,b]$ as a function on $[a,b]$ (def-derivative; at $a$ and $b$ this is the one-sided derivative), let $f:=G^{\mathrm{\prime }}$, and suppose $f$ is integrable on $[a,b]$ (def-darboux-integral). Then

${\int }_a^{b}f=G(b)-G(a)\mathrm{.}$

Both hypotheses are needed and neither is removable. A function may be differentiable everywhere with $G^{\mathrm{\prime }}$ not integrable — then the left-hand side does not exist (an everywhere differentiable function with unbounded derivative) — and an integrable $f$ need not be the derivative of anything (the sign function); both witnesses are on the companion page.

No continuity of $f$ is assumed, which is what makes this the working form: the theorem evaluates ${\int }_a^{b}f$ for every integrable derivative, not only for continuous integrands.

A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral: Assume the Axiom of Countable Choice. Let $a<b$ and let $f:[a,b]\to \mathbb{R}$ be bounded and Riemann integrable. Then $f$ is Lebesgue measurable on $[a,b]$ and is integrable there, and its Lebesgue integral equals its Riemann integral: ${\int }_{[a,b]}fd{\lambda }_1={\int }_a^{b}f(x)dx\mathrm{.}$

This is the point at which the completeness of Lebesgue measure is used essentially: the proof obtains a Borel function equal to $f$ almost everywhere, and measurability of $f$ itself is then a completeness statement.

Monotone convergence for the integral: Let $0\le f_1\le f_2\le \cdots$ be measurable and suppose $f_n(x)\uparrow f(x)$ for every $x$. Then $\int f_{n}d\mu \uparrow \int fd\mu \mathrm{.}$

The indefinite integral of a nonnegative measurable function is a measure: Let $f:X\to [0,+\mathrm{\infty }]$ be measurable and define ${\nu }_f(A):={\int }_{A}fd\mu (A\in \mathcal{A})\mathrm{.}$ Then ${\nu }_f$ is a measure on $(X,\mathcal{A})$.

Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system: Let $\mathcal{P}$ be a $\pi$-system on $X$ generating $\mathcal{A}$, and let $\mu ,\nu$ be measures on $(X,\mathcal{A})$ that agree on $\mathcal{P}$. Suppose there is an increasing sequence $(P_n)$ in $\mathcal{P}$ with

$X={\bigcup }_n{P}_n,\mu (P_n)=\nu (P_n)<+\mathrm{\infty }(n\in \mathbb{N})\mathrm{.}$

Then $\mu =\nu$ on $\mathcal{A}$.

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: For $x>0$, $\log$ is differentiable and ${\log }^{\mathrm{\prime }}(x)=\frac{1}{x},\log x={\int }_1^x\frac{dt}{t}\mathrm{.}$

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: The function $\log :(0,\mathrm{\infty })\to \mathbb{R}$ is continuous and strictly increasing, is onto $\mathbb{R}$, and satisfies, for $x,y>0$, $\log (xy)=\log x+\log y,\log (x\mathrm{/}y)=\log x-\log y,\log (1\mathrm{/}x)=-\log x\mathrm{.}$ Also $\log 1=0$.

Integrability is necessary for an iid finite mean strong law: If IID real $(X_n)$ have $S_n\mathrm{/}n$ converging almost surely to a finite, possibly random, limit $L$, then $\mathbb{E}\mathrm{\mid }{X}_1\mathrm{\mid }<\mathrm{\infty }$ and $L=\mathbb{E}{X}_1$ almost surely.

Independent random elements have product joint law: Let $n\ge 1$, and let $X_i:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (S_i,{\mathrm{\Sigma }}_i)$ for $i<n$ be independent random elements. Define

$X=(X_0,\dots ,X_{n-1}):\mathrm{\Omega }\to {\prod }_{i<n}{S}_i\mathrm{.}$

Then $X$ is a random element of $({\prod }_{i<n}{S}_i,{⨂}_{i<n}{\mathrm{\Sigma }}_i)$, and its law is the finite product of the marginal laws:

${\mathbb{P}}_X={⨂}_{i<n}{\mathbb{P}}_{X_i}\mathrm{.}$

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be $\sigma$-finite measure spaces, and let $f:X\times Y\to [0,\mathrm{\infty }]$ be product-measurable. Then $x\mapsto {\int }_Y{f}_{x}d\nu$ and $y\mapsto {\int }_X{f}^{y}d\mu$ are measurable, and

${\int }_{X\times Y}fd(\mu \times \nu )={\int }_X\left({\int }_Y{f}_{x}d\nu \right)d\mu ={\int }_Y\left({\int }_X{f}^{y}d\mu \right)d\nu \mathrm{.}$

The principal inverse tangent $\arctan :\mathbb{R}\to (-\pi \mathrm{/}2,\pi \mathrm{/}2)$: By lem-tangent-principal-branch-is-bijective, tangent restricts to a continuous strictly increasing bijection

$\tan :(-\pi \mathrm{/}2,\pi \mathrm{/}2)⟶\mathbb{R}\mathrm{.}$

Its inverse is the principal inverse tangent

$\arctan :\mathbb{R}⟶(-\pi \mathrm{/}2,\pi \mathrm{/}2)\mathrm{.}$

Thus $\tan (\arctan y)=y$ for every real $y$, while $\arctan (\tan x)=x$ precisely for $x$ in the displayed principal interval. The inverse is continuous and strictly increasing by thm-continuous-inverse.

A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included: Let $n\ge 1$, assume the Axiom of Countable Choice (def-countable-choice), and let $a_i\le b_i$ be reals for $i<n$. Write

$R^{\circ }:=\{x\in {\mathbb{R}}^n:a_i<x_i<b_i\text{ for every }i<n\},\bar{R}:=[a,b]=\{x\in {\mathbb{R}}^n:a_i\le x_i\le b_i\text{ for every }i<n\}$

(def-multidimensional-rectangle-and-volume). Then $R^{\circ }$ is open and $\bar{R}$ is closed, so both are Borel and Lebesgue measurable, and every set $R$ with $R^{\circ }\subseteq R\subseteq \bar{R}$ is Lebesgue measurable with

${\lambda }_n(R)={\prod }_{i<n}(b_i-a_i)\mathrm{.}$

In particular this covers the four one-dimensional face conventions in each coordinate — the open box, the closed box $[a,b]$, the half-open box $B(a,b)={\prod }_{i<n}(a_i,b_i]$ of def-half-open-box, and every mixture of them, in any combination of coordinates — and it gives measure $0$ to all of them whenever $a_i=b_i$ for some $i<n$. For a half-open box with infinite parameters the value is already ${\lambda }_n(B)=\mathrm{vol}(B)$ (thm-lebesgue-measure-is-a-complete-measure).