The standard normal density has total mass one

The power-series, product-limit, IVP, functional-equation, and Picard definitions agree: The following descriptions give the same function $\mathbb{R}\to (0,\mathrm{\infty })$: the power series $\sum x^n\mathrm{/}n!$; the product limit ${\lim }_n(1+x\mathrm{/}n{)}^n$; the normalized solution of $y^{\mathrm{\prime }}=y,y(0)=1$; the normalized continuous multiplicative function; and the compact-uniform limit of the Picard iterates.

The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$: The real exponential function is $C^{\mathrm{\infty }}$, and for every $m\in \mathbb{N}$, ${\exp }^{(m)}=\exp \mathrm{.}$ In particular $(\exp {)}^{\mathrm{\prime }}=\exp$.

Continuous functions on Euclidean spaces are Borel measurable: Assume the Axiom of Countable Choice. Let $n,m\ge 1$. Every continuous map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is Borel measurable in the sense of def-borel-and-lebesgue-measurable-function-on-rn.

Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$: Let $F$ be a complete ordered field (def-complete-ordered-field). Then every $a\in F$ with $a\ge 0$ has a unique $s\in F$ with $s\ge 0$ and $s^2=a$; we write $s=\sqrt{a}$. Consequently the positive elements of $F$ are exactly the nonzero squares: $x>0$ if and only if $x=y^2$ for some $y\mathrm{\ne }0$.

Substitution: if $\phi$ is differentiable on $[c,d]$ with ${\phi }^{\mathrm{\prime }}$ integrable and $f$ is continuous on an interval containing $\phi ([c,d])$, then ${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }}$: Let $c<d$ be reals and let $\phi :[c,d]\to \mathbb{R}$ be differentiable at every point of $[c,d]$ as a function on $[c,d]$ (def-derivative), with ${\phi }^{\mathrm{\prime }}$ integrable on $[c,d]$ (def-darboux-integral). Let $J\subseteq \mathbb{R}$ be order-convex with at least two elements (def-interval) with $\phi [[c,d]]\subseteq J$, and let $f:J\to \mathbb{R}$ be continuous on $J$ (def-continuity-real).

Then $(f\circ \phi ){\phi }^{\mathrm{\prime }}$ is integrable on $[c,d]$ and

${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }},$

the left-hand integral being the oriented one of def-oriented-integral.

Neither injectivity nor monotonicity of $\phi$ is assumed, and that is exactly why the left-hand side is written with oriented limits: $\phi (d)$ may lie below $\phi (c)$, and $\phi$ may return to the same value many times. The proof runs through a primitive of $f$ and the chain rule, and no inverse function is ever formed.

Continuity of $f$ is a hypothesis and cannot be weakened to integrability. With $f$ merely integrable the composite $f\circ \phi$ need not be integrable at all, so the right-hand side need not exist; that is the false statement that weakens it on the companion page.

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.

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 Gaussian integral ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$: ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }\mathrm{.}$

Monotonicity and nonnegative homogeneity of the nonnegative integral: The nonnegative integral is monotone, and for every positive real scalar $c$, $\int cfd\mu =c\int fd\mu$. The zero-scalar case is separate: the integral of the zero function is $0$.

Under Countable Choice, every Borel subset of $\mathbb{R}$ is Lebesgue measurable (Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable). Thus a Borel real-valued function is Lebesgue measurable, since its inverse images of Borel sets are Lebesgue measurable.