User:Yoni

Me
I'm Yoni.

Pages I've contributed to

 * Merged Weyl's criterion into Equidistributed sequence and improved the latter (before, after)
 * More to come!

Fubini's theorem and Tonelli's theorem
Let X, Y be measure spaces with measures μ, ν respectively.

Let $$f:X \times Y \to \mathbb{R} \cup \left\{ +\infty, -\infty \right\}$$ be a measurable function.

Then it is true that $$ \int_X \left( \int_Y f(x,y)\,\mathrm{d}\nu(y) \right)\,\mathrm{d}\mu(x) = \int_Y \left( \int_X f(x,y)\,\mathrm{d}\mu(x) \right)\,\mathrm{d}\nu(y) = \int_{X \times Y} f(x,y)\,\mathrm{d}\mu\times\nu(x,y) $$

provided one of the following criteria:
 * 1) (Fubini's theorem) The spaces X, Y are complete (all null sets are measurable), and $$f \in L^1\left(\mu \times \nu\right)$$.
 * 2) (Tonelli's theorem) The spaces X, Y are σ-finite (a countable union of finite-measure sets)*, and f ≥ 0.

(*) For probability spaces this is automatic.

Convergence of integrals
Let Ω be a measure space with a measure μ.

Let fn : Ω → ℝ be a sequence of measurable functions that converges pointwise (everywhere, or μ-almost everywhere if μ is a complete measure) to a function f : Ω → ℝ.

Then it is true that $$\int_\Omega f_n\,d\mu \to \int_\Omega f\,d\mu$$ provided one of the following criteria:
 * 1) (Monotone convergence theorem) $$0 \le f_1 \le f_2 \le \ldots$$ μ-almost everywhere in Ω.  Note: If additionally $$f \in L^1(\mu)$$ then $$f_n \to f$$ in L1(μ) by Scheffé’s lemma.
 * 2) (Dominated convergence theorem) $$\left|f_n\right| \le g$$ for some $$g \in L^1\left(\mu\right)$$ (everywhere, or μ-almost everywhere if μ is a complete measure).  Note: This also gives us $$f_n \to f$$ in L1(μ), and $$||f_n||_{L^1(\mu)} \uparrow ||f||_{L^1(\mu)} \le ||g||_{L^1(\mu)}$$.
 * 3) (Bounded convergence theorem) $$\mu(\Omega) < \infty$$ and $$\left|f_n\right| \le M$$.  Note: This also gives us $$f_n \to f$$ in L1(μ), and $$||f_n||_{L^1(\mu)} \uparrow ||f||_{L^1(\mu)} \le M \mu\left(\Omega\right)$$.

Corollary: Differentiation under the integral sign
Let $$F(x) := \int_\Omega f(x,\omega)\,d\mu(\omega)$$, wherein $$x \in \mathbb{R}$$, and if ω is held constant, for all ω (or μ-almost all ω if μ is a complete measure), f is differentiable in x. Suppose F is defined in a neighborhood of 0.

Then it is true that $$F'(0) = \int_\Omega \frac {\partial f} {\partial x} \left(0, \omega\right)\,d\mu(\omega)$$ provided one of the following criteria:
 * 1) $$\frac {\partial f}{\partial x}(0, \omega) \in L^1(\mu)$$.
 * 2) $$\mu(\Omega) < \infty$$ and $$\left|\frac {\partial f}{\partial x}(0, \omega)\right| \le M$$.

A smooth transition from 0 to nonzero

 * $$\varphi(x) = \begin{cases}e^{-\frac 1 x} & \mbox{if }x > 0, \\ 0 &\mbox{if }x \le 0\end{cases}$$

A bump function - a smooth function with compact support

 * $$\psi(x) = \varphi\left(2\left(1+x\right)\right)\varphi\left(2\left(1-x\right)\right) = \begin{cases}e^{-\frac 1 {1-x^2}} & \mbox{if }|x| < 1, \\ 0 &\mbox{if }|x| \ge 1\end{cases}$$

A smooth transition from 0 to 1
This is designed as a partition of unity.
 * $$\eta(x) = \frac {\varphi(x)} {\varphi(x) + \varphi(1-x)} = \begin{cases}0 & \mbox{if } x \le 0, \\ \left(1 + \exp \left({\frac {1 - 2x} {x(1-x)}}\right)\right)^{-1} & \mbox{if } 0 < x < 1, \\ 1 & \mbox{if } x > 1\end{cases}$$

Good-to-know changes of variables
List of canonical coordinate transformations

Let σd-1 be the uniform probability measure on the d-1-dimensional unit sphere and let κd be the volume of the d-dimensional unit ball (so that dκd is the surface area of the sphere). Then:
 * $$\frac 1 {d \kappa_d} \int_{x \in \mathbb{R}^d, R_1 < |x| < R_2} f(x)\, dx = \int_{R_1}^{R_2} r^{d-1} \left(\int_{\zeta\in\mathbb{R}^d, \left|\zeta\right|=1} f(r\zeta)\,d\sigma_{d-1}\left(\zeta\right)\right)\,dr$$

Corollary: If f is radial, that is: f(x) = f(|x|)'', then:
 * $$\frac 1 {d \kappa_d} \int_{x \in \mathbb{R}^d, R_1 < |x| < R_2} f(x)\, dx = \int_{R_1}^{R_2} r^{d-1} f(r)\,dr$$

Integral convergence
This may be proven using the previously-mentioned change of variables.


 * Supposing ε > 0, we have $$\int_{x\in\mathbb{R}^d, |x|<1} \frac 1 {|x|^{d-\varepsilon}}\,dx = \frac {d\kappa_d} \varepsilon$$

In particular, $$\int_{x\in\mathbb{R}^d, |x|<1} \frac 1 {|x|^t}\,dx < \infty \Leftrightarrow t < d$$.

Basics
Let (Ω, P) be a probability space.


 * A real-valued random variable is a Borel-measurable $$X: \Omega \to \mathbb{R}$$.
 * The expected value of X is $$\operatorname{E} [X] = \int_\Omega X(\omega) \, \mathrm{d}P(\omega)$$.

Euclidean balls
Denote by κd the volume of the d-dimensional unit ball. Then


 * $$\kappa_d = \frac {\pi^{d/2}} {\Gamma\left(\frac d 2 + 1\right)} = \begin{cases}\dfrac{\pi^k}{k!}& d=2k \\ \\ \dfrac{2^{k+1} \pi^k }{\left(2k+1\right)!!} & d=2k+1 \end{cases} = 2, \pi, \frac 4 3 \pi, \frac 1 2 \pi^2, \frac 8 {15} \pi^2, \frac 1 6 \pi^3, \frac {16} {105} \pi^3, \ldots$$

Denote by sd-1 the surface area of the d-1-dimensional unit sphere (the boundary of the d-dimensional unit ball). Then


 * $$s_{d-1} = d \kappa_d$$

Proof.

Let Bd(r) be the d-dimensional Euclidean ball centered at the origin with radius r. Then the following inclusion is true:


 * $$\left[-\frac r {\sqrt d}, \frac r {\sqrt d}\right]^d \subset B^d \left(r\right) \subset \left[-r, r\right]^d$$

(TODO: The more general result with Hölder's inequality, inclusions of Lp spaces, etc.)