PFR

If X has identical distribution to X₀, and X₀ has finite range, then X is almost everywhere equivalent to a random variable of finite range.

A version of independent_copies that guarantees that the copies have FiniteRange if the original variables do.

A version of independent_copies3_nondep that guarantees that the copies have FiniteRange if the original variables do.

A version of independent_copies4_nondep that guarantees that the copies have FiniteRange if the original variables do.

A version of independent_copies' that guarantees that the copies have FiniteRange if the original variables do.

If (Z₁, Z₂, Z₃, Z₄) are independent, so are (Z₁, Z₂, φ Z₃ Z₄) for any measurable φ.

Uniform distributions exist.

The image of a uniform random variable under an injective map is uniform on the image.

Uniform distributions exist, version with a Finite set rather than a Finset and giving a measure space

A "unit test" for the definition of uniform distribution.

If $X$ is uniform w.r.t. $\mu$ on $H$, then $X$ is uniform w.r.t. $\mu$ conditioned by $H'$ on $H' \cap H$.

A random variable is uniform iff its distribution is.

Let $H$ be a subgroup of $G \times G'$. Then there exists a subgroup $H_0$ of $G$, a subgroup $H_1$ of $G'$, and a homomorphism $\phi: G \to G'$ such that $H := \{ (x, \phi(x) + y): x \in H_0, y \in H_1 \}.$ In particular, $|H| = |H_0| |H_1|$.

Let $f: G \to G'$ be a function, and let $S$ denote the set $S := \{ f(x+y)-f(x)-f(y): x,y \in G \}.$ Then there exists a homomorphism $\phi: G \to G'$ such that $|\{f(x) - \phi(x)\}| \leq |S|^{10}.$

If $d[X ;X]=0$, and $x,y \in G$ are such that $P[X=x], P[X=y]>0$, then $x-y \in \mathrm{Sym}[X]$.

If d[X # X] = 0, then X - x₀ is the uniform distribution on the subgroup of G stabilizing the distribution of X, for any x₀ of positive probability.

If $d[X ;X]=0$, then there exists a subgroup $H \leq G$ such that $d[X ;U_H] = 0$.

If $d[X_1;X_2]=0$, then there exists a subgroup $H \leq G$ such that $d[X_1;U_H] = d[X_2;U_H] = 0$. Follows from the preceding claim by the triangle inequality.