Boundedness of first moments alone does not give uniform integrability

A uniformly integrable family: Let $(X,\mathcal{A},\mu )$ be a measure space. A family $\mathcal{F}\subseteq L^1(\mu )$ of integrable real-valued functions is uniformly integrable when ${\sup }_{f\in \mathcal{F}}{\int }_{\{\mathrm{\mid }f\mathrm{\mid }>M\}}\mathrm{\mid }f\mathrm{\mid }d\mu \to 0\text{as }M\to \mathrm{\infty }\mathrm{.}$

Equivalently, for every $\epsilon >0$ there is $M>0$ such that $f\in \mathcal{F}⟹{\int }_{\{\mathrm{\mid }f\mathrm{\mid }>M\}}\mathrm{\mid }f\mathrm{\mid }d\mu <\epsilon \mathrm{.}$

This page adopts the tail-integral definition. On finite measure spaces it is equivalent to $L^1$-boundedness plus uniform absolute continuity, proved later on this page.

Weak convergence of borel probability measures: For Borel probability measures ${\mu }_n,\mu$ on a metric space S, write ${\mu }_n\Rightarrow \mu$ if $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function f on S. Continuity is def-metric-continuity. Such f is Borel measurable (inverse images of open sets are open) and $\int \mathrm{\mid }f\mathrm{\mid }d\mu \le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mu (S)<\mathrm{\infty }$, so the integrals are finite in def-integrable-real-and-complex-functions-and-their-integrals. No completeness or coupling is required.

Markov's inequality for random variables: If $X:\mathrm{\Omega }\to [0,+\mathrm{\infty }]$ is a nonnegative random variable on a probability space and $a>0$, then $\mathbb{P}(X\ge a)\le \frac{\mathbb{E}[X]}{a}\mathrm{.}$

Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: Let $n\in \mathbb{N}$ with $n\ge 1$, let ${\mathbb{R}}^n$ be the set of functions $n\to \mathbb{R}$ and let $d_2$ be the Euclidean metric on it (lem-metrics-on-rn). Then:

1. Closed boxes are compact. For reals $a_k\le b_k$ $(k<n)$ the box $Q=\{x\in {\mathbb{R}}^n:a_k\le x_k\le b_k\text{ for every }k<n\}$ is a compact subset of $({\mathbb{R}}^n,d_2)$ (def-metric-compactness).
2. Heine-Borel. A subset $K\subseteq {\mathbb{R}}^n$ is a compact subset of $({\mathbb{R}}^n,d_2)$ if and only if $K$ is closed in ${\mathbb{R}}^n$ (def-metric-topology) and bounded (def-metric-bounded-diameter).
3. The real line. A subset $K\subseteq \mathbb{R}$ is a compact subset of $(\mathbb{R},d_{\mathbb{R}})$, the usual metric $d_{\mathbb{R}}(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ (lem-real-line-is-a-metric-space), if and only if $K$ is closed in $\mathbb{R}$ and bounded.

No choice principle is used. The bisection below halves one coordinate at a time and takes the left half whenever the left half still fails to be finitely covered, the right half otherwise: a rule with two outcomes, decided by a property of the box, not a selection. That is the whole reason the theorem is available in ZF, while the general "complete and totally bounded implies compact" (thm-complete-and-totally-bounded-implies-compact) is not.

The hypothesis $n\ge 1$ is inherited from lem-metrics-on-rn, which defines ${\mathbb{R}}^n$ and its metrics only there; the last remark below records what happens at $n=0$.