Sequence compactness in L1 by finite cell averages

For $\delta >0$, tile ${\mathbb{R}}^n$ by the half-open cubes $Q=\delta (z+[0,1{)}^n)$, $z\in {\mathbb{Z}}^n$, and let $A_{\delta }v$ equal ${\delta }^{-n}{\int }_{Q}v$ on $Q$. The triangle inequality, followed by [F1] and the substitution $y=x+h$ using [F2], gives $\mathrm{\parallel }v-A_{\delta }v{\mathrm{\parallel }}_1\le {\delta }^{-n}{\sum }_Q{\int }_Q{\int }_Q\mathrm{\mid }v(x)-v(y)\mathrm{\mid }dydx\le {\delta }^{-n}{\int }_{[-\delta ,\delta {]}^n}\mathrm{\parallel }v(\cdot +h)-v{\mathrm{\parallel }}_{1}dh$. Thus, writing $\omega (r)={\sup }_{k,\mathrm{\mid }h\mathrm{\mid }\le r}\mathrm{\parallel }{v}_k(\cdot +h)-v_k{\mathrm{\parallel }}_1$, one has ${\sup }_k\mathrm{\parallel }{v}_k-A_{\delta }{v}_k{\mathrm{\parallel }}_1\le 2^n\omega (\sqrt{n}\delta )\to 0$ as $\delta \downarrow 0$.

For each $m\ge 1$, only finitely many cubes of side $2^{-m}$ meet a fixed ball containing the common supports; all other cell averages vanish. Enumerate all these cells at all levels by first increasing $m$ and then lexicographic order of their integer coordinates. Each cell-average coefficient $a_j(k)$ is a bounded real sequence, with bound $M$ divided by that cell's volume. Set $S_0=\mathbb{N}$ and initialize $k_0=0$. Recursively, at stage $m\ge 1$, partition the bounded ranges of the first $m$ coefficients into finitely many half-open dyadic bins of width $2^{-m}$, adding an endpoint bin if necessary. Among the finitely many vector bins containing infinitely many indices in $S_{m-1}$, take the lexicographically first; let $S_m$ be exactly that infinite set of indices. The finite pigeonhole principle guarantees existence and the specified order makes the rule unique. Then let $k_m$ be the least member of $S_m$ larger than $k_{m-1}$; such a member exists because $S_m$ is infinite. This deterministic successor rule and initial pair $(S_0,k_0)$ define the sequence recursively by [F3], with no dependent-choice instance. For fixed $j$ and $m\ge j$, all selected indices $k_q$ with $q\ge m$ lie in $S_m$, so their $j$th coefficients differ by at most $2^{-m}$. Consequently every coefficient converges and, at each fixed cell level $\mathrm{\ell }$, $(A_{2^{-\mathrm{\ell }}}{v}_{k_m}{)}_m$ is Cauchy in $L^1$, being a finite weighted sum of these coefficients.

Given $\epsilon >0$, choose a cell level $\mathrm{\ell }$ for which $2^n\omega (\sqrt{n}{2}^{-\mathrm{\ell }})<\epsilon \mathrm{/}3$, using step 1.1. For sufficiently large $a,b$, step 1.2 gives $\mathrm{\parallel }{A}_{2^{-\mathrm{\ell }}}{v}_{k_a}-A_{2^{-\mathrm{\ell }}}{v}_{k_b}{\mathrm{\parallel }}_1<\epsilon \mathrm{/}3$. The two approximation errors from step 1.1 then show $\mathrm{\parallel }{v}_{k_a}-v_{k_b}{\mathrm{\parallel }}_1<\epsilon$. Thus the subsequence is Cauchy.

For completeness at exactly the stated axiom strength, choose representatives of this countable sequence by [A1]. Select a further subsequence $g_j$ with $\mathrm{\parallel }{g}_{j+1}-g_j{\mathrm{\parallel }}_1<2^{-j}$, always choosing the least sufficiently late index, which exists by the Cauchy property. By [F1], ${\sum }_j\mathrm{\mid }{g}_{j+1}-g_j\mathrm{\mid }$ is integrable and finite almost everywhere. Hence $g_j$ has a measurable almost-everywhere limit $g$; moreover $\mathrm{\mid }g\mathrm{\mid }\le \mathrm{\mid }{g}_1\mathrm{\mid }+{\sum }_j\mathrm{\mid }{g}_{j+1}-g_j\mathrm{\mid }$, so $g\in L^1$, and $\mathrm{\parallel }g-g_j{\mathrm{\parallel }}_1\le {\sum }_{i\ge j}{2}^{-i}\to 0$. The original Cauchy subsequence also converges to this $L^1$ class by the triangle inequality. This proves the assertion using only [A1].