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I am trying to understand the proof provided on wikipedia that there is no Lebesgue measure on infinite-dimensional, separable Banach spaces. This proof starts by saying that if $(X,\|\cdot\|)$ is such a Banach space, and $\delta > 0$, and $B$ is an open ball of radius $\delta$, then there is a sequence $(B_n)^\infty_{n=1}$ in $B$ of pairwise disjoint open balls of radius $\delta/4$. Why is that?

For convenience, suppose that $B$ is centered around $0$. I understand that if $X$ is the Hilbert space $\ell^2$ of square summable sequences, then it has a basis $(e_n)^\infty_{n=1}$ defined by \begin{equation*} \begin{aligned} e_1&=(1,0,0,0,\ldots)\\ e_2&=(0,1,0,0,\ldots)\\ e_3&=(0,0,1,0,\ldots)\\ \text{etc.} \end{aligned} \end{equation*} and one can obtain a sequence $(B_n)^\infty_{n=1}$ of pairwise disjoint open balls in $B$, each of the same strictly positive radius, by centering each $B_n$ around a certain multiple of $e_n$.

But how to show this for general infinite-dimensional, separable Banach spaces?

Bart
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  • Perhaps construct a basis with your Hilbert space example as a guide. You have a countable basis by assumption, and it looks to me like angle doesn't matter here as much as distance. – ZxJx Dec 01 '19 at 21:33

1 Answers1

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There is no translation invariant measure $\mu$, such that $0<\mu\big(B_r(x)\big)<\infty$, for every ball, with $r>0$.

In the case of $\ell^2(\mathbb N)$, observe that $$ B_{½}(e_i)\cap B_{½}(e_j)=\varnothing, \quad i,j\in\mathbb N,\,\,i\ne j, $$ and $$ \bigcup_{i\in\mathbb N}B_{½}(e_i)\subset B_2(0), $$ and hence, if such a measure existed, then we would have that $$ \mu\big(B_{½}(e_i)\big)=\mu\big(B_{½}(e_j)\big), \quad \text{for all $i,j\in\mathbb N$,} $$ and $$ \infty=\sum_{i\in\mathbb N}\mu\big(B_{½}(e_i)\big)\le \mu\big(B_{2}(0)\big) $$

Yiorgos S. Smyrlis
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