There is no Lebesgue measure in infinite dimensions—this slogan is familiar to every student interested in analysis. One possible, precise statement of this result may be as follows: if $X$ is an infinite dimensional Banach space there is no Borel, translation invariant measure with the property that each point has a neighborhood with finite measure. However people tried some constructions weakening some of the assumptions. One approach is described in Gill, Pantsulaia, and Zachary - Constructive analysis in infinitely many variables. The authors claim that they were able to construct an analogue of Lebesgue measure on the space of all sequences $\mathbb{R}^{\infty}$ (of course this space is not Banach). I wonder whether there is someone familiar with this paper: unfortunately I got lost (and I'm not quite sure whether the authors put everything correctly).
Sketch of the strategy: Denote $I_n=\prod_{k=n+1}^{\infty}[-\frac{1}{2},\frac{1}{2}]$: they use the symmetric closed interval of length $1$ to "stabilize" the construction: they split the space $\mathbb{R}^{\infty}$ into two disjoint pieces $X_1=\bigcup_{k=1}\mathbb{R}^{k}_I$ where $\mathbb{R}^k_I=\mathbb{R}^k \times I_{k}$ and $X_2=\mathbb{R}^{\infty} \setminus X_1$.
On $X_1$ they consider the product topology, on $X_2$ they consider discrete topology (sic!) and finally they consider the topology of disjoint union on $\mathbb{R}^{\infty}$. Then they define an analogue of Lebesgue measure, denoted by $\lambda_{\infty}$ in several steps.
First they define (on page 113) this measure for sets of the form $A \times I_1$ (where $A$ is Borel in $\mathbb{R}$ with finite measure) as $\lambda(A)$. I'm guessing that they proceed similarly for $A \times I_n$ where $A \subset \mathbb{R}^n$ (they didn't state it explicitly but seem to be using this definition in the proof of theorem 2.5). Note that it is possible that $A$ is of measure $0$.
Then they define this measure on the class $\Delta_0$ consisting of sets of the form $K \times I_n$ where $K$ is compact with nonzero, finite (which is automatic) measure, then for the class $\Delta$ of finite, (almost) disjoint sums of sets from $\Delta_0$.
After that they define $\lambda_{\infty}$ for open sets as $\lim_{N \to \infty} \sup \lambda_{\infty}(P_N)$ where $P_N \subset G$, $P_N \in \Delta$ (I guess that they meant just supremum together with convention that it is $0$ if there is no $P_N \subset G$: which occurs quite often since $X_2$ was discrete).
Then they define $\lambda_{\infty}$ for compact sets (as I understand this was not necessarily covered before since for example $[0,\frac{1}{2}]^{\infty}$ is compact but does not belong to $\Delta$).
Furthermore having defined this measure for compact and for open sets they define outer and inner measures as infimum over open sets/supremum over compact sets. The inner measure is defined only for sets with finite outer measure. They then call a set bounded measurable if these two measures coincide. Finally the set is called measurable it its intersection with any bounded measurable set is again bounded measurable-then $\lambda_{\infty}(A)$ is defined as the supremum of $\lambda_{\infty}(A \cap M)$ over all bounded measurable sets $M$.
Question How to understand this measure well enough to be able to compute measures of simple sets? For example I would be happy if I could compute $\lambda_{\infty}(J_1 \times J_2 \times J_3 \times \dotsb)$ where each $J_k$ is either an interval (could be closed, could be open) or the real line or a point or a half line.
There are plenty of possibilities: some of them seem understandable for me, some are not. Here are some examples:
- $\lambda_{\infty}([a_1,b_1] \times \dots \times [a_n,b_n] \times I_n)=(b_1-a_1)\cdot ... \cdot (b_n-a_n)$—this set belongs to $\Delta_0$,
- $\lambda_{\infty}([-\frac{1}{2},\frac{1}{2}]^{\infty})=1$—this set is not open but is compact and $(-\frac12-\delta,\frac12+\delta) \times I_1$ is open and its measure is $1+2\delta$ which could be arbitrarly close to $1$
- $\lambda_{\infty}((-\frac{1}{2},\frac{1}{2})^{\infty})=1$—this set is not open but one can take $\delta_n \to 0^+$ which converges quickly enough to ensure that the open set $\prod_n (-\frac12-\delta_n,\frac12+\delta_n)$ has measure arbitrarily close to $1$: similarly one can take compact $\prod_n [-\frac12+\delta_n,\frac12-\delta_n]$ with measure arbitrary close to 1 (which by the way requires considering again open sets, this time containing this compact set)
- $\lambda_{\infty} (\prod_n(-\frac12-a_n,\frac12+a_n))=\prod_n(1-2a_n)$ (here $a_n$ are positive)—this set is open, as a union of open sets $\prod_k^n(-\frac12-a_k,\frac12+a_k) \times I_n$.
Now let me list some sets for which I'm not sure how to proceed:
- $A=[1,x]^{\infty}$ (where $x>1$)—this set is contained in $X_2$ therefore it is open but there is no $P_N \in \Delta$ such that $P_N \subset A$ so its measure should be $0$.
- $A=\mathbb{R} \times \{0\} \times I_3$: on the one hand if $\lambda_{\infty}$ has to be consistent with finite dimensional Lebesgue measure then it should have measure $0$: on the other hand any open set containing $A$ would have infinite measure while any compact set contained in $A$ would have measure $0$—so we get into trouble with unbounded measurable sets.
- $A=\mathbb{R} \times \{0\} \times \{0\} \times \dotsb$: intuition tells me that this set should have measure $0$. However tracking all the definition leads to the same trouble as in the example above.
I'm not sure whether this question is appropriate for this site for it asks about computing measure of very simple sets. However even if this measure should serve as an analogue of Lebesgue measure its definition is quite involved—so I'm looking for simple formulas.
EDIT: I've adressed the comment below. Let me also mention that the authors state that they are using the following convention: $0 \times \infty=0$ but $0 \times \infty^{\infty}=\infty$.
