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One may find a counting measure or trivial probability measure on $L^p$. However, is there any meaningful measure constructed on $L^p$, so that one may try to find a size of subsets of $L^p$? The same question goes for the $W^{k,p}(\Omega)$ spaces as well.

My attempt: $L^p$ (constructed on any measure space) is a topological vector space so we can construct Borel $\sigma$-algebra. We can think of next measure for $r$-balls: $$\mu_0(B_r(f))=r \quad(\text{or }r^2,r^3,e^r,\text{etc})$$ where $$B_r(f)=\{g\in L^p:\|f-g\|_{L^p}<r\}.$$ Then $\mu_0$ can be extended to a pre-measure on the ring $$\mathcal{B}:=\bigg\{\bigcup_{i\in I}B_{r_i}(f_i):r_i\geq 0,f_i\in L^p,\text{$B_{r_i}(f_i)$ are pairwise disjoint},\,\text{$I$ any index set}\bigg\}.$$ So we can use Carathéodory's extension theorem to find a measure $\mu$ on Borel $\sigma$-algebra on $L^p$. Is there any study on such $\mu$? Here I just needed that $L^p$ is a topological vector space. So it seems quite a general result, thus I suspect there is already a name for it, but I could not find any.

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    $\begingroup$ I think the fact that there's no intuitively correct reason suggests that either we shouldn't be trying to measure infinite-dimensional spaces, or that we need to use some extra structure other than the bare normed structure. // I meant to say not that we couldn't measure every subset of $L^p$ but that the pre-measure didn't measure $L^p$ itself, but I see, on checking my definitions, that that doesn't seem to be part of the definition, so it's of no consequence. (One still needs some sort of requirement to guarantee that $L^p$ can be covered by countably many balls, but that's often true.) $\endgroup$
    – LSpice
    Nov 19 '20 at 22:44
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    $\begingroup$ The measure you're trying to construct would be translation invariant, and it's a famous and not-too-difficult fact that there can be no nontrivial translation invariant Borel measure on an infinite-dimensional Banach space. This is often summarized as "there is no infinite-dimensional Lebesgue measure". There are lots of interesting measures without this property; for instance there is a deep theory of Gaussian probability measures on Banach spaces. But no one of them is truly "canonical". $\endgroup$ Nov 19 '20 at 23:40
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    $\begingroup$ For example, a one-dimensional Brownian motion run until time 1 produces a random continuous path, which in particular is in $L^p([0,1])$ for any $p$. So this induces a (Gaussian) probability measure on $L^p$; the pushforward of the Wiener measure on $C([0,1])$ under the inclusion map $C([0,1]) \hookrightarrow L^p([0,1])$ $\endgroup$ Nov 19 '20 at 23:42
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    $\begingroup$ @JingeonAn: What's shown in the proof is that, for instance, each ball of radius 1 contains infinitely many disjoint balls of radius 1/4. So your "premeasure" isn't a premeasure because it isn't monotone. Anyway your collection $\mathcal{B}$ isn't a ring because it isn't closed under relative complement (take one ball and remove a smaller ball from its inside, the resulting set is not a disjoint union of balls, e.g. by connectedness). $\endgroup$ Nov 20 '20 at 0:02
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    $\begingroup$ @JingeonAn: Well, if $\mathcal{B}$ were a ring, then monotonicity of $\mu$ would follow from finite additivity. $\endgroup$ Nov 20 '20 at 0:09

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