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Let $X$ be a set of $N>0$ elements (with the counting measure) and consider a family of measurable functions $f_i:X\to [0,1]$, for $i\in \mathbb N$.

Any function $f_i$ can be seen as a point in the hypercube $[0,1]^N$, therefore by using a mesh of sub-hypercubes (all of the same diameter) it is possible to find finitely many points $g_j\in[0,1]^N$ such that: for any $i \in\mathbb N$ there exists a $j$ with the property that: $$ 0<g_j-f_i<\delta $$

where $\delta$ is a fixed number depending on the mesh. Indeed it is enough to take the corners of the small hypercubes. Clearly $\delta$ can be made arbitrarily small by refining the mesh, i.e. increasing the number of $g_j$.

In other words we can approximate the functions $f_i$ with finitely many functions $g_j$ and with arbitrary precision. The key assumption is that $X$ is a finite set.


Now assume that $X$ is a measure space (of finite measure), do we have a similar approximation result for a family of measurable functions $f_i:X\to [0,1]$? This problem seems very tricky but I wonder if there is some known result (even partial), in this direction. probably one needs to put some more properties on the $f_i$

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  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Commented Aug 3, 2023 at 21:54

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$\newcommand\de\delta\newcommand\N{\mathbb N}$Such an approximation is impossible in such generality.

Indeed, let $f_i:=(1+r_i)/2$ for $i\in\mathbb N$, where $(r_i)_{i\in\mathbb N}$ is the Rademacher system (of real-valued functions on $X:=[0,1]$). Then the $f_i$'s are measurable functions from $X$ to $[0,1]$ and $\|f_i-f_j\|=1$ for all distinct natural $i$ and $j$, where $\|\cdot\|$ is the uniform norm. So, for any $\de\in(0,1/2)$ there is no finite set $G$ of real-valued functions on $[0,1]$ such that for each $i\in\N$ there is some $g_i\in G$ such that $\|f_i-g_i\|\le\de$ (because then, by the norm inequality, the $g_i$'s would have to be pairwise distinct, and hence the set $G$ would have to be infinite).

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  • $\begingroup$ Are you aware of any conditions that one can impose in order to achieve the approximation? $\endgroup$
    – manifold
    Commented Aug 6, 2023 at 2:36
  • $\begingroup$ I think being finite dimensional or something close to it is essential. The key phrase here is "totally bounded". E.g. the infinite-dimensional ellipsoid $\{(x_n)\in\ell^2\colon\sum_n n|x_n|^2\le1\}$ is totally bounded. $\endgroup$ Commented Aug 6, 2023 at 12:55

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