[proofplan] We argue by contradiction. A uniformly bounded $P$-Donsker class must be [totally bounded](/page/Totally%20Bounded) in its $L^2(P)$ canonical semimetric, so it is enough to build infinitely many indicators that remain a fixed positive distance apart. Non-atomicity gives a dyadic splitting tree of measurable sets, from which we construct countably many Bernoulli-$1/2$ sets with pairwise symmetric-difference measure $1/2$. Their indicators are therefore pairwise separated by $L^2(P)$ distance $2^{-1/2}$, contradicting [total boundedness](/page/Total%20Boundedness). [/proofplan]

[step:Construct a dyadic tree of measurable sets by repeatedly splitting mass in half]We use the following standard consequence of non-atomicity: if $E\in\mathcal A$ and $0\le t\le P(E)$, then there exists $F\in\mathcal A$ such that $F\subset E$ and $P(F)=t$. For later reference, write $\mathbb N:=\{1,2,\dots\}$, and for sets $A,B\subseteq S$ write $A\triangle B:=(A\setminus B)\cup(B\setminus A)$ for the symmetric difference. Let $\{0,1\}^m$ denote the set of binary strings of length $m\in\mathbb N\cup\{0\}$, with $\{0,1\}^0=\{\varnothing\}$ consisting of the empty string. We construct measurable sets $E_\sigma\in\mathcal A$ for $\sigma\in\bigcup_{m=0}^{\infty}\{0,1\}^m$, such that $E_\varnothing=S$, and for every $\sigma\in\{0,1\}^m$ the sets $E_{\sigma 0}$ and $E_{\sigma 1}$ are disjoint measurable subsets of $E_\sigma$ satisfying \begin{align*} E_\sigma=E_{\sigma 0}\cup E_{\sigma 1}, \end{align*} and \begin{align*} P(E_{\sigma 0})=P(E_{\sigma 1})=2^{-(m+1)}. \end{align*} Indeed, once $E_\sigma$ has been constructed with $P(E_\sigma)=2^{-m}$, the splitting property applied with $t=2^{-(m+1)}$ gives a measurable subset $E_{\sigma 0}\subset E_\sigma$ of measure $2^{-(m+1)}$; set $E_{\sigma 1}:=E_\sigma\setminus E_{\sigma 0}$.[/step]

[guided]The purpose of this step is to manufacture the probability-space analogue of binary digits. We want many sets that behave like independent fair coin flips, and the cleanest way to obtain them from non-atomicity is to split the space in half, then split each half in half, and continue. We use the following standard splitting fact for non-atomic probability measures: if $E\in\mathcal A$ and $0\le t\le P(E)$, then there is a measurable subset $F\subset E$ with $P(F)=t$. This is exactly where non-atomicity is used. It says not merely that no point has positive mass, but that every positive-mass measurable set can be cut to any prescribed smaller mass. Let $\{0,1\}^m$ be the set of binary strings of length $m$, and let $\varnothing$ denote the unique string of length $0$. We define measurable sets $E_\sigma\in\mathcal A$ for $\sigma\in\bigcup_{m=0}^{\infty}\{0,1\}^m$. Start with \begin{align*} E_\varnothing:=S. \end{align*} Since $P(S)=1$, this set has the desired mass $2^0=1$. Suppose now that $\sigma\in\{0,1\}^m$ and that $E_\sigma$ has already been constructed with \begin{align*} P(E_\sigma)=2^{-m}. \end{align*} Apply the non-atomic splitting fact to $E_\sigma$ with \begin{align*} t:=2^{-(m+1)}. \end{align*} Because $0\le 2^{-(m+1)}\le 2^{-m}=P(E_\sigma)$, there exists $E_{\sigma 0}\in\mathcal A$ such that $E_{\sigma 0}\subset E_\sigma$ and \begin{align*} P(E_{\sigma 0})=2^{-(m+1)}. \end{align*} Define the complementary child inside $E_\sigma$ by \begin{align*} E_{\sigma 1}:=E_\sigma\setminus E_{\sigma 0}. \end{align*} Then $E_{\sigma 1}\in\mathcal A$, the sets $E_{\sigma 0}$ and $E_{\sigma 1}$ are disjoint, and their union is $E_\sigma$. By finite additivity of the probability measure $P$, \begin{align*} P(E_{\sigma 1})=P(E_\sigma)-P(E_{\sigma 0})=2^{-m}-2^{-(m+1)}=2^{-(m+1)}. \end{align*} This recursively constructs a full dyadic tree of measurable sets.[/guided]

[step:Define countably many half-mass sets with pairwise symmetric difference of mass one half] For each $j\in\{1,2,\dots\}$, define the measurable set \begin{align*} A_j:=\bigcup_{\substack{\sigma\in\{0, 1\}^j, \sigma_j=1}}E_\sigma, \end{align*} where $\sigma_j$ denotes the $j$-th coordinate of the binary string $\sigma$. Since the sets $(E_\sigma)_{\sigma\in\{0,1\}^j}$ form a measurable partition of $S$ and each has $P$-measure $2^{-j}$, there are $2^{j-1}$ strings of length $j$ whose $j$-th coordinate is $1$. Hence \begin{align*} P(A_j)=2^{j-1}2^{-j}=\frac{1}{2}. \end{align*} Let $i,j\in\{1,2,\dots\}$ with $i<j$. At level $j$, the set $A_i\cap A_j$ is the union of those $E_\sigma$ with $\sigma\in\{0,1\}^j$, $\sigma_i=1$, and $\sigma_j=1$. There are $2^{j-2}$ such strings, so \begin{align*} P(A_i\cap A_j)=2^{j-2}2^{-j}=\frac{1}{4}. \end{align*} Therefore \begin{align*} P(A_i\triangle A_j)=P(A_i)+P(A_j)-2P(A_i\cap A_j)=\frac{1}{2}. \end{align*} [/step]

[step:Convert symmetric-difference separation into $L^2(P)$ separation of indicators] For each $j\in\{1,2,\dots\}$, define the measurable indicator function \begin{align*} f_j:S \to \mathbb R,\qquad x \mapsto \mathbb 1_{A_j}(x). \end{align*} Then $f_j\in\mathcal I_{\mathcal A}$ for every $j\in\{1,2,\dots\}$. For $i\ne j$, the pointwise identity \begin{align*} |\mathbb 1_{A_i}-\mathbb 1_{A_j}|^2=\mathbb 1_{A_i\triangle A_j} \end{align*} gives \begin{align*} \|f_i-f_j\|_{L^2(P)}^2=\int_S |f_i(x)-f_j(x)|^2\,dP(x)=P(A_i\triangle A_j)=\frac{1}{2}. \end{align*} Thus \begin{align*} \|f_i-f_j\|_{L^2(P)}=\frac{1}{\sqrt 2} \end{align*} for every distinct $i,j\in\mathbb N$. [/step]

[step:Contradict the total boundedness necessary for a Donsker class] Assume, toward a contradiction, that $\mathcal I_{\mathcal A}$ is $P$-Donsker. The class is uniformly bounded by $1$, since \begin{align*} |\mathbb 1_A(x)|\le 1 \end{align*} for every $A\in\mathcal A$ and every $x\in S$. For any [measurable function](/page/Measurable%20Function) $h:S\to\mathbb R$ satisfying $\int_S |h(x)|\,dP(x)<\infty$, write \begin{align*} P h:=\int_S h(x)\,dP(x). \end{align*} By [citetheorem:9862], the effective index space of a uniformly bounded $P$-Donsker class is totally bounded in the canonical semimetric \begin{align*} d_P(f,g)^2:=P(f-g)^2-(Pf-Pg)^2. \end{align*} For the indicators $f_i=\mathbb 1_{A_i}$ and $f_j=\mathbb 1_{A_j}$, we have \begin{align*} P f_i=P f_j=\frac{1}{2}. \end{align*} Hence \begin{align*} d_P(f_i,f_j)^2=P(f_i-f_j)^2=\frac{1}{2} \end{align*} whenever $i\ne j$. Therefore the sequence $(f_j)_{j\ge 1}$ is pairwise separated by distance $2^{-1/2}$ in $d_P$. A totally bounded semimetric space cannot contain infinitely many points separated by a fixed positive distance: if $\varepsilon:=2^{-3/2}$, then each ball of radius $\varepsilon$ contains at most one point of a $2^{-1/2}$-separated family, while total boundedness would cover the space by finitely many such balls. This contradiction shows that $\mathcal I_{\mathcal A}$ is not $P$-Donsker. [/step]