= Dudley entropy integral
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Let $(X_t)_{t\in T}$ be a centered <Gaussian process> with a <separable stochastic process>[separable] version: its relevant suprema are captured by a countable dense index subset. Give $T$ the <canonical pseudometric of a Gaussian process> $d(s,t)=\|X_s-X_t\|_2$, and assume it is <totally bounded>. With <metric covering number> $N(\varepsilon,T,d)$ and diameter $D$, the <Dudley entropy integral> bounds
$$
\mathbb E\sup_{t\in T}X_t\leq C\int_0^D\sqrt{\log N(\varepsilon,T,d)}\,d\varepsilon
$$
for a universal constant $C$. Finiteness of the <integral> gives a finite expected <supremum>. For an absolute <supremum>, include an anchor: $\sup_t|X_t|\leq|X_{t_0}|+\sup_t|X_t-X_{t_0}|$. The anchor term cannot generally be omitted.
The proof builds successively finer nets. Write each increment as a telescoping sum between neighboring net scales, bound each finite Gaussian maximum by its scale times the square root of the logarithmic net size, and sum over scales. This becomes the displayed entropy <integral>.<Probability in High Dimension by Ramon van Handel>[Theorem 5.24, chaining bound]{ref}
The version assumption is substantive. If $U$ is uniform on $[0,1]$ and $X_t=\mathbf1_{\{U=t\}}$, every finite <vector> is zero <almost surely> and is therefore a degenerate Gaussian <vector>; the canonical <pseudometric> is identically zero. Yet $\sup_tX_t=1$ for every outcome. Its all-zero separable modification has <supremum> zero. Finite-dimensional laws alone do not control the <supremum> of an arbitrary uncountable version.
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