Completeness of the weighted continuous trajectory space

Let $(X_n)$ be Cauchy in $d_{\mathcal{X}}$. By [A1], choose measurable representatives and, for each representative, a measurable conull set on which its paths are continuous. Countable Choice gives these sets simultaneously; their intersection $C\in \mathcal{A}$ is conull. For a fixed countable dense set $D\subseteq [0,T]$, continuity on $C$ gives ${\sup }_{0\le t\le T}\mathrm{\mid }{X}_m(t,z)-X_n(t,z)\mathrm{\mid }={\sup }_{q\in D}\mathrm{\mid }{X}_m(q,z)-X_n(q,z)\mathrm{\mid }(z\in C)\mathrm{.}$ Each fixed-time section of each coordinate is measurable by [F6], since coordinate projections are continuous, and the right side is measurable by [F5]. Define it to be zero on $C^c$; this is a measurable representative of the time-supremum difference, as prescribed in the Statement. Changing the conull set or the representative does not change its integral. Symmetry and the triangle inequality follow pointwise on a common conull set, and a zero distance forces the paths to agree almost everywhere: for a nonnegative measurable supremum $g$ with zero integral, $\mu \{g>1\mathrm{/}j\}\le j\int g=0$ for every $j\ge 1$. It is integrable because on $C$ it is bounded by the sum of the two given integrable path suprema.

Choose increasing indices $n_k$ successively as the least indices large enough that $\mathrm{\parallel }{X}_m-X_n{\mathrm{\parallel }}_{\mathcal{X}}\le 2^{-k}$ whenever $m,n\ge n_k$. Put $g_k(z)={\sup }_t\mathrm{\mid }{X}_{n_{k+1}}(t,z)-X_{n_k}(t,z)\mathrm{\mid }$ on $C$ and $g_k(z)=0$ off $C$. By step 1.1 each $g_k$ is measurable and ${\int }_E{g}_{k}d\mu \le 2^{-k}$. Thus $G={\sum }_{k=1}^{\mathrm{\infty }}{g}_k$ is measurable by [F5], and Tonelli gives ${\int }_{E}Gd\mu \le {\sum }_{k=1}^{\mathrm{\infty }}{2}^{-k}<\mathrm{\infty }$. By [F7], $A:=C\cap \{G<\mathrm{\infty }\}$ is measurable and conull. For every $z\in A$, the paths $X_{n_k}(\cdot ,z)$ are uniformly Cauchy on $[0,T]$.

Replace each selected representative by ${\tilde{X}}_{n_k}(t,z)=X_{n_k}(t,z)$ on $A$ and $0$ outside $A$. These maps remain jointly measurable and represent the same equivalence classes. By [F1], for each $z\in A$ their uniform limit is a continuous path $X(\cdot ,z)$; set $X(\cdot ,z)=0$ off $A$. The coordinate projections of ${\tilde{X}}_{n_k}$ are measurable, so their pointwise limits are measurable by [F5]; [F6] makes the resulting map into ${\mathbb{R}}^d$ jointly measurable. Its time-supremum function is measurable because its paths are continuous and the supremum equals the countable supremum over $D$ of measurable sections. On $A$, ${\sup }_t\mathrm{\mid }{X}_t-{\tilde{X}}_{n_k,t}\mathrm{\mid }\le {\sum }_{j\ge k}{g}_j$; off $A$ this difference is zero. Hence the integrable series tail bounds $\mathrm{\parallel }X-X_{n_k}{\mathrm{\parallel }}_{\mathcal{X}}$ and tends to zero. Since $X_{n_k}$ has finite norm, [F4] gives $\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathcal{X}}<\mathrm{\infty }$.

Since the original sequence is Cauchy and its subsequence converges to $X$, the triangle inequality gives $\mathrm{\parallel }{X}_n-X{\mathrm{\parallel }}_{\mathcal{X}}\to 0$: for any $ϵ>0$, first take $k$ with $\mathrm{\parallel }{X}_{n_k}-X\mathrm{\parallel }<ϵ\mathrm{/}2$, then take $n$ beyond the Cauchy index so $\mathrm{\parallel }{X}_n-X_{n_k}\mathrm{\parallel }<ϵ\mathrm{/}2$. Thus every Cauchy sequence converges, which is completeness by [F3].