Reason: Carried onto thm:lqg-separation-extended-2026b. Metric continuity convention stated inline in place of the redacted c54 continuity definition; entry bounds rerouted to thm:extreme-value-closed-interval-2026a; in-cluster references bumped to standing successors. No mathematical change.
Proof
Throughout, a real-valued function on a subinterval I of the real numbersR is called continuous on I when it is continuous relative to I, both I and the codomain R carrying the metric of the real line.
Fix an extended admissible Ξ± and an approximating sequence ((Ξ±(n)),D). Write X(n):=X(Ξ±(n)) for the controlled estimator of Ξ±(n) and X:=X(Ξ±) for the extended controlled estimator, and set, componentwise,
where Anβ(t):=βΞΊββ₯vt(n),ΞΊββvtΞΊββ₯2β and Bnβ(t):=βΞΊβ(β₯vt(n),ΞΊββ₯2β+β₯vtΞΊββ₯2β). By (1), Anβ(t)β0 for tβD while Bnβ(t) converges, so Οnβ(t)βΟΞ±β(t) for every tβD. Since each Οnβ is continuous, hence measurable, claim 6 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval shows 1DβΟΞ±β is B[0,T]β-measurable.
The functions 1DβAnβ and 1DβBnβ are also measurable, by the same claim 6: for fixed n and each jβN, the function tβ¦βΞΊββ₯vt(n),ΞΊββwt(j),ΞΊββ₯2β with wt(j)β:=Ξ±t(j)ββΞ(t)Xtβ is continuous (every ingredient is componentwise mean-square continuous), and for tβD it converges to Anβ(t) as jββ because β₯Ξ±t(j),ΞΊββΞ±tΞΊββ₯2ββ0 along the full sequence; similarly for Bnβ using β₯vtΞΊββ₯2β=limjββ₯wt(j),ΞΊββ₯2β on D.
Step 2 (uniform bounds). Let K1β bound maxtββiββ₯Xt(n),iββ₯2β uniformly in n: indeed βiββ₯Xt(n),iββ₯2ββ€βiββ₯Xt(n),iββXtiββ₯2β+βiββ₯Xtiββ₯2β, where the first summand is bounded uniformly in n and t by claim 4 of Conditional Expectation and Estimation Error of the Extended Controlled State (a convergent sequence of maxima is bounded) and the second is a continuous function of t (claim 1 there and the triangle inequality), bounded by Extreme Value Theorem on a Closed Real Interval; write K:=maxtββiββ₯Xtiββ₯2β for the latter bound. Next, Nn2β:=β«0TβgΞ±(n),0βdt, with 0 the zero control, is bounded uniformly in n: choosing N0β with d(Ξ±(n),Ξ±(N0β))β€1 for nβ₯N0β, the pointwise bound gΞ±(n),0ββ€2gΞ±(n),Ξ±(N0β)β+2gΞ±(N0β),0β (triangle inequality and (x+y)2β€2x2+2y2) gives Nn2ββ€2+2maxmβ€N0ββNm2β.
Since Ξ(t) is a kΓl matrix with entries bounded by MΞβ, for any tuple Y=(Y1,β¦,Yl) of square-integrable random variables one has βΞΊββ₯(Ξ(t)Y)ΞΊβ₯2ββ€kMΞββiββ₯Yiβ₯2β; combined with (x+y)2β€2x2+2y2 and (βΞΊβaΞΊβ)2β€kβΞΊβaΞΊ2β this yields the pointwise bounds
Integrating with claims 3 and 6 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval gives β«1Dβ(βΞΊββ₯v(n),ΞΊβ₯2β)2dΞ»β€2kNn2β+2k2MΞ2βTK12β and β«1Dβ(βΞΊββ₯w(j),ΞΊβ₯2β)2dΞ»β€2ksupmβNm2β+2k2MΞ2βTK2 for every j. On D, (βΞΊββ₯vtΞΊββ₯2β)2=limjβ(βΞΊββ₯wt(j),ΞΊββ₯2β)2, so by Fatou's Lemma,
Now, for every n, the two-sided pointwise bounds 1DβΟnββ€1DβΟΞ±β+MRβ1DβAnβBnβ and 1DβΟΞ±ββ€1DβΟnβ+MRβ1DβAnβBnβ, integrated with the same tools, give (all integrals now being finite real numbers)
by (3) and (4). Hence β«1DβΟΞ±βdΞ»=limnβpnβ=J[Ξ±]βVβ, which is the displayed representation. If ((Ξ²(m)),Dβ²) is any other approximating sequence for Ξ±, the entire argument applies verbatim to that sequence and yields β«1Dβ²βΟΞ±βdΞ»=J[Ξ±]βVβ as well, since J[Ξ±] does not depend on the approximating sequence by The Linear-Quadratic-Gaussian Cost of an Extended Admissible Control; so the integral is independent of the choice of approximating sequence.
Step 4 (claim 2). Since 1DβΟΞ±ββ₯0, claim 1 gives J[Ξ±]β₯Vβ. Suppose J[Ξ±]=Vβ, so β«1DβΟΞ±βdΞ»=0. By claim 5 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, Ξ»({1DβΟΞ±β>0})=0. Set D0β:=Dβ{1DβΟΞ±β>0}; its complement is the union of two null sets, hence null by the subadditivity argument of claim 5 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval. For tβD0β: E[vtββ (R(t)vtβ)]=0 with the integrand nonnegative pointwise, so by Markov's inequality (Markov's and Chebyshev's Inequalities) vtββ (R(t)vtβ)=0almost surely; on that event, positive definiteness of R(t) forces vtβ=0, that is, Ξ±tβ=Ξ(t)Xtβ componentwise almost surely. Conversely, suppose such a co-null D0β exists. For tβD0β, vtβ=0 almost surely componentwise, so ΟΞ±β(t)=0 (the expectation of a random variable almost surely equal to 0). Applying claim 6 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval with the co-null set D0β and g:=1DβΟΞ±β gives β«1DβΟΞ±βdΞ»=β«1DβΟΞ±β1D0ββdΞ»=0, since the integrand vanishes identically. By claim 1, J[Ξ±]=Vβ.