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Write ${\tau }_{\epsilon }=T_{\epsilon }\wedge T_M$. The stopped Bessel process has the same law as the weighted stopped Brownian motion, so for every bounded measurable functional $F$ of a continuous path,The Bessel process hits $M$ in finite time and never hits zero. Its minimum before that hitting time is strictly positive. Thus for sufficiently small $\epsilon$, depending on the path, its stopped path $R^{{\tau }_{\epsilon }}$ equals $R^{T_M}$ exactly. Under Wiener measure, the events $\{T_M<T_{\epsilon }\}$ increase to $\{T_M<T_0\}$, and on that event the stopped paths eventually equal $X^{T_M}$. The remaining term is bounded by $\epsilon \Vert F{\Vert }_{\infty }/x$. The dominated convergence theorem therefore yields the precise limiting identityThe bounded stopped martingale $X_{t\wedge T_0\wedge T_M}$ givesHence the limiting identity saysThis establishes Brownian conditioning by a stopped Bessel density.

The displayed density in the question requires a stopping convention. On the full, unstopped path space, the probability measureis well-defined, but its paths continue as Brownian motion after $T_M$. Its pushforward by the stopping map $\omega \mapsto \omega (\cdot \wedge T_M(\omega ))$ is the law ${\mathbb{Q}}^M$ of the stopped Bessel process. Equivalently, if ${\mathbb{W}}_x^{0,M}$ denotes the law of Brownian motion stopped at $T_0\wedge T_M$, then the actual stopped laws satisfyLiteral absolute continuity of the stopped law with respect to unstopped Wiener measure is false: paths that become permanently constant at $M$ have probability one for the stopped law and zero under Wiener measure. The same density is also valid on the stopped sigma-algebra of the original coordinate space. Thus the corrected stopped-law formulation proves the intended conclusion without equating these different full-path measures.