Let $W_{2}=\sum_{j}m_{j}X_{j}$ be the decomposition of $W_{2}$ into prime cycles and for each $j$ let $(X_{ij})_{i\in I_{j}}$ be the irreducible components of $W_{1}$ with $X_{ij}\subseteq X_{j}\cap W_{1}$. Then $W_{1}=\sum_{j,i}m_{j}X_{ij}$ is the decomposition of $W_{1}$ into prime cycles. Furthermore, the intersection of any two irreducible components of $W_{1}$ does not contain a Shilov point as it is of lower dimension and hence does not meet the support of the measures of interest. By linearity in the irreducible components we may therefore assume that $W_{1}$ and $W_{2}$ are irreducible and reduced. Let $\mathfrak{X}_{2}$ be a formal model of $W_{2}$ with reduced special fibre on which there exist formal models of $\overline{L}_{1},...,\overline{L}_{n}$. Let $\mathfrak{X}_{1}$ be a formal model of $W_{1}$ which exists by paracompactness of $W_{1}$, see Remark 3.2. After possibly blowing up, the inclusion $W_{1}\hookrightarrow W_{2}$ induces a morphism $\iota:\mathfrak{X}_{1}\rightarrow\mathfrak{X}_{2}$ ([Bos14, Theorem 8.4.3]). Let $x\in\overset{\circ}{W_{1}}$. As both measures are discrete it is enough to show that they have the same mass at $x$. Let $\Int(W_{i})$ denote the relative interior of $W_{i}$ over $K$ in the sense of [Ber93, 1.5]. If $x\in\Int(W_{2})$ then $x\in\Int(W_{1})$ by [Ber93, Proposition 1.5.5 (ii)]. Conversely if $x\in\Int(W_{1})$ then there exists an affinoid neighbourhood $V$ of $x$ in $W_{1}$ such that $x$ is in the relative interior of $V$ over $K$. But $V$ is also a neighbourhood of $x$ in $W_{2}$ as $x\in\overset{\circ}{W_{1}}$ and therefore $x\in\Int(W_{2})$. Hence $x\in\Int(W_{1})$ if and only if $x\in\Int(W_{2})$. If this is not the case then by definition of the measures and Corollary A.4, both of them are zero at $x$. So assume that $x\in\Int(W_{1})$. Choose a locally finite cover $(\mathfrak{U}_{i})_{i\in I}$ of $\mathfrak{X}_{1}$ by open affine formal subschemes and let $\mathfrak{U}$ be the union of all $\mathfrak{U}_{i}$ which contain $\red(x)$. Then $\mathfrak{U}$ is an open and quasi-compact formal subscheme of $\mathfrak{X}_{1}$. Analogously choose a cover $(\mathfrak{V}_{i})_{i\in J}$ of $\mathfrak{X}_{2}$ by open affine formal subschemes. As $\iota(\mathfrak{U})$ is quasi-compact, there is a finite subcover of it. Let $\mathfrak{V}$ be the union of the sets in this subcover and add all $\mathfrak{V}_{i}$ with $\red(x)\in\mathfrak{V}_{i}$. Then also $\mathfrak{V}$ is an open and quasi-compact formal subscheme of $\mathfrak{X}_{2}$ and $\iota$ induces a morphism $\mathfrak{U}\rightarrow\mathfrak{V}$. By [BL93, Corollary 5.4] there is an admissible formal blowing up $\mathfrak{V}^{\prime}\rightarrow\mathfrak{V}$ such that the induced morphism $\mathfrak{U}^{\prime}\rightarrow\mathfrak{V}^{\prime}$ is an open immersion.
Let $Y$ be an irreducible component of $\tilde{\mathfrak{X}}_{2}$ with corresponding divisorial point $\zeta_{Y}=x$. Then $Y\subseteq\tilde{\mathfrak{V}}$ by definition and hence we may calculate the mass of $c_{1}(\overline{L}_{1})\wedge...\wedge c_{1}(\overline{L}_{n})$ at $x$ using $\mathfrak{V}$. By Proposition 4.5 iii) we may also use $\mathfrak{V}^{\prime}$. So let $Y^{\prime}$ be the irreducible component of $\tilde{\mathfrak{V}}^{\prime}$ corresponding to $x$. Since $\red(x)\in\tilde{\mathfrak{U}}^{\prime}$ we see that $Y\cap\tilde{\mathfrak{U}}^{\prime}$ is an irreducible component of $\tilde{\mathfrak{U}}^{\prime}$. Additionally, by Corollary A.4, $Y^{\prime}$ and $Y^{\prime}\cap\tilde{\mathfrak{U}}^{\prime}$ are proper and hence $Y^{\prime}=Y^{\prime}\cap\tilde{\mathfrak{U}}^{\prime}$ and it is an irreducible component of $\tilde{\mathfrak{U}}^{\prime}$. It’s image in $\tilde{\mathfrak{U}}$ is a proper irreducible component of $\tilde{\mathfrak{U}}$ and hence also an irreducible component of $\tilde{\mathfrak{X}}_{1}$. By the same argumentation as above we may use $\mathfrak{U}^{\prime}$ instead of $\mathfrak{X}_{1}$ to calculate the mass of $c_{1}\left(\overline{L}_{1}\Big|_{W_{1}}\right)\wedge...\wedge c_{1}\left(\overline{L}_{n}\Big|_{W_{1}}\right)$ at $x$. This shows that the mass of the two measures is equal at $x$ in this case.
Conversely, if $Y$ is an irreducible component of $\tilde{\mathfrak{X}}_{1}$ with corresponding divisorial point $\zeta_{Y}=x$ then $Y\subseteq\tilde{\mathfrak{U}}$ by definition. Again we may use $\mathfrak{U}^{\prime}$ to calculate the mass at $x$ and we denote the corresponding irreducible component by $Y^{\prime}$. Then the closure $\overline{Y}^{\prime}$ of $Y^{\prime}$ in $\tilde{\mathfrak{V}}^{\prime}$ is an irreducible component of $\tilde{\mathfrak{V}}^{\prime}$ with corresponding divisorial point $\zeta_{\overline{Y}^{\prime}}=x$ and hence by the above $\overline{Y}^{\prime}=Y^{\prime}$. Therefore $c_{1}(\overline{L}_{1})\wedge...\wedge c_{1}(\overline{L}_{n})$ and $c_{1}\left(\overline{L}_{1}\Big|_{W_{1}}\right)\wedge...\wedge c_{1}\left(\overline{L}_{n}\Big|_{W_{1}}\right)$ coincide at $x$. ∎