Pick a point $q\in\tilde{V}$. Since $\pi$ is a log resolution, there exists an open set $Z\subset\tilde{U}$ with a coordinate system $(w_{1},\dots,w_{n})$ centered at $q$ such that $\tilde{V}=\{w_{1}=0\}$ and $\pi^{*}\exp(\varphi_{1}),\pi^{*}\exp(\varphi_{2})$ are of the form

	$\displaystyle\pi^{*}\exp(\varphi_{1})$	$\displaystyle=U_{1}(w_{1},\dots,w_{n})\prod_{i=2}^{n}|w_{i}|^{2\alpha_{i}}$
	$\displaystyle\pi^{*}\exp(\varphi_{2})$	$\displaystyle=U_{2}(w_{1},\dots,w_{n})\prod_{i=2}^{n}|w_{i}|^{2\beta_{i}},$

where $U_{j}$ are smooth, positive functions on $\overline{Z}$, and $\alpha_{i},\beta_{i}$ are nonnegative real numbers. That $w_{1}$ does not appear in the product follows from the fact that $\varphi_{1},\varphi_{2}\not\equiv-\infty$ on $V\cap U$. By definition, we have

$$\pi^{-1}(\mathcal{S})\cap Z=\left\{w\in Z\bigg|U_{1}(w)\prod_{i\geqslant 2}|w_{i}|^{2\alpha_{i}}<e^{\eta}U_{2}(w)\prod_{i\geqslant 2}|w_{i}|^{2\beta_{i}}\right\}.$$

Now, since $\varphi_{1}|_{V}=\varphi_{2}|_{V}$, and $w_{i}|_{\tilde{V}}\neq 0$ for $i\geqslant 2$, we clearly have that $\alpha_{i}=\beta_{i}$, and that $U_{1}|_{\tilde{V}}=U_{2}|_{\tilde{V}}$. Since $e^{\eta}>1$, the lemma is proved. ∎