Corollary 6.11.6 (Strong openness).EA6067858516 Let $\varphi $ and $\chi $ be weights on $X$, and let $x\in X$. Then
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6.11.7
\begin{equation} \mathcal I(\varphi )_ x=\bigcup _{\varepsilon >0} \mathcal I((1+\varepsilon )\varphi )_ x \end{equation}
and, more generally,
\[ \mathcal I(\varphi )_ x=\bigcup _{\varepsilon >0} \mathcal I(\varphi +\varepsilon \chi )_ x. \]
Moreover, there is a single $\varepsilon >0$ with $\mathcal I((1+\varepsilon )\varphi )_ x=\mathcal I(\varphi )_ x$, and similarly a single $\varepsilon >0$ with $\mathcal I(\varphi +\varepsilon \chi )_ x=\mathcal I(\varphi )_ x$.
Proof.
The assertions are local. On a relatively compact neighborhood of $x$ the weights are bounded above, and subtracting constants we may assume $\varphi \leq 0$ and $\chi \leq 0$ there; this changes $(1+\varepsilon )\varphi $ and $\varphi +\varepsilon \chi $ only by bounded amounts and hence changes none of the ideals. Then $(1+\varepsilon )\varphi $ and $\varphi +\varepsilon \chi $ decrease as $\varepsilon $ increases, so the corresponding ideals decrease with $\varepsilon $ and are contained in $\mathcal I(\varphi )_ x$. The sequences $(1+1/j)\varphi $ and $\varphi +\chi /j$ increase with $j$ and converge to $\varphi $ wherever $\varphi $ and $\chi $ are finite, hence almost everywhere. By theorem 6.11.4, $\mathcal I((1+1/j)\varphi )_ x=\mathcal I(\varphi )_ x$ and $\mathcal I(\varphi +\chi /j)_ x=\mathcal I(\varphi )_ x$ for all large $j$, which proves both assertions.
$\square$
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