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Let $G$ be a compact Hausdorff topological group. Let $U$ be a neighbourhood of the identity $e$, and let $V = U \cap U^{-1}$ where $U^{-1} = \{x^{-1} : x \in U\}$. Apparently there always exists a continuous function $v$ on $V$ such that (1) $v$ is supported in $V$, (2) $v \geq 0$ and $\int_G v = 1$ (with our normalized Haar measure).

Why is this the case?

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Pick a point $g \in V$. Because $G$ is locally compact, $x$ has a closed neighborhood $C \subseteq V$ which is disconnected from $V^c$. Define a continuous function $f : C \cup V^c \to [0, 1]$ to have value $1$ on $C$ and $0$ on $V^c$. Then by the Tietze extension theorem, $f$ has a continuous extension to a function $\widetilde{f} : G \to [0, 1]$.

$\widetilde{f}$ is nonnegative and still has value $1$ on $C$ and $0$ on $V^c$, so it is supported in $V$. Now divide it by $\int_G \widetilde{f}$, which is positive since it must be at least the measure of $C$.

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