Let $G$ be a non-discrete (LCH) group. How can one find a sequence $(V_n)$ of compact, symmetric neighbourhoods of the identity element $1$ such that $\mu(V_n)\to 0$, where $\mu$ denotes the Haar measure of $G$?
It is clear that the sequence can be taken from a neighbourhood base at $1$, but I don't see how one can make sure that we get $\mu(V_n)\to 0$, even though I am sure that's where the non-discrete part comes in.
Let $V_1$ be a compact symmetric neighbourhood of $1$. Then $V_1\neq\{1\}$ since $G$ is non-discrete, so there exists $x_2\in V_1\setminus\{1\}$. As $G$ is Hausdorff, there exist disjoint neighbourhoods $U_2\subseteq V_1$ and $U_2'\subseteq V_1$ of $1$ and $x_2$, respectively. Then $W_2=\overline{x_2^{-1}U_2'\cap U_2}$ and $W_2'=\overline{U_2'\cap x_2U_2}$ are disjoint compact neighbourhoods of $1$ and $x_2$, respectively, and $$2\mu(W_2)=\mu(W_2\cup W_2')\leq\mu(V_1).$$ By defining $V_2=W_2\cap W_2^{-1}$, we obtain a compact symmetric neighbourhood $V_2$ of $1$ with $V_2\subseteq V_1$ and $\mu(V_2)\leq\frac{1}{2}\mu(V_1)$. Continue inductively.