I am trying to work through the proof of the following theorem:
Let $f$ be a measurable function on the locally compact group $G$, which is integrable with respect to a Haar measure. Then the support of $f$ is contained in a $\sigma$-compact open subgroup of $G$.
There's only one part of the proof which I don't understand. Let $\mu$ be a Haar measure on a locally compact group $G$, let $U$ be any open subset with finite measure, and let $H$ be an $\sigma$-compact open subgroup. Why does $U$ intersect only countably many cosets $xH$? The author claims that this must be the case, but I don't see how to prove it.
$U$ is open, and each $xH$ is open, so the intersection $U\cap xH$ is open for every $x$. Therefore (by definition of a Haar measure) the intersection $U\cap xH$ is of positive measure for every $x$ for which $U\cap xH$ is nonempty. Since $U$ has finite measure, this means there can only be countably many such $x$'s, by the next lemma.
Lemma. Let $X$ be a measure space. If $I$ is an uncountable index set and $U_i\subseteq X$ is a subset of positive measure for every $i\in I$, then the union $\bigcup_{i\in I} U_i$ has infinite measure.
Proof. For each $n\in\Bbb{N}$, consider the subset $I_n$ of $I$ of $i$'s for which $U_i$ has measure at least $1/n$. Since $I$ is uncountable and $\Bbb{N}$ is countable, there is some $n$ for which $I_n$ is infinite. Therefore $\bigcup_{i\in I_n} U_i$ has infinite measure, because it has measure at least $1/n\cdot\infty=\infty$. Therefore, $\bigcup_{i\in I} U_i$ has infinite measure too.