$\DeclareMathOperator\intr{int}\DeclareMathOperator\meas{meas}$This is one—probably not the best—way of thinking about it. (I am used to the $p$-adic world, where one can think about (real-valued) measures much less subtly than in the real world.) I hope someone will come along and write something more elegant.

The way we know we have the correct measure on $G(F)\backslash G(\mathbb A)$ is that $\int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f(h g)\mathrm dh\,\mathrm d\dot g = \int_{G(\mathbb A)} f(g)\mathrm dg$ for all $f \in \operatorname C_c^\infty(G(\mathbb A))$. Now take $\mathfrak S$ and $\mathfrak S'$ such that $G(\mathbb A) = G(F)\mathfrak S'$ and $\mathfrak S' \subseteq \intr(\mathfrak S)$. Write $\mathfrak S'$ and $\mathfrak S$ as countable, increasing unions of compacts $\mathfrak S'_k$ and $\mathfrak S_k$ such that $\mathfrak S'_k \subseteq \intr(\mathfrak S_k)$ for all $k$; and, for each $k$, let $f_k \in \operatorname C_c^\infty(G(\mathbb A))$ be a non-negative function that is $1$ on $\mathfrak S'_k$ and $0$ outside $\mathfrak S_k$. Then \begin{multline*} \meas_{\mathrm d\dot g}(G(F)\backslash G(\mathbb A)) = \lim_{k \to \infty} \meas_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'_k) \le \lim_{k \to \infty} \int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f_k(h g)\mathrm dh\,\mathrm d\dot g \\ = \lim_{k \to \infty} \int_{G(\mathbb A)} f_k(g)\mathrm dg \le \lim_{k \to \infty} \meas_{\mathrm dg}(\mathfrak S_k) = \meas_{\mathrm dg}(\mathfrak S) < \infty. \end{multline*}

$\DeclareMathOperator\intr{int}\DeclareMathOperator\meas{meas}$This is one—probably not the best—way of thinking about it. (I am used to the $p$-adic world, where one can think about (real-valued) measures much less subtly than in the real world.) I hope someone will come along and write something more elegant.

Since $G(F)$ is discrete, so that its Haar measure may be taken to be the counting measure, the way we know we have the correct measure on $G(F)\backslash G(\mathbb A)$ is that $$ \int_{G(F)\backslash G(\mathbb A)} \sum_{h \in G(F)} f(h g)\,\mathrm d\dot g = \int_{G(\mathbb A)} f(g)\mathrm dg $$ for all $f \in \operatorname C_c^\infty(G(\mathbb A))$. Now take $\mathfrak S$ and $\mathfrak S'$ such that $G(\mathbb A) = G(F)\mathfrak S'$ and $\mathfrak S' \subseteq \intr(\mathfrak S)$. Write $\mathfrak S'$ and $\mathfrak S$ as countable, increasing unions of compacts $\mathfrak S'_k$ and $\mathfrak S_k$ such that $\mathfrak S'_k \subseteq \intr(\mathfrak S_k)$ for all $k$; and, for each $k$, let $f_k \in \operatorname C_c^\infty(G(\mathbb A))$ be a non-negative function that is $1$ on $\mathfrak S'_k$ and $0$ outside $\mathfrak S_k$. Notice that, for each $k$ and each $g \in \mathfrak S'_k$, we have that $$ \sum_{h \in G(F)} f_k(h g) \ge f_k(g) = 1, $$ so that \begin{multline*} \meas_{\mathrm d\dot g}(G(F)\backslash G(\mathbb A)) = \lim_{k \to \infty} \meas_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'_k) \le \lim_{k \to \infty} \int_{G(F)\backslash G(\mathbb A)} \sum_{h \in G(F)} f_k(h g)\,\mathrm d\dot g \\ = \lim_{k \to \infty} \int_{G(\mathbb A)} f_k(g)\mathrm dg \le \lim_{k \to \infty} \meas_{\mathrm dg}(\mathfrak S_k) = \meas_{\mathrm dg}(\mathfrak S) < \infty. \end{multline*}

$\DeclareMathOperator\intr{int}\DeclareMathOperator\meas{meas}$This is one—probably not the best—way of thinking about it. (I am used to the $p$-adic world, where one can think about (real-valued) measures much less subtly than in the real world.) I hope someone will come along and write something more elegant.

The way we know we have the correct measure on $G(F)\backslash G(\mathbb A)$ is that $\int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f(h g)\mathrm dh\,\mathrm d\dot g = \int_{G(\mathbb A)} f(g)\mathrm dg$ for all $f \in \operatorname C_c^\infty(G(\mathbb A))$. Now take $\mathfrak S$ and $\mathfrak S'$ such that $G(\mathbb A) = G(F)\mathfrak S'$ and $\mathfrak S' \subseteq \intr(\mathfrak S)$. Write $\mathfrak S'$ and $\mathfrak S$ as countable, increasing unions of compacts $\mathfrak S'_k$ and $\mathfrak S_k$ such that $\mathfrak S'_k \subseteq \intr(\mathfrak S_k)$ for all $k$; and, for each $k$, let $f_k \in \operatorname C_c^\infty(G(\mathbb A))$ be $1$ on $\mathfrak S'_k$ and $0$ outside $\mathfrak S_k$. Then \begin{multline*} \meas_{\mathrm d\dot g}(G(F)\backslash G(\mathbb A)) = \lim_{k \to \infty} \meas_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'_k) \le \lim_{k \to \infty} \int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f_k(h g)\mathrm dh\,\mathrm d\dot g \\ = \lim_{k \to \infty} \int_{G(\mathbb A)} f_k(g)\mathrm dg \le \lim_{k \to \infty} \meas_{\mathrm dg}(\mathfrak S_k) = \meas_{\mathrm dg}(\mathfrak S) < \infty. \end{multline*}

$\DeclareMathOperator\intr{int}\DeclareMathOperator\meas{meas}$This is one—probably not the best—way of thinking about it. (I am used to the $p$-adic world, where one can think about (real-valued) measures much less subtly than in the real world.) I hope someone will come along and write something more elegant.

The way we know we have the correct measure on $G(F)\backslash G(\mathbb A)$ is that $\int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f(h g)\mathrm dh\,\mathrm d\dot g = \int_{G(\mathbb A)} f(g)\mathrm dg$ for all $f \in \operatorname C_c^\infty(G(\mathbb A))$. Now take $\mathfrak S$ and $\mathfrak S'$ such that $G(\mathbb A) = G(F)\mathfrak S'$ and $\mathfrak S' \subseteq \intr(\mathfrak S)$. Write $\mathfrak S'$ and $\mathfrak S$ as countable, increasing unions of compacts $\mathfrak S'_k$ and $\mathfrak S_k$ such that $\mathfrak S'_k \subseteq \intr(\mathfrak S_k)$ for all $k$; and, for each $k$, let $f_k \in \operatorname C_c^\infty(G(\mathbb A))$ be a non-negative function that is $1$ on $\mathfrak S'_k$ and $0$ outside $\mathfrak S_k$. Then \begin{multline*} \meas_{\mathrm d\dot g}(G(F)\backslash G(\mathbb A)) = \lim_{k \to \infty} \meas_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'_k) \le \lim_{k \to \infty} \int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f_k(h g)\mathrm dh\,\mathrm d\dot g \\ = \lim_{k \to \infty} \int_{G(\mathbb A)} f_k(g)\mathrm dg \le \lim_{k \to \infty} \meas_{\mathrm dg}(\mathfrak S_k) = \meas_{\mathrm dg}(\mathfrak S) < \infty. \end{multline*}