proof that$ L^1 (G)$ is a subspace of $M(G)$ Let G be a locally compact group, and let $M(G)$ be the space of complex Radon measures on G. I identify the function f with the measure $f(x) \rm dx$ . but How do I prove this inclusion?؟ .
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 A: The precise answer to this question depends on your definition of a (complex) Radon measure. I assume in the following that this means that the total variation $|\mu|$ is Radon. I also assume that you know $|f(x) \,dx| = |f(x)| \,dx$, where the absolute value on the left-hand side denotes the total variation. Thus, we can assume $f\geq 0$ in the following.
Let us first show that $f(x) \,dx$ is indeed Radon.
Let $f \in L^1(G)$, $f \geq 0$. By approximating $f$ with simple functions, you can show that for every $\varepsilon > 0$, there is some $\delta > 0$ such that
$$
\int_E f(x) \, dx < \frac{\varepsilon}{2}
$$
holds for all measurable $E \subset G$ with $|E| < \delta$.
Furthermore, there is a countable sequence $E_n \subset G$ of sets of finite measure with $f \equiv 0$ on the complement of $\bigcup_n E_n$, take e.g. $E_n = \{x \in G \mid |f(x)| \geq 1/n\}$. We can assume w.l.o.g. that $E_n \subset E_{n+1}$ for all $n$.
These are the main ingredients for showing that $f(x) \,dx \in M(G)$ (i.e. is Radon).
Indeed, let $F \subset G$ be measurable. Then
$$
\int_{F\cap E_n} f(x) \,dx \rightarrow \int_F f(x) \,dx,
$$
(for example by monotone convergence), i.e. $\int_{F \cap E_n} f(x) \,dx > \int_F f(x) \,dx - \frac{\varepsilon}{2}$ for $n$ large enough. By inner regularity of $dx$ on sets of finite measure, there is a compact set $K \subset F \cap E_n \subset F$ with $|(F \cap E_n) \setminus K| < \delta$, where $\delta$ is chosen as above. This yields
$$
\int_K f(x) \,dx \leq \int_F f(x) \,dx < \int_{F \cap E_n} f(x) \,dx + \frac{\varepsilon}{2} = \int_K f(x) \,dx + \int_{(F\cap E_n) \setminus K} f(x) \,dx + \frac{\varepsilon}{2} < \varepsilon.
$$
This establishes inner regularity of $f(x) \,dx$.
For outer regularity, let again $F \subset G$ be measurable. As shown above, there is a compact set $K \subset F^c$ with
$$
\int_K f(x) \,dx \leq \int_{F^c} f(x) \,dx < \int_K f(x) \,dx + \varepsilon.
$$
This entails that $U := K^c \supset F$ is open and
$$
\begin{eqnarray*}
\int_{F}f\left(x\right)\, dx & \leq & \int_{U}f\left(x\right)\, dx\\
 & = & \int_{G}f\left(x\right)\, dx-\int_{U^{c}}f\left(x\right)\, dx\\
 & = & \int_{G}f\left(x\right)\, dx-\int_{K}f\left(x\right)\, dx\\
 & < & \int_{G}f\left(x\right)\, dx-\left[\int_{F^{c}}f\left(x\right)\, dx-\varepsilon\right]\\
 & = & \int_{F}f\left(x\right)\, dx+\varepsilon
\end{eqnarray*}
$$
This also proves outer regularity of $f(x) \,dx$.
Now, if you know (as used above) $|f(x) \,dx| = |f(x)|\,dx$ for the total variation, you see
$$
\Vert f(x) \,dx \Vert_{M(G)} = (|f(x)| \, dx )(G) = \int_G |f(x)| \,dx = \Vert f \Vert_1,
$$
so that the (by the above) well-defined map
$$
L^1(G) \rightarrow M(G), f \mapsto f(x) \,dx
$$
is isometric, and hence injective. It is clearly linear, completing the proof.
