proof that$ L^1 (G)$ is a subspace of $M(G)$

Question

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?؟ . .

Answer 1

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.

Answer 2

If one takes a look at the proof of the construction of a Haar measure $m$ for $G$, one finds that $m$ is a Radon measure ($m$ arrises from a linear functional on $C_c(G)$ by invoking the Riesz representation theorem). The following post proves that $f\mu$ where $\mu$ is a Radon measure is again a Radon measure. By Theorem C.8 in Conway's Functional Analysis, we know that $\vert f m\vert=\vert f\vert m$, and hence $\Vert fm\Vert=\Vert f\Vert_1$ (details spelled out by PhoemueX already). Therefore we can consider $L^1(G)$ as a closed subspace of $M(G)$.