Let $G$ be a locally compact group. Let $X$ be a Banach space. Let $C_{c}(G)$ be the space of continuous functions with compact support on $G$. Suppose $j\in C_{c}(G; X)$. Does this imply that $\int_G j(x) dx \in X$ (with respect to the left Haar measure)?
I think the statement is true. I think that there's an argument of simple functions with compact support $f: G \rightarrow X$ will show it. But I'm not sure how exactly that will work.
Any help will be appreciated!
Best Answer
I think the integration that you are dealing with is Bochner integration. $G$ is equipped with Haar measure on Borel sigma algebra. A function $f:G\to X$ is said to be strongly measurable if there exists a sequence of simple measurable functions converging to $f$ pointwise almost everywhere. By Pettis's measurability theorem, this is equivalent to $f$ being Borel measurable ($f^{-1}(O)$ is Borel set for each open subset $O$ of $X$) and essentially separably valued. A strongly measurable function $f$ is said to be Bochner integrable if there exists a sequence of measurable simple functions $\{f_n\}$ converging pointwise almost everywhere to $f$ such that $$\lim_n\int||f(x)-f_n(x)||d\mu=0.$$ And the Bochner integral over a measurable set $E$ is defined as $\int_E fd\mu=\lim_n\int_E f_nd\mu$ (where integration of simple functions is defined in usual sense). Further it can be shown that a strongly measurable function $f$ is Bochner integrable if and only if $||f(x)||$ is integrable.
Now, in your case $j$ is a compactly supported continuous function and hence it is essentially separably valued and Borel measurable and thus stongly measurable function. Further $||j(x)||\in C_c(G)$ is integrable and hence $ j$ is Bochner integrable and $\int_G j(x)d\mu\in X$ exists.