Let $G$ be a compact topological group with Haar measure $\mu$. A finite dimensional representation $(\pi,V)$ of $G$ is a continuous homomorphism $G \rightarrow \textrm{GL}(V)$. Equivalently, the map $G \times V \rightarrow V, (g,v) \mapsto \pi(g)v$ is continuous.
Let $H$ be a closed subgroup of finite index in $G$, and $(\rho,W)$ a representation of $H$. J.P. Serre writes in Linear Representations of Finite Groups (pg. 34), "...the representation (of $G$) induced by $(\rho,W)$ is defined as the Hilbert space of square integrable functions $f$ on $G$, with values in $W$, such that..."
My question is:
What does it mean for a function $f: G \rightarrow W$ to be integrable?
My attempt at an answer: If $w_1, ... , w_n$ is a basis for $W$, and $\pi_i: W \rightarrow \mathbb{C}$ is the map $c_1w_1 + \cdots + c_nw_n \mapsto c_i$, it makes sense for
$$\int\limits_G f d\mu$$ to be defined in terms of the integrals $\int\limits_G (\pi_i \circ f) d\mu$. Moreover, this definition should be independent of the choice of basis for $W$. However, if we take another basis $v_1, ... , v_n$ of $W$, and let $p_i$ be the corresponding projections, and if
$$v_j = \sum\limits_i r_{ij}w_i$$
Then $$\int\limits_G (p_i \circ f) d\mu = \sum\limits_{j=1}^n \int\limits_G r_{ij}(\pi_j \circ f) d\mu$$
I do not see any reasonable way to define the integral $\int_G f d\mu$ without first fixing a basis. My guess as to what should be done is we define $f: G \rightarrow W$ to be square integrable if for a basis $w_1, ... , w_n$ of $W$ with corresponding projections $\pi_i$, each function $\pi_i \circ f:G \rightarrow \mathbb{C}$ is measurable and square integrable. But the actual integral of $f$ is undefined. This definition of square-integrable is clearly independent of the choice of basis.