Proof that the induced class function $\theta^G$ is a character if $\theta$ is a representation on subgroup In these lecture notes by Daniel Bump on Induced Characters I have a question on the proof of Theorem 2.5.1. If $H$ is a subgroup of the finite group $G$ and $(\pi, V)$ a representation of $H$, i.e. a homomorphism $\pi : H \to GL(V)$, then the theorem states a way to get a representation $(\theta^G, V^G)$ on the whole of $G$.
Here if $\theta$ is a function on $H$, the extension to $G$ is done by defining
$$
 \dot \theta(g) = \left\{ \begin{array}{ll} \theta(g) & \mbox{ if } g \in H,\\
                 0 & \mbox{ otherwise} \end{array}\right.
$$
and then
$$
 \theta^G(g) = \frac{1}{|H|} \sum_{x\in G} \dot \theta(xgx^{-1}).
$$
The theorem states:
Theorem: Let $\theta$ be the character of the representation $V$ of $H$. The function $\theta^G$ is the character of a representation of $G$. Its degree is $[G : H]\theta(1)$. If $\chi$ is any character of $G$, and $\chi_H$ is its restriction to $H$, then
$$
 \langle \theta^G, \chi \rangle_G = \langle \theta, \chi_H \rangle_H.
$$
In its proof almost everything is clear, how the above formula is verified and so on, but then in the part where he proves that $\theta^G$ is a character I do not understand whats happening. There he states:

It remains to be proved that $\theta^G$ is a character. Let $(\pi_1, V_1),
  \cdots, (\pi_h, V_h)$ be representatives of the isomorphism classes of the
    irreducible $G$-modules, and $\chi_1, \cdots, \chi_h$ be their characters.
    Since the $\chi_i$ are a basis of the complex vector space of class
    functions, and since $\theta^G$ is a class function, we can write
    $$ \theta^G = \sum_{i = 1}^h a_i \chi_i, $$
    where the $a_i$ are complex numbers, and if we can show that $a_i$ are
    nonnegative integers, it will follow that $\theta^G$ is a character, namely
    it will be the character of the representation on the module
    $$ a_1 V_1 \oplus \cdots \oplus a_h V_h . $$

I do not understand, why if the $a_i$ are nonnegative integers, then $\theta^G$ is a character? And why is it the character of the representation on the module $a_1 V_1 \oplus \cdots \oplus a_h V_h$?
 A: Because the trace is an additive invariant of a representation.  If $V= \bigoplus a_iV_i$ is a $G$-module, then for each $g\in G$ $$\chi_V(g) = Tr(\pi(g)) = Tr(\sum_i a_i\pi_i(g)) = \sum a_i Tr(\pi_i(g)) = \sum a_i\chi_i(g). $$
In view of the comments below, let me supplant my answer with the rest of the story (I am working over $\mathbb{C}$ from now on).
The character of a $G$-representation is the class function associated to $\pi:G\curvearrowright V$ given by $\chi_V(g) = Tr(\pi(g))$.  We saw above that a non-negative integer linear combination of the $\chi_i$ associated to the irreducible representations is a character.
Conversely, every character is a non-negative integer linear combination of the $\chi_i$.  To show this: let $V$ be a $G$-space.  By complete reducibility, $V$ is a direct sum of irreducibles.  Gathering multiplicities, $V\simeq \oplus a_i V_i$ where $a_i$ are the multiplicities, possibly zero, and $V_i$ runs through irreps.  Then the same computation as above shows that $\chi_V$ is a non-negative integer linear combination of the $\chi_i$.  The $\chi_i$ are independent because they are orthogonal with respect to the usual inner product, so the expression of $\chi_V$ is unique.
This shows that the set $C$ of characters of the various complex representations of $G$ are precisely the non-negative cone in the integer lattice spanned by the $\chi_i$: $$C=\bigoplus_i \mathbb{Z}_{\geq 0}\chi_i.$$
