A little hint: You need to apply the definition :-) If you say you cannot to that, I have to agree with Theo and Qiaochu that this is a strong indication that you either don't know, or don't understand, the definition.
In that case, you should try to understand the most simple example that was mentioned in your class. For every definition in every math class, you should come up with the most simple example possible, and some nontrivial example.
The first step in applying the definition is a choice of a partition of the integration interval. Choose some numbers $x_1, x_2, x_3 \in [0, 2]$ and write the sum that approximates the integral. Then choose a refinement of this partition and write down the sum again. Maybe you'll notice that it is possible to choose a sequence of partitions such that you can evaluate the sums and determine the limit that is the integral...
Addendum: Ok, here is another hint: For $\alpha$ differentiable, we know that \int f(x) d \alpha(x) = \int f(x) \alpha'(x) dx So, if $\alpha$ is a jump function, that is constant with the notable exeption of a jump, it is possible to simplify the integral. For example, define $ \alpha(x) := 0 \; \text{for} \; x \lt 0 $ and $ \alpha(x) := 1 \; \text{for} \; x \ge 0 $
Then we can rewrite the integral \int_{-a}^{a} f(x) d\alpha(x) = \int_{-a}^{-\epsilon} f(x) \alpha'(x) dx + \int_{-\epsilon}^{\epsilon} f(x) d\alpha(x) + \int_{\epsilon}^{a} f(x) \alpha'(x) dx for an arbitrarily small $\epsilon \gt 0, \epsilon \lt a$. Since the differential of $\alpha$ vanishes for the first and the last term, we get $ \int_{-a}^{a} f(x) d\alpha(x) = \int_{-\epsilon}^{\epsilon} f(x) d\alpha(x) $ So, it is actually enough to consider what happens in an $\epsilon$ neighborhood of 0, in order to calculate the integral.
Now, any summand in U(P) and in L(P) will be zero except the one containing the difference of $\alpha(\epsilon) - \alpha(-\epsilon)$, so we have $ U(P) = (\alpha(\epsilon) - \alpha(-\epsilon)) \sup_{ x \in [-\epsilon, \epsilon]} f(x) = \sup_{ x \in [-\epsilon, \epsilon]} f(x) $ and $ L(P) = (\alpha(\epsilon) - \alpha(-\epsilon) ) \inf_{ x \in [-\epsilon, \epsilon]} f(x) = \inf_{ x \in [-\epsilon, \epsilon]} f(x) $ Since $f$ is continuous, we get in the limit $\epsilon \to 0$: $ U(P) = L(P) = f(0) $