limit of Integral of product of convergent functions Let $f,g$ be real functions that are Riemann-integrable over $[0,t] \, \forall t \in {\mathbb R^+}$. With: 
$$\lim_{x \to \infty}(f(x))= p$$
$$\lim_{x \to \infty}(g(x))= q$$
Show that:
$$\lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t f(x) \cdot g(t-x) \,dx \Bigr) = pq$$
Hint use $f(x) \cdot g(t-x)=(f(x)-p)g(t-x)+p \cdot g(t-x)$ and the fact that when $f,g$ are integrable so are $|f|, f^2, fg$ and $|\int_a^b f \,dx| \le \int_a^b |f| \,dx$.
What I tried so far was the following:
$$\lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t f(x) \cdot g(t-x) \,dx \Bigr)$$
$$= \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x)+p \cdot g(t-x) \,dx \Bigr)$$
$$= \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x)+p(g(t-x)+q-q) \,dx \Bigr)$$
$$= \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x)\,dx \Bigr) + \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t pq \,dx \Bigr) + p\cdot \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t g(t-x)-q \,dx \Bigr)$$
$$= pq + \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x)\,dx \Bigr) + p\cdot \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t g(t-x)-q \,dx \Bigr)$$
Then I'd only have to show that the remaining limits converge to 0. And since $\frac{1}{t} \to 0$ already converges to 0 it is sufficient when the Integrals are bounded.
Show that $\int_0^t (f(x)-p)g(t-x)\,dx$ and $\int_0^t g(t-x)-q \,dx$ are bounded.
So my question is: How do I show that these integrals are bounded? Or if it doesn't work this way how would one show the limit instead?
Thanks in advance :)
 A: Maybe I found a solution, though I'm not sure it really works:
$$\lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t f(x) \cdot g(t-x) \,dx \Bigr)$$
$$= \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x)+p \cdot g(t-x) \,dx \Bigr)$$
$$= \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x) \,dx+p \cdot \frac{1}{t} \int_0^t g(t-x) \,dx \Bigr)$$
$$= \lim_{t \to \infty} \Bigl( \frac{1}{t} \int_0^t (f(x)-p)g(t-x) \,dx -p \cdot \frac{1}{t} \int_t^0 g(x) \,dx \Bigr)$$
We know from the hint that $h(x)=(f(x)-p)g(t-x)$ is also an integrable function.
Let the antiderivative of $h(x)$ be $H(x)$ and of $g(x)$ be $G(x)$.
$$= \lim_{t \to \infty} \biggl( \frac{1}{t} \Bigl(H(t)-H(0)-p\bigl(G(0)-G(t)\bigr)\Bigr) \biggr)$$
and since we have a fraction over $t$ and $t \to \infty$ we can use L'Hospital's rule.
$$= \lim_{t \to \infty} \bigl(h(t) + p\cdot g(t) \bigr) = \lim_{t \to \infty} \bigl((f(t)-p)g(0) + p\cdot g(t) \bigr)$$
$$= g(0) \cdot\lim_{t \to \infty} f(t) - p\cdot g(0) + p \cdot\lim_{t \to \infty} g(t) = g(0)\cdot p - p\cdot g(0) +pq = pq $$
Could anyone comment on that? The part that sticks out to me is mostly when applying L'Hospital's rule.
