Given $f(x)$ is positive and $\int_{0}^{\infty}f(x)dx$ converges, prove $\int_{a}^{\infty}h(x)dx$ converges where $h(t)=\int_{t-a}^{t+a}f(x)dx$ Please help me with the following excercise:

Given $f(x)$ is positive in $[0,\infty)$ and $\int_{0}^{\infty}f(x)dx$ converges, prove $\int_{a}^{\infty}h(x)$ ($a>0$) exists where $h(t)=\int_{t-a}^{t+a}f(x)dx$ ($t\ge a$). 

Thanks! :)
Attempt
I tried calculating the limit directly, but without much luck. Comparison test doesn't seem to work.... Also, I think representing $h(x)$ as $F(x+a)-F(x-a)$ might help. (where $F(x)=\int_{0}^{x}f(x)dx$)
What I know
We've only just learned the definition of an integral going to infinity and some theorems (Dirchlet, Comparison, Cauchy). [I don't know any useful theorems about double integrals, (edit: such as Fubini's) and the such.]
 A: This can be solved with a smart use of Fubini's theorem:
$$
\begin{align*}
 \int_a^{+ \infty} h(x)\, dx  
&= \int_a^{+ \infty} \int_{x-a}^{x+a} f(t) \, dt dx \\
&= \int_a^{+ \infty} \int_0^{+ \infty} 1_{[x-a, x+a]} (t) f(t)\, dt \, dx \\
&= \int_a^{+ \infty} \int_0^{+ \infty} 1_{[t-a, t+a]} (x) f(t)\, dt \, dx \\ 
&= \int_0^{+ \infty} f(t) \int_a^{+ \infty} 1_{[t-a, t+a]} (x) dx dt.
\end{align*}
$$
Now, you can check that $\displaystyle\int_a^{+ \infty} 1_{[t-a, t+a]} (x)\, dx = \min \{t, 2a\}$. Hence, 
$$
\begin{align*}
\int_a^{+ \infty} h(x)\, dx 
&= \int_0^{+ \infty} f(t) \min \{t, 2a\}\, dt\\ 
&\leq 2a \int_0^{+ \infty} f(t)\, dt.
\end{align*}
$$
Edit : the method above is very useful and general. You can adapt it to see what happens of one puts $h(x) = \int_{x-a}^{x+a} g(x+t) f(t) dt$, where the function $g$ is, say, continuous. However, in your problem $g$ is equal to $1$, which is easier to deal with. Let $F$ be a primitive of $f$. Then :
$$
\begin{align*}
\int_a^X h(x) dx & = \int_a^X F(x+a)-F(x-a) \, dx \\
&= \int_{2a}^{X+a} F(x)\, dx - \int_0^{X-a} F(x)\, dx \\
&= \int_{X-a}^{X+a} F(x)\, dx - \int_0^{2a} F(x)\, dx.
\end{align*}
$$
Since $f$ is integrable, $F$ converges monotonically to $\int_0^{+ \infty} f(x)\,dx$, so that $\int_{X-a}^{X+a} F(x) dx$ converges to $2a \int_0^{+ \infty} f(x)\,dx$ (if you are not sure about that, you can write down the argument : for any $\varepsilon > 0$, there exists a $X> 0$ such that, for all $x > X$, we have $\int_0^{+ \infty} f(t)\,dt - \varepsilon \leq F(x) \leq \int_0^{+ \infty} f(t)dt$, and then...). Hence, 
$$\int_a^{+ \infty} h(x)\, dx = \lim_{X \to + \infty} \int_a^X h(x)\, dx = 2a \int_0^{+ \infty} f(x)\,dx - \int_0^{2a} F(x)\, dx,$$
which is finite.
