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.