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

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.]

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(If you've read this question during the first 4 minutes I posted it - please reread, I corrected some typos and tried making the question clearer) – ro44 Dec 8 '11 at 19:18
First of all, your $h$ is only well defined if $t \ge a$ (since $f$ is only defined for $x >0$ -- make sure you get statement well defined). Then write $h$ as $\int \int \chi_{[t-a, t+a]} f(x) dx dt$, $\chi$ being the characteristic function of a set, i.e. $=1$ if in the set, $0$ otherwise. – user20266 Dec 8 '11 at 19:29
Oh right, I forgot to mention that about $h$! Also, I don't understand the notation in the second sentence you wrote... I don't think I've learned anything that let me use a "characteristic function" :)... We've only just learned the definition of an integral going to infinity and some theorems (Dirchlet, Comparison, Cauchy). – ro44 Dec 8 '11 at 19:31
For a given set $A$, $\chi_A(x) = 1$ if $x\in A$ and $\chi_A(x) = 0$ if $x\not \in A$. – user20266 Dec 8 '11 at 19:34
If $A\subseteq\mathbb{R}$, the indicator (or characteristic) function of $A$ is the function $\chi_A$ defined by $$\chi_A(x)=\begin{cases}1,&x\in A\\0,&x\ne A\;.\end{cases}$$ – Brian M. Scott Dec 8 '11 at 19:35

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.

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