TITLE: If $\int_0^\infty x f(x) \ dx < \infty$, is $\int_0^\infty \int_0^\infty f(x+y) \ dx \ dy< \infty$?
QUESTION [7 upvotes]: Question: Let $f : [0, \infty) \to [0,\infty)$ be a measurable function which, more than being integrable, satisfies $\int_0^\infty x f(x) < \infty$. Does it follow that  $\int_0^\infty \int_0^\infty  f(x+y) \ dx \ dy< \infty$?
Remark: The conclusion does not hold if you only assume $\int_0^\infty f(x) \ dx <\infty$. For example, the sequence $a_n = \frac{1}{n} - \frac{1}{n+1}$ is such that $\sum_{n=1}^\infty a_n < \infty$, but $\sum_{n=1}^\infty \sum_{m=1}^\infty a_{n+m} = \infty$.  It's not difficult to adapt this example to the present setting.  
Remark: Things also fall apart if the integral $\int_0^\infty \int_0^\infty  f(x+y) \ dx \ dy$ isn't taken over the first quadrant only. If $f : \mathbb{R} \to [0,\infty)$ is measurable and not zero almost everywhere, then assuming $\int_0^\infty x f(x) \ dx < \infty$ doesn't change the fact that the integral of $f(x+y)$ over every, e.g., horizontal line will be the same positive number so that $\int_{-\infty}^\infty \int_{-\infty}^\infty f(x+y) \ dx \ dy = \infty$. 

Added: In retrospect, a nice geometric way to see that the two integrals should be equal is to note that $(x,y) \mapsto f(x+y)$ is constant along lines of slope $-1$, the value of $f$ on $x+y = a$  being $f(a)$.  Meanwhile, the length of the intersection of $x+y = a$ with the 1st quadrant grows linearly with $a$.

REPLY [5 votes]: This is Tonelli's theorem:
$$\eqalign{
\int_{[0,+\infty)^2} f(x+y)dydx&=\int_0^\infty\left(\int_0^\infty f(x+y)dy\right)dx\cr
&=\int_0^\infty\left(\int_x^\infty f(t)dt\right)dx\cr
&=\int_0^\infty f(t)\left(\int_0^t dx\right)dt\cr
&=\int_0^\infty tf(t)dt<+\infty}
$$
which is the desired conclusion.