Show there exists a positive, strictly increasing measurable function Let $f$ be a nonnegative measurable function on $[0,\infty)$ such that $\int_0^\infty f(x)dx < \infty$ is finite. Show that there is a positive, strictly increasing measurable function $a(x)$ on $[0,\infty)$ with $\lim_{x\to\infty} a(x)=\infty$ and such that
$$\int_0^\infty a(x)f(x)dx<\infty.$$
I am just looking for a hint, since I'm pretty stuck. I thought about arguing by contradiction and assuming all such functions cause the last integral to diverge, but I'm not sure that's correct. Maybe I could use Chebychev's inequality to arrive at a contradiction?
 A: Ok, no solution, just 2 hints:
1) Since $\int_0^\infty f$ is finite you can choose any strictly decreasing sequence $a_n>0$ you like, tending to $0$, and  there will be $R_n>0$ with $\int_{R_n}^\infty f < a_n/n^2$. So $1/a_n \int^\infty_{R_n}f \le 1/n^2 $. Wlog $R_n<R_{n+1}$. Then you can obviously also estimate $\int_{R_n}^{R_{n+1}} f  $
2) it suffices to find $a$ as a step function, increasing strictly with each step, then interpolate.
Edit: Ok, the OP allowed to post a spoiler: let $0 < a_n < a_{n-1}$ be any strictly decreasing sequence tending to zero. Since for any $\varepsilon >0 $ we can find $R$ such that $\int_R^\infty f(x)\, dx < \varepsilon$ we can find a sequence $R_n$, wlog increasing, such that the inequality in 1) holds -- simply choose $\varepsilon = a_n/n^2$
Now define $\bar a(x) = 1/a_n$ for $R_n \le x < R_{n+1}$. Clearly $\bar a$ is increasing and tends to $\infty$ when $x$ does. Then
$$\int_{R_1}^\infty \bar a(x) f(x) dx = \sum_i \int_{R_i}^{R_{i+1}}\frac{f(x)}{a_i} < \sum_i\frac{1}{i^2}  <\infty $$
(Exchanging sum and integral can be easily justified by looking at finite sums first).
Now define $a$ as a piecewise affine linear function such that $a(0)=0$, $a(R_1) = 1/a_0$, furthermore $a(R_2)= \bar a(R_1)$ and in general $a(R_i)=\bar a(R_{i-1})$. Because $a_n$ is strictly decreasing, $ a$ will be strictly increasing, and clearly $a\le \bar a$ for $x\ge R_1$ (draw a picture). Since $f\ge 0$,
$$ \int_{R_1}^\infty  a(x) f(x) dx \le \int_{R_1}^\infty \bar a(x) f(x) dx < \infty$$
and $\int_0^{R1}af dx$ is clearly finite.
