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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?

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    I reckon you could construct $a$ explicitly. I don't know if this works, but it feels like it might: break the integral up into pieces, set $a$ equal to 1 on the first piece, 2 on the second piece, 4 on the third piece, $2^n$ on the $n^\mathrm{th}$ piece, and then choose your piece cuttings so that the terms still decay faster than geometrically. (the first piece contains three quarters of the integral's value, the second contains three quarters of the remaining quarter, that sort of thing). (if this does work, fill in the details and post an answer yourself, or someone else can.)2012-06-29

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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. 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.

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    @Thomas I've been staring at your answer, and I think I understand it all now. I hope I can apply this type of analysis to more problems. Thank you very much :)2012-06-29