# 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?

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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.) –  Ben Millwood Jun 29 '12 at 14:00

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.

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I remember I read the same question for series: If the positive series $\sum_n a_n$ converges, then there exists a positive sequence $\{b_n\}_n$, strictly increasing to $+\infty$, such that $\sum_n a_n b_n$ still converges. I can't find a reference, now. –  Siminore Jun 29 '12 at 14:11
@Siminore Yes, this is a common exercise in several analysis 1 lectures in Germany (and probably all over the world, usually under the heading 'there is no slowest growing series), so I'd definitely expect that this is answered here. But how to find it? Anyway, $\sum$ is nothing but $\int$ if you choose the appropriate measure :-) –  user20266 Jun 29 '12 at 14:15
@Siminore: as another remark, the OP asked for a hint, not a solution. Any reference is likely to spoil his fun. And I would get no upvotes ;-) –  user20266 Jun 29 '12 at 14:26
I guess that the "theorem" for series is more clear, and might be useful to understand the strategy for the continuous case :-) –  Siminore Jun 29 '12 at 14:28
@Thomas thank you for your answer. I was also thinking about approximating $a(x)$ with simple functions, but I didn't have a clear idea on how to use that to my advantage. If you don't mind, would you help me understand why your hint in #1 is justified? I actually don't mind a full solution (I have kind of lost hope) if you wouldn't mind providing one. :) –  ae0709 Jun 29 '12 at 15:59