measure theory - Asymptotics of integrals with respect to asymptotically equivalent distribution functions

Let us assume that $f\colon[0,\infty)\to[0,\infty)$ is a non-decreasing continuous function with

$$ \frac{f(t)}{t}\xrightarrow{t\to\infty}1. \tag{A}\label{A}$$

We then have the following elementary Lemma:

For all $k\in\mathbb N_0$ we have
$$ \frac{\int_0^t s^k df(s)}{\int_0^t s^k ds}\xrightarrow{t\to\infty}1. $$

For the sake of completeness, a detailed elementary proof is given below which makes explicit use of the very simple nature of the integrand. However, it seems plausible that such a property should hold for much more general functions in place of $s^k$. In particular, the following statement seems reasonable to me.

Conjecture: If $h\colon[0,\infty)\to[0,\infty)$ is measurable and
$$ \liminf_{t\to\infty}h(t)>0$$
(and maybe something else), then we have
$$ \frac{\int_0^t h(s) df(s)}{\int_0^t h(s) ds}\xrightarrow{t\to\infty}1. $$

Since $h$ cannot "lose mass at infinity", this property would give meaning to the idea that if a distribution function of a measure $\mu$ on $\mathcal B([0,\infty))$ is asymptotically equivalent to the identity (which is the distribution function of Lebesgue's measure), then "$\mu$ is asymptotically equivalent to Lebesgue's measure".

Of course, further natural steps would be to replace Lebesgue's measure by any other locally finite measure $\nu$ on $\mathcal B([0,\infty))$ with a continuous distribution function and then compare $\mu$ and $\nu$ asymptotically. Then one could ask if continuity can be dropped. But for the moment, I guess the above conjecture is nice enough.

Does anyone know a reference where something like this is stated (and proved)? Or does anyone have an idea how to prove it? Thanks a lot in advance!

---

Proof of the elementary Lemma: Using the Stieltjes product formula, we can write
\begin{align*}
\frac{\int_0^t s^k df(s)}{\int_0^t s^k ds}&=\frac{k+1}{t^{k+1}} \int_0^t s^k df(s) \\
&=\frac{k+1}{t^{k+1}} \left( t^kf(t)-\int_0^t f(s)ds^k \right) \\
&=(k+1)\frac{f(t)}{t}-\frac{k+1}{t^{k+1}} \int_0^t f(s)ks^{k-1}ds \\

&=(k+1)\frac{f(t)}{t}-k-\frac{k(k+1)}{t^{k+1}} \int_0^t (f(s)-s)s^{k-1}ds.
\end{align*}
By \eqref{A}, the first summand converges to $k+1$ for $t\to\infty$, so it remains to prove that
$$g(t):=\frac{k+1}{t^{k+1}} \int_0^t (f(s)-s)s^{k-1}ds$$
vanishes for $t\to\infty$. Let $\epsilon>0$. According to \eqref{A}, there is some $t_0\in(0,\infty)$ such that
$$\left|\frac{f(s)}{s}-1\right|\le\epsilon \quad \text{for all $s\in[t_0,\infty)$.}$$
Then
\begin{align*}
|g(t)|&\le \frac{k+1}{t^{k+1}} \int_0^{t_0} |f(s)-s|s^{k-1}ds + \frac{k+1}{t^{k+1}} \int_{t_0}^t (s\epsilon)s^{k-1}ds \\
&\le \frac{k+1}{t^{k+1}} \int_0^{t_0} |f(s)-s|s^{k-1}ds + \epsilon

\end{align*}
for all $t\in[t_0,\infty)$. As the integral in the last step is finite and does not depend on $t$, this shows that
$$\limsup_{t\to\infty}|g(t)|\le\epsilon,$$
which completes the proof, since $\epsilon$ was chosen arbitrarily.

Answer

Here are some counter-examples: Let $f(t) = t+\sin t$. Then clearly $f(t)/t \to 1$ as $t \to \infty$.

• If $h(s) = e^s$, then

  
  
  
  

  $$ \frac{\int_{0}^{t}h(s)\,f(\mathrm{d}s)}{\int_{0}^{t}h(s)\,\mathrm{d}s} = 1 + \frac{\cos t+\sin t}{2} + \mathcal{O}(e^{-t}). $$




• If $h(s) = 2+\cos s$, then

  
  
  

  $$ \frac{\int_{0}^{t}h(s)\,f(\mathrm{d}s)}{\int_{0}^{t}h(s)\,\mathrm{d}s} = \frac{5}{4} + \mathcal{O}\left(\frac{1}{t}\right). $$