Upper bound on integrand that converges to zero? Let $H(x)$ be the cumulative distribution function for a continuous nonnegative random variable $X$ that has a finite mean.
Let $\tilde{H}(x) = 1 - H(x)$ (the tail probability).
For $r \geq 1$, suppose that the integral
$\int_{0}^{\infty} x^r \tilde{H}(x) \, dx$ is convergent. (Remark: This condition arises when the $(r+1)$th moment of $X$ is finite. While the integral is indeed convergent for $r=0$ by assumption of $X$ having a finite mean, this case is not being considered here.)
Does there necessarily exist a function $B(x)$ such that $\tilde{H}(x) \leq \displaystyle\frac{B(x)}{x^{r+1}}$ where $B(x) \to 0$ (monotonically decreasing) as $x \to \infty$?
Remarks
As this is my third attempt to formulate a proposition that can do what I’m seeking, here is what I have tried:

*

*I am aware that I cannot necessarily claim that $x^r \tilde{H}(x) \to 0$ as this does not necessarily hold for improper integrals, even if they converge.  I can make that conclusion if it is established that $x^r \tilde{H}(x) \to \ell$ for some finite limit $\ell$, but the existence of $\ell$ does not necessarily follow solely from $x^r$ being monotone increasing to infinity and $\tilde{H}(x)$ being monotone decreasing to zero.


*But if  $\tilde{H}(x) = \displaystyle\frac{B(x)}{x^{r+1}}$ with $B(x) = c>0$ constant then the integral
$\int_{0}^{\infty} x^r \tilde{H}(x) \, dx$ is divergent.
That $\tilde{H}(x)$ decreases monotonically to zero (from being a tail probability) seems to make all the difference. I tried constructing a counterexample for $r = 1$ in which
$$
  \displaystyle
  \int_{0}^{\infty} x \tilde{H}(x) \, dx
  =
  \int_{0}^{\infty} D(x) \, dx
$$
is convergent but $D(x)$ is nondecreasing. I thought to construct
$$
  \tilde{H}(x) = \frac{D(x)}{x}
$$
where $D$ is mostly zero except that at each integer $n$ there is a triangle of height 1 and area $1/n^2$ (so $\int_{0}^{\infty} D(x) \, dx$ converges but $D(x)$ does not decrease to zero). However
$$
  \tilde{H}'\!(x) = \frac{x D'\!(x) - D(x)}{x^2}
=
  \frac{1}{x}\left( D'\!(x) - \tilde{H}(x) \right)
$$
whenever $\tilde{H}(x)$ is differentiable.  Thus as $x$ increases and hence $\tilde{H}(x)$ decreases, the gradient that I can introduce via $D'\!(x)$ must also decrease to preserve $\tilde{H}'\!(x) \leq 0$.  So in particular, if I tried to specify that $D(n) = 1$ for each positive integer $n$ then I would be restricted to $D'\!(x) \leq 1/n$ for the triangle at $n$, which leads to a triangle of area $\geq 1/2n$.  Doing so causes $\int_{0}^{\infty} D(x) \, dx$ to diverge.
(Many thanks again.)
 A: This answer requires that $\tilde{H}(x)$ is differentiable on an interval $(a,\infty)$.  In the following analysis, $\tilde{H}(x)$ has been replaced with $f(x)$ to make the assumptions more explicit and traceable.  I have refactored the logic into small units to increase reusability.
Lemma 1. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r  > -1$ the integral
$\int_a^\infty x^r f(x) \, dx$ is convergent.
Write
$$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$
Suppose further that $g(x) \to c$ as $x \to \infty$ where $c$ is finite. Then $c = 0$.
Proof of Lemma 1. For $x > a$, $f(x)$ is nonnegative so $g(x)$ is nonnegative hence $c \geq 0$.
Now suppose that $c > 0$ and let $\epsilon = \tfrac{1}{2}c$. Then $\epsilon > 0$, so from $g(x) \to c$:
$$
 \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ |g(x) - c| < \epsilon
$$
or equivalently
$$
 \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ c - \epsilon < g(x) < c + \epsilon
$$
so in particular
$$
 \text{there exists}\ x' > a \ \text{such that if}\ x > x' \ \text{then}\ g(x) > c - \epsilon = \tfrac{1}{2}c
$$
Therefore
$$\begin{align*}
  \int_a^\infty x^r f(x) \, dx
&=
  \int_a^\infty \frac{g(x)}{x} \, dx
\\&=
  \int_a^{x'} \frac{g(x)}{x} \, dx
  +
  \int_{x'}^\infty \frac{g(x)}{x} \, dx
\\&\geq
  \int_a^{x'} \frac{g(x)}{x} \, dx
  +
  \int_{x'}^\infty \frac{c}{2x} \, dx
\end{align*}$$
Now the second term on the righthand side diverges to infinity but the lefthand side is a convergent integral by assumption. Contradiction, therefore $c = 0$. $\blacksquare$
Lemma 2. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r  > -1$ the integral
$\int_a^\infty x^r f(x) \, dx$ is convergent.
Write
$$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$
Suppose further that $f(x)$ is differentiable and nonincreasing on its domain.
Then there exists $c \geq 0$ (finite) such that $g(x) \to c$.
Proof of Lemma 2.
Let $\left\{ x_n \right\}_n$ be any strictly increasing sequence of numbers drawn from $(a,\infty)$. Construct $\left\{ g_n \right\}_n$ by putting $g_n = g(x_n)$ for each $n$. Proceeding in four steps:

*

*It is sufficient to show that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. The function $f(x)$ is nonincreasing so its limit as $x \to \infty$ is either finite or $-\infty$. Furthermore $f(x)$ is nonnegative so the option of $-\infty$ is excluded and the limit must be nonnegative. Meanwhile the function $d(x) = x^{r+1} \to \infty$ as $x \to \infty$. Now $g(x) = f(x)d(x)$ therefore $g(x)$ either has a finite, nonnegative limit or it diverges to infinity. So suppose there exist sequences $\left\{ x_n \right\}_n$
and $\left\{ x'_m \right\}_m$ such that if $g_n = g(x_n)$ for each $n$ and $g'_m = g(x'_m)$ for each $m$ then $g_n \to c$ as $n \to \infty$ and $g'_m \to c'$ as $m \to \infty$ but $c \neq c'$. Construct sequence $\left\{ x''_k \right\}_k$ by merging $\left\{ x_n \right\}_n$ and $\left\{ x'_m \right\}_m$ in increasing order.  Then as $k \to \infty$, $g(x''_k)$ does not converge. Contradiction hence $c = c'$. Therefore if $g_n \to c$ finite as $n \to \infty$ for any given sequence $\left\{ x_n \right\}_n$ then $g(x) \to c$ as $x \to \infty$.


*$\displaystyle g'(x) \leq (r+1)\frac{g(x)}{x}$ for all $x > a$. The function $f(x)$ is differentiable on $(a,\infty)$. Indeed
$$\begin{align*}
  f'(x) &= -\frac{r+1}{x^{r+2}}g(x) + \frac{1}{x^{r+1}}g'(x)
\\&=
  \frac{1}{x^{r+1}}\left(
    g'(x) - (r+1)\frac{g(x)}{x}
  \right)
\end{align*}$$
on that interval. Now for $x > a$, $f(x)$ is nonincreasing so $f'(x) \leq 0$. Meanwhile $1/x^{r+1}$ is nonnegative. Therefore
$\displaystyle g'(x) - (r+1)\frac{g(x)}{x} \leq 0$
for $x > a$ and result follows immediately.


*For all $\epsilon > 0$ there exists $t > 0$ such that if $t \leq x_i \leq x_j$ then $g_j - g_i < \epsilon$.  Let $\epsilon > 0$.  The integral
$\int_a^\infty x^r f(x) \, dx$
is convergent so there exists $t$ such that
$$
  \int_{t}^\infty x^r f(x) \, dx < \frac{\epsilon}{r+1}
$$
noting that $r+1 > 0$ by assumption.
Now if $t \leq x_i \leq x_j$ then
$$\begin{align*}
  \int_{t}^\infty x^r f(x) \, dx
&\geq
  \int_{x_i}^{x_j} x^r f(x) \, dx  
  & \text{as $f(x)$ is nonnegative}
\\&=
  \int_{x_i}^{x_j} \frac{g(x)}{x} \, dx
\\&\geq
  \int_{x_i}^{x_j} \frac{g'(x)}{r+1} \, dx
  & \text{by step 2}
\\&=
  \frac{g_j - g_i}{r+1}
\end{align*}$$
and result follows immediately.


*$\left\{ g_n \right\}_n$ is a Cauchy sequence.  We need to show that for any $\epsilon > 0$ there exists $N$ such that if $m,n > N$ then $|g_n - g_m| < \epsilon$.  So let $\epsilon > 0$.  By step 3, there exists $t > 0$ such that
$$
  \text{if}\ t \leq x_i \leq x_j \ \text{then}\
  g_j - g_i < \frac{\epsilon}{2}
  \tag{1}\label{eqn:boundjumpup}
$$
Let $\ell = \inf\left\{ g_n : x_n \geq t \right\}$.
Now $g(x)$ is bounded below (by zero) so $\ell$ is finite.
Hence by the definition of infimum, there exists $N$ such that $x_N \geq t$ and
$\displaystyle g_N < \ell + \frac{\epsilon}{2}$.
Indeed
$\displaystyle g_N - \frac{\epsilon}{2} < \ell$.
Moreover, to emphasize the construction of $\ell$, if $k \geq N$ then $\ell \leq g_k$. Consequently
$$
  \text{if}\ k \geq N \ \text{then}\
  g_N - \frac{\epsilon}{2} < g_k
  \tag{2}\label{eqn:boundbelowgN}
$$
Meanwhile, applying (\ref{eqn:boundjumpup}) with $i = N$ and $j = k$  yields
$$
  \text{if}\ k \geq N \ \text{then}\
  g_k - g_N < \frac{\epsilon}{2}
$$
or equivalently
$$
  \text{if}\ k \geq N \ \text{then}\
  g_k < g_N + \frac{\epsilon}{2}
  \tag{3}\label{eqn:boundabovegN}
$$
Now suppose $m,n > N$. Then (\ref{eqn:boundbelowgN}) and (\ref{eqn:boundabovegN}) yield
$$\begin{align*}
  g_N - \frac{\epsilon}{2} &< g_m < g_N + \frac{\epsilon}{2} \quad \text{and}
\\
  g_N - \frac{\epsilon}{2} &< g_n < g_N + \frac{\epsilon}{2}
\end{align*}$$
therefore $|g_n - g_m| < \epsilon$ as required.
Having established that $\left\{ g_n \right\}_n$ is a Cauchy sequence of real numbers, it follows immediately that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. Result follows from step 1.  $\blacksquare$
Proposition. Let function $f(x)$ be real-valued and nonnegative on $(a,\infty)$, and suppose that for $r  > -1$ the integral
$\int_a^\infty x^r f(x) \, dx$ is convergent.
Write
$$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$
Suppose further that $f(x)$ is differentiable and nonincreasing on its domain. Then $g(x) \to 0$ as $x \to \infty$.
Proof of Proposition. Apply Lemma 2 and then Lemma 1. $\blacksquare$
