Integrand of convergent integral - absolute continuity is necessary?

I think the following proposition is true (proof offered below)

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 absolutely continuous on its domain except at a countable set of discontinuities and nonincreasing on its domain. Then $g(x) \to 0$ as $x \to \infty$.

Can the assumption '$f(x)$ is absolutely continuous on its domain except at a countable set of discontinuities' be weakened?

Remarks:

• This question generalizes an earlier question & answer in which $f(x)$ was assumed to be differentiable and nonincreasing.

• The condition of nonincreasing is necessary - as noted in the discussion of the earlier question and its predecessors, if $f(x)$ is allowed to increase then it could intermittently shoot up in 'spikes': the area of the spikes can be contrived to be finite, but the spikes prevent convergence of $f(x)$ to zero.

• As $f$ is monotone, its discontinuities form a set that is at most countable (Froda's Theorem, see also Shirali & Vasudeva (2019, Theorem 5.2.5) and this question & answer). Indeed $f$ is differentiable almost everywhere (Lebesgue's Theorem for the Differentiability of Monotone Functions, see also Shirali & Vasudeva (2019, Theorem 5.4.6)).

• The assumption '$f(x)$ is absolutely continuous on its domain except at a countable set of discontinuities' comes into the proof at Lemma 2, step 5. The overall strategy is to show that for any sequence $\left\{ x_n \right\}_n$ of strictly increasing numbers drawn from $(a,\infty)$, the sequence $\left\{ g(x_n) \right\}_n$ is a Cauchy sequence. Lemma 2, step 5 establishes that $g(x_j) - g(x_i)$ can be made arbitrarily small by making $x_i$ and $x_j$ sufficiently large ('upward jumps' of $g(x)$ become smaller as $x$ becomes large). Absolute continuity is invoked to obtain$$ \int_t^\infty x^r f(x) \, dx \geq \int_{[x_i, x_j]} \frac{g'(x)}{r+1} \, dx \geq g(x_j) - g(x_i)$$for $t > 0$, so taking $t \to \infty$ provides the requisite upper bound.

• The crucial step 5 cannot be applied without absolute continuity because it does not hold for a function $g(x)$ that increases but has $g'(x) = 0$ almost everywhere (for example the Cantor function). However this only means that the proof strategy won't apply, not that absolute continuity is necessary for the Proposition. But I haven't been able to generate a counterexample that affirms the requirement for absolute continuity.

Any suggestions or improvements would be gratefully received. Many thanks in advance.

Proof of Proposition

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 is a divergent integral but the lefthand side is a convergent integral by assumption. Contradiction, therefore $c = 0$. $\blacksquare$

Technical Lemma. Let function $h(x)$ be real-valued, nonnegative, and absolutely continuous on $[a,b]$ except for a discontinuity at $c \in (a,b)$. Suppose that $h(c^-) \geq h(c) \geq h(c^+)$ are all finite. Then for all $x_1, x_2 \in [a,b]$ where $x_1 \leq x_2$,$$h(x_2) - h(x_1) \leq \displaystyle\int_{[x_1,x_2]} h'(x) \, dx$$

Proof of Technical Lemma. Throughout the following, note that if$h(x)$ is absolutely continuous on $[s,t]$ then$$h(t) - h(s) = \displaystyle\int_{[s,t]} h'(x) \, dx$$There are five cases to check:

1. $x_1 \leq x_2 < c$. The function $h(x)$ is absolutely continuous on $[x_1,x_2]$ so$$h(x_2) - h(x_1) = \displaystyle\int_{[x_1,x_2]} h'(x) \, dx$$and the inequality follows trivially.
2. $x_1 < c = x_2$. Observe$$\begin{align*} h(x_2) - h(x_1) &= h(x_2) - h(x_1) - h(c^-) + h(c^-)\\ &= h(c^-) - h(x_1) + h(x_2) - h(c^-)\\ &= \int_{[x_1, c^-]} h'(x) \, dx + h(c) - h(c^-) & \text{as $x_2 = c$}\\ &= \int_{[x_1, x_2]} h'(x) \, dx - \left( h(c^-) - h(c) \right) & \text{noting the discontinity at $x = c$}\\ &\leq \int_{[x_1, x_2]} h'(x) \, dx & \text{as $h(c^-) \geq h(c)$}\end{align*}$$
3. $x_1 < c < x_2$. Observe$$\begin{align*} h(x_2) - h(x_1) &= h(x_2) - h(x_1) - h(c^+) + h(c^+) - h(c^-) + h(c^-)\\ &= h(x_2) - h(c^+) + h(c^-) - h(x_1) + h(c^+) - h(c^-)\\ &= \int_{[c^+, x_2]} h'(x) \, dx + \int_{[x_1, c^-]} h'(x) \, dx + h(c^+) - h(c^-)\\ &= \int_{[x_1, x_2]} h'(x) \, dx - \left( h(c^-) - h(c^+) \right) & \text{noting the discontinity at $x = c$}\\ &\leq \int_{[x_1, x_2]} h'(x) \, dx & \text{as $h(c^-) \geq h(c^+)$}\end{align*}$$
4. $x_1 = c < x_2$. Observe$$\begin{align*} h(x_2) - h(x_1) &= h(x_2) - h(x_1) - h(c^+) + h(c^+)\\ &= h(x_2) - h(c^+) - h(x_1) + h(c^+)\\ &= \int_{[c^+, x_2]} h'(x) \, dx - h(c) + h(c^+) & \text{as $x_1 = c$}\\ &= \int_{[x_1, x_2]} h'(x) \, dx - \left( h(c) - h(c^+) \right) & \text{noting the discontinity at $x = c$}\\ &\leq \int_{[x_1, x_2]} h'(x) \, dx & \text{as $h(c) \geq h(c^+)$}\end{align*}$$
5. $c < x_1 \leq x_2$. The function $h(x)$ is absolutely continuous on $[x_1,x_2]$ so$$h(x_2) - h(x_1) = \displaystyle\int_{[x_1,x_2]} h'(x) \, dx$$and the inequality follows trivially.

The inequality therefore holds in all cases. $\blacksquare$

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 absolutely continuous on its domain except at a countable set of discontinuities 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 six steps:

1. 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 $x \to \infty$.

2. If $f(x)$ is discontinuous at $c > a$ then $g(c^-) \geq g(c) \geq g(c^+)$. Since $f(x)$ is nonincreasing, if $x = c$ then$$ f(c^-) \geq f(c) \geq f(c^+)$$so$$ \lim_{x \to c^-} \frac{g(x)}{x^{r+1}} \geq \frac{g(c)}{c^{r+1}} \geq \lim_{x \to c^+} \frac{g(x)}{x^{r+1}} $$Let $d(x) = x^{r+1}$. Now $d(x)$ is continuous so$ \displaystyle \lim_{x \to c^-} d(x) = \lim_{x \to c^+} = d(c) = c^{r+1}$and $c > a \geq 0$ so $c^{r+1} \geq 0$so$$ c^{r+1} \left( \lim_{x \to c^-} \frac{g(x)}{x^{r+1}} \right)\geq c^{r+1} \left( \frac{g(c)}{c^{r+1}} \right)\geq c^{r+1} \left( \lim_{x \to c^+} \frac{g(x)}{x^{r+1}} \right)$$so$$ \left( \lim_{x \to c^-} x^{r+1} \right) \left( \lim_{x \to c^-} \frac{g(x)}{x^{r+1}} \right)\geq g(c)\geq \left( \lim_{x \to c^+} x^{r+1} \right) \left( \lim_{x \to c^+} \frac{g(x)}{x^{r+1}} \right)$$so$$ \lim_{x \to c^-} x^{r+1}\frac{g(x)}{x^{r+1}}\geq g(c)\geq \lim_{x \to c^+} x^{r+1}\frac{g(x)}{x^{r+1}}$$so$$ g(c^-)\geq g(c)\geq g(c^+)$$as required.

3. If $a < x_i \leq x_j$ then $g_j - g_i \leq \int_{[x_i, x_j]} g'(x) \, dx$. The function $f(x)$ is absolutely continuous on its domain except at a countable set of discontinuities, and the function $d(x) = x^{r+1}$ is absolutely continuous on any interval of the real line so $g(x) = f(x)d(x)$ is also absolutely continuous except at the discontinuities of $f(x)$. Write$$ \int_{[x_i, x_j]} g'(x) \, dx = \sum_{k} \int_{[s_k, t_k]} g'(x) \, dx$$where $\{ [s_k, t_k] \}_k$ is a countable set of intervals that partition $[x_i, x_j]$ and each interval $[s_k, t_k]$ contains one discontinuity of $g(x)$ on its interior. By step 2, applying the Technical Lemma to $g(x)$ yields$$\begin{align*} \int_{[x_i, x_j]} g'(x) \, dx &\geq \sum_{k} \bigl( g(t_k) - g(s_k) \bigr) \\ &= g(x_j) - g(x_i)\end{align*}$$where the last step comes from noting that the intervals partition $[x_i, x_j]$ so i) the first interval starts at $x_i$ and the last intervals finishes at $x_j$ and ii) the intervals meet at points where $g(x)$ is continuous.

4. $g'(x)$ exists almost everywhere on $x > a$ and $\displaystyle g'(x) \leq (r+1)\frac{g(x)}{x}$ whenever $g'(x)$ is defined. The function $f(x)$ is monotone on $(a,\infty)$ so it is differentiable almost everywhere on that interval. Indeed $f(x)$ is nonincreasing so if $f'(x)$ exists then $f'(x) \leq 0$. Moreover if $f'(x)$ exists then$$\begin{align*} f'(x) &= -\frac{r+1}{x^{r+2}}g(x) + \frac{1}{x^{r+1}}g'(x)\\&= \frac{1}{x^{r+1}}\left( g'(x) - (r+1)\frac{g(x)}{x} \right)\end{align*}$$Meanwhile $1/x^{r+1}$ is nonnegative. Therefore$\displaystyle g'(x) - (r+1)\frac{g(x)}{x} \leq 0$wherever $f'(x)$ exists and result follows immediately.

5. 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$$\begin{align*} \int_{t}^\infty x^r f(x) \, dx&\geq \int_{x_i}^{x_j} x^r f(x) \, dx & \text{as $f(x)$ is nonnegative}\\&= \int_{x_i}^{x_j} \frac{g(x)}{x} \, dx\\&= \int_{[x_i, x_j]} \frac{g(x)}{x} \, dx& \text{Lebesgue integral equals Riemann integral}\\&\geq \int_{[x_i, x_j]} \frac{g'(x)}{r+1} \, dx& \text{by step 4 and monotone property of Lebesgue integrals}\\&= \frac{g_j - g_i}{r+1}& \text{by step 3}\end{align*}$$and result follows immediately.

6. $\left\{ g_n \right\}_n$ 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 5, 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 absolutely continuous on its domain except at a countable set of discontinuities and nonincreasing on its domain. Then $g(x) \to 0$ as $x \to \infty$.

Proof of Proposition. Apply Lemma 2 and then Lemma 1. $\blacksquare$

References

Shirali S., Vasudeva H.L. (2019) Differentiation. In: Measure and Integration. Springer Undergraduate Mathematics Series. Springer, Cham. https://doi.org/10.1007/978-3-030-18747-7_5