Integrand of convergent integral - absolute continuity is necessary?

Question

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

Answer 1

It appears that nonnegative and nonincreasing is sufficient! (Note slight change in domain of $f$)

Proposition. Let function $f(x)$ be real-valued, nonnegative, and nonincreasing on $[a,\infty)$, and suppose that for $r > -1$ the integral $\int_a^\infty x^r f(x) \, dx$ is convergent. Write $$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$ Then $g(x) \to 0$ as $x \to \infty$.

The strategy is:

• Apply the Lebesgue Decomposition Theorem for functions to decompose $f = f_A + f_Z$ where $f_A$, $f_Z$ are both nonnegative and nonincreasing, $f_A$ is absolutely continuous, and $f_Z$ has zero derivative almost everywhere. (In terms of the canonical decomposition $f = f_A + f_C + f_J$ where $f_C$ is singular and $f_J$ is a jump function, we have $f_Z = f_C + f_J$. Nonnegativity is assured by $f_C(a) = 0$ and $f_J(a) = 0$.)

• Write $f_A(x) = g_A(x)/x^{r+1}$. Show that the 'upward jumps' of $g_A$ have a decreasing bound: For all $\epsilon > 0$ there exists $t \geq a$ such that if $t \leq x_i \leq x_j$ then $g_A(x_j) - g_A(x_i) < \epsilon$. The bound comes from the integral $\int_a^\infty x^r f(x) \, dx$ being convergent.

• Write $f_Z(x) = g_Z(x)/x^{r+1}$. Show that 'upward jumps' of $g_Z$ have a decreasing bound: For all $\epsilon > 0$ there exists $t \geq a$ such that if $t \leq x_i \leq x_j$ then $g_Z(x_j) - g_Z(x_i) < \epsilon$. The bound again comes from the integral $\int_a^\infty x^r f(x) \, dx$ being convergent, but the manifestation of the bound is different.

• Let $g = g_A + g_Z$ and use all the existing machinery to show that $g(x) \to 0$ as $x \to \infty$.

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 $x \geq a$, $f(x)$ 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 $g(x) \to c$: $$ \text{there exists}\ x' \geq a \ \text{such that if}\ x > x' \ \text{then}\ |g(x) - c| < \epsilon $$ or equivalently $$ \text{there exists}\ x' \geq a \ \text{such that if}\ x > x' \ \text{then}\ c - \epsilon < g(x) < c + \epsilon $$ so in particular $$ \text{there exists}\ x' \geq a \ \text{such that if}\ x > x' \ \text{then}\ g(x) > c - \epsilon = \tfrac{1}{2}c $$ 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 $c = 0$. $\blacksquare$

Lemma 2. Let function $f(x)$ be real-valued, nonnegative, and nonincreasing on $[a,\infty)$. Write $$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$ Suppose further that 'upward jumps' of $g(x)$ have a decreasing bound: for all $\epsilon > 0$ there exists $t \geq a$ such that if $t \leq x_i \leq x_j$ then $g(x_j) - g(x_i) < \epsilon$. Then there exists $c$ finite such that $g(x) \to c$ as $x \to \infty$.

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 two 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. $\left\{ g_n \right\}_n$ 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 assumption, there exists $t > 0$ such that $$ \text{if}\ t \leq x_i \leq x_j \ \text{then}\ g_j - g_i < \frac{\epsilon}{2} \tag{1}\label{ans:eqn:boundjumpup} $$ 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 $$ \text{if}\ k \geq N \ \text{then}\ g_N - \frac{\epsilon}{2} < g_k \tag{2}\label{ans:eqn:boundbelowgN} $$ Meanwhile, applying (\ref{ans: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 $$ \text{if}\ k \geq N \ \text{then}\ g_k < g_N + \frac{\epsilon}{2} \tag{3}\label{ans:eqn:boundabovegN} $$ Now suppose $m,n > N$. Then (\ref{ans:eqn:boundbelowgN}) and (\ref{ans: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 (from the completeness of the real numbers) that there exists $c$ finite such that $g_n \to c$ as $n \to \infty$. Result follows from step 1. $\blacksquare$

Lemma 3. Let function $f(x)$ be real-valued, nonnegative, and nonincreasing 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 $[a,\infty)$. Then 'upward jumps' of $g(x)$ have a decreasing bound: for all $\epsilon > 0$ there exists $t \geq a$ such that if $t \leq x_i \leq x_j$ then $g(x_j) - g(x_i) < \epsilon$.

Proof of Lemma 3. In three steps:

1. $g(x)$ is absolutely continuous on $[a,\infty)$. The function $f(x)$ is absolutely continuous on $[a,\infty)$, 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 on $[a,\infty)$.

2. $\displaystyle g'(x) \leq (r+1)\frac{g(x)}{x}$ almost everywhere on $[a,\infty)$. The function $f(x)$ 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. Finally $f'(x)$ exists almost everywhere on $[a,\infty)$ as $f(x)$ is absolutely continuous on $[a,\infty)$. Result follows immediately.

3. 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 $$\begin{align*} \int_{t}^\infty x^r f(x) \, dx &\geq \int_{x_i}^{x_j} x^r f(x) \, dx & \text{as $f(x)$ is nonnegative} \\&= \int_{x_i}^{x_j} \frac{g(x)}{x} \, dx \\&= \int_{[x_i, x_j]} \frac{g(x)}{x} \, dx & \text{Lebesgue integral equals Riemann integral} \\&\geq \int_{[x_i, x_j]} \frac{g'(x)}{r+1} \, dx & \text{by step 2 and monotone property of Lebesgue integrals} \\&= \frac{g_j - g_i}{r+1} & \text{by $g(x)$ being absolutely continuous - step 1} \end{align*}$$ and result follows immediately. $\blacksquare$

Lemma 4. Let function $f(x)$ be real-valued, nonnegative, and nonincreasing 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 $f'(x) = 0$ for $x$ almost everywhere on $[a,\infty)$. Then 'upward jumps' of $g(x)$ have a decreasing bound: for all $\epsilon > 0$ there exists $t \geq a$ such that if $t \leq x_i \leq x_j$ then $g(x_j) - g(x_i) < \epsilon$.

Proof of Lemma 4. By contradiction: We show that if the converse is true then $\int_{a}^\infty x^r f(x) \, dx$ is divergent. In two steps:

1. Let $\epsilon > 0$ and $t \geq a$. If there exists $t \leq x_i \leq x_j$ with $g(x_j) - g(x_i) \geq \epsilon$ then $\int_{t}^\infty x^r f(x) \, dx \geq \epsilon/(r+1)$. Let $m = f(x_j^+)$ and $\ell = m x_i^{r+1}$. Then $f(x) \geq m$ for all $x \in [x_i, x_j]$ with equality if and only if $f'(x) = 0$ for all $x \in [x_i, x_j]$ (not merely $x$ almost everywhere), and in particular $\ell \leq g(x_i)$. Moreover $$\begin{align*} \int_{t}^\infty x^r f(x) \, dx &\geq \int_{x_i}^{x_j} x^r f(x) \, dx & \text{as $f(x)$ is nonnegative} \\ &\geq \int_{x_i}^{x_j} m x^r \, dx & \text{as $f(x) \geq m$ on $[x_i, x_j]$} \\&= \left[ m \frac{x^{r+1}}{r+1} \right]_{x_i}^{x_j} \\&= \frac{f(x_j^+) x_j^{r+1} - m x_i^{r+1}}{r+1} & \text{by $m = f(x_j^+)$} \\&= \frac{g(x_j) - \ell}{r+1} \\&\geq \frac{g(x_j) - g(x_i)}{r+1} & \text{as $\ell \leq g(x_i)$} \\&\geq \frac{\epsilon}{r+1} \end{align*}$$ as required.
2. Suppose there exists $\epsilon > 0$ such that for all $t \geq a$ there exists $t \leq x_i \leq x_j$ with $g(x_j) - g(x_i) \geq \epsilon$. Then $\int_{a}^\infty x^r f(x) \, dx$ is divergent. For the given $\epsilon$ and any $t \geq a$, step 1 establishes that $\int_{t}^\infty x^r f(x) \, dx \geq \epsilon/(r+1)$. This contradicts the convergence of $\int_{t}^\infty x^r f(x) \, dx \to 0$ as $t \to \infty$. $\blacksquare$

Proposition. Let function $f(x)$ be real-valued, nonnegative, and nonincreasing on $(a,\infty)$, and suppose that for $r > -1$ the integral $\int_a^\infty x^r f(x) \, dx$ is convergent. Write $$\displaystyle f(x) = \frac{g(x)}{x^{r+1}}$$ Then $g(x) \to 0$ as $x \to \infty$.

Proof of Proposition. Apply the Lebesgue Decomposition Theorem for functions to decompose $f = f_A + f_Z$ where $f_A$, $f_Z$ are both nonnegative and nonincreasing, $f_A$ is absolutely continuous, and $f_Z$ has zero derivative almost everywhere. (In terms of the canonical decomposition $f = f_A + f_C + f_J$ where $f_C$ is singular and $f_J$ is a jump function, we have $f_Z = f_C + f_J$. Nonnegativity is assured by $f_C(a) = 0$ and $f_J(a) = 0$.) Write $$ f_A(x) = \frac{g_A(x)}{x^{r+1}} \quad f_Z(x) = \frac{g_Z(x)}{x^{r+1}} $$ and note $g = g_A + g_Z$. Applying Lemma 3, then Lemma 2, and then Lemma 1 to $f_A(x)$ establishes that $g_A(x) \to 0$ as $x \to \infty$. Applying Lemma 4, then Lemma 2, and then Lemma 1 to $f_Z(x)$ establishes that $g_Z(x) \to 0$ as $x \to \infty$. Result follows. $\blacksquare$