Improper integral convergence implies the existence of an infinite series which its partial sum converges

Question

Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a Riemann-integrable function at $[0, \beta]$ for each $\beta \in (0, \infty)$.
Suppose that $\forall x \in [0, \infty): \space f(x) \geq 0$.

If the improper integral $\int_{0}^{\infty}f(x) \space dx$ converges, then there exists a sequence $(\xi_{k})_{k\in \mathbb{N}}$ s.t $\lim_{k\to\infty}\xi_k=\infty$ and $\sum_{k=1}^{\infty}f(\xi_{k})$ converges.

If for each $k \in \mathbb{N}$ the infimum at $f([k,k+1])$ is a minimum (.e.g $f$ is continues or monotonic) this was pretty straight for me to reason. I followed my intuition and this is what I came to so far:

Let $k \in \mathbb{N}$ and consider the interval $[k, k+1]$.
Since $f$'s integrable, there exists a partition $\mathscr{P} = \{x_0, \dots, x_n \}$ of $[k,k+1]$ s.t $\sum_{j=0}^{n}(x_{j}-x_{j-1})(M_{j}-m_{j}) < \frac{1}{k^2}$
where $M_j, m_j$ are the supremum and infimum of $f([x_{j-1}, x_j])$ correspondingly and $j = 1,\dots, n$.

Then I argue that there exists some $l \in \{1,\dots, n\}$ s.t $M_l - m_l \leq \frac{1}{k}$ (otherwise we get $k^2 > k$ and choose $\xi_k = x_l$.
In this I get a sequence $(\xi_k)_{k\in\mathbb{N}}$ but I'm struggling to reason why it converges, since I've only relied intuition to build it.

In case I'm completely wrong with the intuition too I'd like to know why, and anyways, I'd like to have some guidance.
Thanks.

Answer 1

You said:

If for each $k \in \mathbb{N}$ the infimum at $f([k,k+1])$ is a minimum [...] this was pretty straight for me to reason.

With a small modification, this idea can be used even if $f$ does not attain its infimum on those intervals.

Let $(a_k)_{k \ge 1}$ be an arbitrary sequence of positive real numbers such that $\sum_{k=1}^\infty a_k$ is convergent (for example, $a_k = 1/k^2$). In every interval $[k-1, k]$ there is a $\xi_k$ with $$ f(\xi_k) \le \inf \{ f(x) \mid k-1 \le x \le k \} + a_k \, . $$ Also $$ \inf \{ f(x) \mid k-1 \le x \le k \} \le \int_{k-1}^k f(x) \, dx \, . $$ Then $\xi_k \to \infty$. For every positive integer $n$ is $$ \sum_{k=1}^{n} f(\xi_k) \le \sum_{k=1}^{n} a_k + \int_0^n f(x) \, dx $$ and since the right-hand side is convergent for $n \to \infty$, the convergence of $\sum_{k=1}^\infty f(\xi_k)$ follows.

Remark: $\sum_{k=1}^\infty a_k$ can be made arbitrarily small, so the proof shows that for every $\epsilon > 0$ there is a sequence $(\xi_k)$ with $\xi_k \to \infty$ and $\sum_{k=1}^\infty f(\xi_k) < \int_0^\infty f(x) \, dx + \epsilon$.

Answer 2

If $\int f$ converges, then for any $\epsilon >0$ and any $x$ ($>0$) there must exist some $\xi \ge x$ such that $f(\xi) < \epsilon$.

(If not, then $f(\xi) \ge \epsilon$ for all $\xi \ge x$ which would contradict convergence.)

Now choose $\xi_k$ such that $\xi_{k+1} \ge \xi_k +1$ and $f(\xi_k) < {1 \over 2^k}$.