Riemann Sums of Improper Integrals on Unbounded Domain

Question

Assume that $f:[0,\infty)\rightarrow[0,\infty)$ is decreasing (nonincreasing), continuous, and in $L^{1}[0,\infty)$, it is not hard to see that the following holds:

\begin{align*} \lim_{n\rightarrow\infty}\dfrac{1}{n}\sum_{i=1}^{\infty}f\left(\dfrac{i}{n}\right)=\int_{0}^{\infty}f(t)dt. \end{align*}

The problem that I have is that, if we let $\{0=x_{0}^{N}<x_{1}^{N}<\cdots\}$ be a partition of $[0,\infty)$ such that $x_{i}^{N}-x_{i-1}^{N}<1/N$, $i=1,2,...$, here $N=1,2,...$ is fixed and we consider the (infinite) Riemann sum $S_{N}$ at the stage $N$ of the form

\begin{align*} S_{N}=\sum_{i=1}^{\infty}f(x_{i}^{N})(x_{i}^{N}-x_{i-1}^{N}), \end{align*} must $S_{N}\rightarrow\displaystyle\int_{0}^{\infty}f(t)dt$ as $N\rightarrow\infty$?

Obviously $S_{N}\leq\displaystyle\int_{0}^{\infty}f(t)dt$, but how to proceed further then? For the special case that all $x_{i}^{N}$ are equal distributed with length $1/n$, we can easily get the lower bound as $\displaystyle\int_{1/n}^{\infty}f(t)dt$ and then Squeeze Theorem just finishes the job. But now I have no idea how to estimate if $x_{i}^{N}$ are not necessarily equal distributed.

Further thoughts: Writing that \begin{align*} S_{N}(x)=\sum_{i=1}^{\infty}f(x_{i}^{N})\chi_{[x_{i-1}^{N},x_{i}^{N})}(x), \end{align*} we have $S_{N}(x)\leq f(x)$ and it seems that $S_{N}(x)\rightarrow f(x)$ because of the continuity, if so, then we are ready to conclude.

Edit: The continuity implies the result. Now the question is, what if continuity is not given?

Answer 1

Since $\int_0^{\infty}f(t)dt<\infty$ for given any $\epsilon>0$ we can choose $a>0$ such that $\int_a^{\infty}f(t)dt<\epsilon$. Now for each $N$ choose integer $j_N$ such that $x_{j_{N}-1}≤a<x_{j_N}$. Notice that $$\sum_{i=j_N+1}^{\infty}f(x_i^N)(x_i^N-x_{i-1}^N)≤\int_a^{\infty}f(t)dt<\epsilon$$

Now $P_N:=\{x_0,....,x_{j_N-1}\}\cup \{a\}$ is a partition of $[0,a]$ such that norm of $P_n$ tends to zero as $N\rightarrow \infty$. Therefore $$\sum_{i=1}^{j_{N}-1}f(x_i^N)(x_j^N-x_{j_N-1}^N)+f(a)(a-x_{j_N-1})=L(P_N)\rightarrow \int_0^af(t)dt\ as \ N\rightarrow \infty$$

Next $(a-x_{j_{N}-1})\rightarrow 0\ as \ N\rightarrow \infty$. Therefore $$\sum_{i=1}^{j_{N}-1}f(x_i^N)(x_j^N-x_{j_N-1}^N)\rightarrow \int_0^af(t)dt, N\rightarrow \infty$$

Now we can choose $m\in \Bbb N$ such that whenever $N≥m$ we have $$|\sum_{i=1}^{\infty}f(x_i^N)(x_i^N-x_{i-1}^N)-\int_0^{\infty}f(t)dt|≤|\sum_{i=0}^{j_N-1}f(x_i)(x_i-x_{i-1})-\int_0^af(t)dt|+\sum_{i=j_N+1}^{\infty}f(x_i^N)(x_i^N-x_{i-1}^N)+\int_a^{\infty}f(t)dt≤\epsilon+\epsilon+\epsilon$$

Hence $lim_{N\rightarrow \infty}S_N =\int_0^{\infty}f(t)dt$

Main Tool used in this answer is that If $g:[c,d]\rightarrow \Bbb R$ is a Riemann integrable function on the compact interval $[c,d]$ and $\{P_n\}_n$ is a sequence of partitions of $[c,d]$ such that norm of $P_n$ (maximal length of subintervals corresponding to $P_n$) tends to zero as $n\rightarrow \infty$ , then $lim_{n\rightarrow \infty}U(P_n)=lim_{n\rightarrow \infty}L(P_n)=\int_c^dg(x)dx$, where $U(P_n),L(P_n)$ denote upper and lower Riemann sums of $g$ for $P_n$.