An alternative way to define improper integrals Improper Riemann integrals are usually defined via limits. 
Standard Definition: Let $f:[0, \infty) \to \mathbb{R}$. We say $f$ is improper Riemann integrable on $[0, \infty)$ if it is proper Riemann integrable on compact intervals and the following limit exists
$$ \lim_{t \to \infty} \int_{0}^{t} f(x) \ dx$$
Instead of this, can we define the improper integral with "partitions" of $[0, \infty)$ as we do with the normal Riemann integral? 
Preliminaries: A partition  of $[0, \infty)$ is a strictly increasing sequence $p:\mathbb{N}_{\geq 1} \to [0, \infty)$ with $p(1) = 0$ and $$\lim_{n \to \infty} p(n) = +\infty$$
A tagging of a given partition $p$ is any sequence $t:\mathbb{N}_{\geq 1} \to [0, \infty)$ such that $t(n) \in [p(n), p(n+1)]$ for all $n \geq 1$. 
A refinement of a partition $p$ is a partition $p'$ which contains $p$ as a subsequence. 
The mesh of a partition $p$ is the quantity $\sup\{p(n+1) - p(n): n \in \mathbb{N}_{\geq 1}\}$. It is denoted as $||p||$. $||p|| = +\infty$ is possible.
There's two ways we can go about our definition. 
Definition 1: Let $f:[0, \infty) \to \mathbb{R}$. We say $f$ is improper Riemann integrable on $[0, \infty)$ if there is a real number $L$ such that for all $\epsilon>0$ there is a partition $p_{\epsilon}$ such that for every refinement $p_{\epsilon}'$ of $p_{\epsilon}$ and any tagging $t$ of $p_{\epsilon}'$ the sum 
$$S(f, p_{\epsilon}', t) \stackrel{\text{def}}{=} \sum_{n=1}^{\infty} [p_{\epsilon}'(n+1) - p_{\epsilon}'(n)]f(t(n))$$ converges and 
$$|L - S(f, p_{\epsilon}', t)| < \epsilon$$
Definition 2: Let $f:[0, \infty) \to \mathbb{R}$. We say $f$ is improper Riemann integrable on $[0, \infty)$ if there is a real number $L$ such that for all $\epsilon>0$ there is a $\delta>0$ such that for every partition $p_{\delta}$ with $||p_{\delta}||<\delta$ and any tagging $t$ of $p_{\delta}$, the sum 
$$S(f, p_{\delta}, t) \stackrel{\text{def}}{=} \sum_{n=1}^{\infty} [p_{\delta}(n+1) - p_{\delta}(n)]f(t(n))$$ converges and 
$$|L - S(f, p_{\delta}, t)| < \epsilon$$

Problem: Are all of these definitions equivalent? 

Partial answers are fine. 
 A: Neither definition is valid.  
Even on a finite interval, a convergent improper integral is not guaranteed to be computable as the limit (in any sense) of Riemann sums.  Only if restrictions like monotonicity and carefully selected tags are imposed is there any hope of this working.  A good example is the convergent improper integral 
$$\int_0^1 \frac{1}{x} \sin \frac{1}{x} \, dx = \int_1^\infty \frac{\sin x }{x} \, dx \approx 0.627143$$
discussed here and here.
This is essentially due to a double limiting process where limits are not interchangeable.
For a counterexample pertaining to an infinite interval and Definition 1, consider a sequence of dyadic partitions 
$$P_{n,m} = (0,1/2^n, 2/2^n, \ldots,1, \ldots, m-1, m-1 + 1/2^n, \ldots,m)$$
With $m$ fixed, successive partitions are refinements as $n$ is increased.
Take a function $f$ where $f(k) = 1$ at any integer $k$ and $f(x) = 0$ otherwise. It is true that
$$0 = \int_0^\infty f(x) \, dx = \lim_{m \to \infty}\int_0^mf(x) \, dx =\lim_{m \to \infty} \lim_{n \to \infty} \frac{1}{2^n} \sum_{k=1}^{m2^n} f(k/2^n) = \lim_{m \to \infty} \lim_{n \to \infty}\frac{m}{2^n},$$
but
$$\lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{2^n} \sum_{k=1}^{m2^n} f(k/2^n) = \lim_{n \to \infty} \frac{1}{2^n}\sum_{k=1}^{\infty} f(k/2^n) = +\infty$$ 
A: I think certain smoothness conditions on $f$ could help us.
For each $N$, break the interval $[0,N]$ into $N^2$ sub-intervals of length $1/N$. I want the difference $$\int_{0}^{\infty}f(x)dx-\sum_{m=1}^{N^2}\frac{1}{N}f(x_m)$$ to converge to $0$ as $N\rightarrow\infty$.
(Instead of taking a double limit in $2$ variables I'm taking a single limit in $N$.)
This sum can be written as $$\int_{0}^{N}f(x)dx-\sum_{m=1}^{N^2}\frac{1}{N}f(x_m)+\int_{x>N}f(x)dx$$
Now for integrable functions, the last term would converge to $0$. So we are interested only in $$\int_{0}^{N}f(x)dx-\sum_{m=1}^{N^2}\frac{1}{N}f(x_m)$$
which can be broken up as sums of $$\int_{k/N}^{(k+1)/N}f(x)dx-\frac{1}{N}f(x_m)$$
$$=\int_{k/N}^{(k+1)/N}[f(x)-f(x_m)]dx$$
$$=\int_{k/N}^{(k+1)/N}f'(y_m)[x-x_m]dx$$
$$\leq\int_{k/N}^{(k+1)/N}|f'(y_m)|\frac{1}{N}dx$$
Suppose $f'$ satisfies the bound $$|f'(x)|\leq\frac{M}{1+x^2}$$
Then if we choose the tags $x_m$ to be the largest points in our intervals we get $$x<y_m<x_m$$ In that case the above integral is bounded by $$\frac{1}{N}\int_{k/N}^{(k+1)/N}\frac{M}{1+x^2}dx$$
Then our original sum is bounded by $$\frac{1}{N}\int_0^{\infty}\frac{M}{1+x^2}dx=\frac{M\pi}{2N}\rightarrow 0$$ as $N\rightarrow\infty$.
