Improper integrals and right-hand Riemann sums

Suppose that $\forall \varepsilon\in(0,1)$ there exists Riemann integral $\int\limits_{\varepsilon}^1f(x)dx$. Is it true, that the improper integral $\int\limits_{0}^1f(x)dx$ can be defined as $\lim\limits_{\Delta\to+0}\sum\limits_{k=1}^nf(x_{k})\left(x_k-x_{k-1}\right)$, where $0=x_0 < x_1 < \cdots < x_n=1$ and $\Delta=max(x_k-x_{k-1}), 1\le k\le n$? I mean, will the limit of right-hand Riemann sums be equivalent to the standart definition of improper Riemann integral in this case? It is easy to show, that these two definitions are equivalent when $f$ is a monotonic function, but I could not find such a statement for arbitrary $f$, along with any counterexample. The positive answer in case, when $f$ is Riemann-integrable on $[0,1]$, was given here: "Ito-Riemann" integration, but I'm interested in improper integration.

In general, the limit of the right-hand Riemann sums need not exist.

Consider for a counterexample $f(x) = \frac{1}{x}\sin \frac{1}{x}$. It is clear that $\int_\varepsilon^1 f(x)\,dx$ exists for all $0 < \varepsilon < 1$, and the substitution $u = \frac{1}{x}$ shows that the improper Riemann integral

$$\int_0^1 f(x)\,dx := \lim_{\varepsilon \downarrow 0} \int_\varepsilon^1 f(x)\,dx$$

exists.

But for any partition $0 = x_0 < x_1 < \dotsc < x_n = 1$ with mesh $\Delta < \delta$, we can insert a further partition point $y$ between $x_0$ and $x_1$, and the corresponding sum is

$$ y\cdot f(y) + (x_1-y)f(x_1) + \sum_{k=2}^n f(x_k)(x_k-x_{k-1}) = y\left(f(y) - f(x_1)\right) + \sum_{k=1}^n f(x_k)(x_k-x_{k-1}).$$

The difference between the two right-hand Riemann sums is

$$y\left(f(y) - f(x_1)\right) = \sin \frac{1}{y} + \frac{y}{x_1}\sin \frac{1}{x_1},$$

which can be made arbitrarily close to $\pm 1$ by choosing $y$ close to $0$ with $\sin \frac{1}{y} = \pm 1$.

And here is a counterexample for a non-negative function: let $g(\displaystyle\frac {1}{2^n}):=1, g(x):=0$ $ \forall x\ne \displaystyle\frac {1}{2^n}, n\in N;$ and $f(x):=\displaystyle{\frac{g(x)}{x}}$. It is obvious, that there exists improper Riemann integral $\int\limits_0^1 f(x)\,dx =0$, but $y(f(y)-f(x_1))=g(y)-yf(x_1)$ can be made arbitrarily close to $1$ by choosing $y=\displaystyle\frac {1}{2^k}$ close to $0$.