Replacement of the tagged partition displacement of the Riemann Integral In specific, my question is if I can make the following by the definition on Riemann integral...
$$\int_a^bf(x)dx = \lim_{\lambda \rightarrow 0}\sum_{i=1}^n f(c_i)\Delta x_i = \lim_{\lambda \rightarrow 0}\sum_{i = 1}^n f(c_i)\lambda$$

That is, replacing the displacement $\Delta x_i$ with $\lambda$, where $\; \lambda = \max{\Delta x_i} \;\; (i = \overline{1,n})$

Where $f$ is a real-valued function defined on the interval $[a, b]$.
And the Riemann sum of $f$ with respect to the tagged partition $x_0, ..., x_n$ together with $t_0, ..., t_{n − 1}$
I think succeeded to prove it but I am not sure if this is right...
$$\forall_{(i = \overline{1, n})} \quad \lim_{\lambda \rightarrow 0} \Delta x_i = \lim_{\max{\Delta x_i \rightarrow 0}}\Delta x_i \leq \lim_{\max{\Delta x_i \rightarrow 0}}\max{\Delta x_i} = 0$$
$$\therefore \forall_{(i = \overline{1, n})} \quad 0 \leq \lim_{\max{\Delta x_i \rightarrow 0}}\Delta x_i \leq 0 \quad \Rightarrow \lim_{\lambda \rightarrow 0} \Delta x_i = \lim_{\lambda \rightarrow 0} \lambda$$
 A: This is obviously true if the partition is uniform and $\Delta x_i = \lambda$, but not true in general.
For a counterexample, consider the function $f : [1,e] \to \mathbb{R}$ where $f(x) \equiv 1$ and the partition with points $1 < e^{1/n} < e^{2/n} < \ldots <  e^{(n-1)/n}< e$. The partition norm is
$$\lambda= \max_{1 \leqslant i \leqslant n}\Delta x_i = \max_{1 \leqslant i \leqslant n}(e^{i/n} - e^{(i-1)/n}) = \max_{1 \leqslant i \leqslant n}e^{(i-1)/n}(e^{1/n} - 1) =e^{(n-1)/n}(e^{1/n} - 1) $$
Note that $\lim_{n \to \infty} e^{(n-1)/n}(e^{1/n} - 1) = e\cdot 0 = 0$ and $\lambda \to 0$ if and only if $n \to \infty$.
For any choice of intermediate points $c_i$, we have
$$\sum_{i=1}^n f(c_i) \lambda = \sum_{i=1}^n [1 \cdot e^{(n-1)/n}(e^{1/n} - 1)] = ne^{(n-1)/n}(e^{1/n} - 1),$$
and
$$\lim_{\lambda \to 0}\sum_{i=1}^n f(c_i) \lambda = \lim_{n \to \infty} ne^{(n-1)/n}(e^{1/n} - 1) = \lim_{n \to \infty} e^{(n-1)/n}\cdot \lim_{n \to \infty}\frac{e^{1/n} - 1}{1/n} = e$$
However,
$$\int_1^e f(x) \, dx = e-1\neq \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i) \lambda = e$$

Since $\lambda \geqslant \Delta x_i$, it is true in general for $f \geqslant 0$ that
$$ \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i) \lambda = \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i)\, \Delta x_i \frac{ \lambda}{\Delta x_i}\geqslant \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i)\, \Delta x_i = \int_a^b f(x) \, dx$$
The counterexample clearly satisfies that condition.
