Arithmetic Mean is defined by: $$ \bar{x} = \frac{\sum_{i=1}^{n} f(x_i)}{n} = \sum_{i=1}^{n} \frac{f(x_i)}{n} $$ Now, let's define the $x_i$ as, $x_{i+1}-x_{i} = \Delta x$ with $x_1 = a, x_n = b$ where $[a, b]$ is an interval in which $f(x)$ is integrable. Then clearly, $$x_n = x_1 + (n-1)\Delta x \Rightarrow b = a + (n-1)\Delta x \Rightarrow n = \frac{b-a+\Delta x}{\Delta x}$$
Plugging this value of n into the Arithmetic Mean formula gives: $$\Rightarrow \bar{x} = \sum_{i=1}^{n} \frac{f(x_i) \Delta x}{b-a+\Delta x}$$
Setting the limit $\Delta x \rightarrow 0$: $$ \Rightarrow \bar{x} = \lim_{\Delta x\to 0} \sum_{i=1}^{n} \frac{f(x_i)\Delta x}{b - a + \Delta x} $$ $$\Rightarrow \bar{x} = \int_{a}^{b} \frac{f(x)}{b - a}dx$$
This is how I convince myself why continuous arithmetic mean is defined this way. But I'm still bothered by the fact that, after applying the limit, I actually dismissed the $\Delta x$ in the denominator $b-a+\Delta x$ while still keeping it as $dx$ in the numerator.
What could be a valid argument to dismiss the $\Delta x$ in the denominator? How to prove that the sums $\sum \frac{f(x)\Delta x}{b-a+\Delta x}$ and $\sum \frac{f(x)\Delta x}{b-a}$ are the same for a sufficiently small $\Delta x$?
We aim to prove that a Riemann sum of the form $\displaystyle \sum_{i=0}^{n-1}\frac{hf\left(a+ih\right)}{\left(b-a\right)+h}$ converges to the same integral regardless of the $+h$ term in the denominator, where $h=\dfrac{b-a}{n}$. Observe that the denominator is independent of the index $i$ and can therefore be separated out of the sum and expanded as a geometric series in $h$. Hence the sum can be written as $\displaystyle \dfrac1{(b-a)+h}S=\frac1{b-a}S+\frac{-h/(b-a)}{b-a}S+\ldots$ The higher order terms vanish as $h\to 0$.