Applying discrete calculus to prove the results of the classical calculus Can the discrete calculus be used as a way of proving theorems of the classical calculus? For example, isn't the following a proof of the fundamental theorem of calculus by means of the discrete calculus?
The discrete derivative can be defined as:
$\Delta f(x) = {f(x+h)-f(x) \over h}$
Similarly, the discrete integral can be defined as:
$\sum\limits_{x:a \to b} f(x) = hf(a) + hf(a+h) + hf(a+2h) + ... + hf(b-h)$
And here's a some kind of the fundamental theorem of calculus in the discrete world:
$\sum\limits_{x:a \to b} \Delta f(x) = (f(a+h)-f(a)) + (f(a+2h)-f(a+h)) + ... + (f(b)-f(b-h))=f(b) - f(a)$
Isn't the following true in such case?
$\int\limits_a^b f(x) dx = \lim\limits_{\lambda(P) \to 0} \sum\limits_{k=1}^n f(\alpha_k) \Delta x_k = \lim\limits_{h \to 0} \sum\limits_{x:a \to b} f(x)$
Another example. Isn't the following a proof of $(e^x)'=e^x$?
$\Delta a^x=a^x$
${a^{x+h}-a^x \over h}=a^x$
$a^x a^h - a^x=ha^x$
$a^h-1=h$
$a=(1+h)^{1 \over h}$
$\lim\limits_{h \to 0} (1+h)^{1 \over h} = e$
$f'(x)=\lim\limits_{h \to 0} \Delta f(x)$
$(e^x)'=e^x$
Edit.
As was noted, the Dirichlet function gives a counterexample:
$\lim\limits_{h \to 0} \sum\limits_{x:a \to b} \chi_\mathbb{Q}=0$
While the Dirichlet function doesn't have a Riemann integral. However, the Dirichlet function does have a Lebesgue integral equal to zero. Would the above formulas be correct if we'll restrict $f$ to only functions continuous over $[a,b]$?
 A: I think you got it backwards. Because claiming that something is equal to something in a limit is weaker then assume it to be true even without the limit.  So starting out with $\Delta a^x = a^x$ is a stronger assumption then saying $(a^x)' = a^x$. Like $x_n = 0$ is much stronger then $\lim_{n\to \infty} x_n = 0$. Because assuming the former you will say, well $x_n = 0$ thus $\lim_{n\to \infty} x_n = 0$. But that is not the general case as you know. Am I wrong with that? 
For the Riemann integral I think you can see that it can't be true in general, because you just claim you can look at one particular case of a tagged partition. But that is not true. You must have it for any partition and any points. Take the Dirichlet function as a famous counterexample. 
So assuming you know the classical results you can also say something meaningful also in the discrete case. The opposite is not true in general.
What do you think? I am still a little bit fuzzy but my "spider senses" just tell me something need to be off.
edit:
Here, let it make me precise in case of the indicator function on $\mathbb{Q}$ denote $\chi_\mathbb{Q}$. If we would have
$$
\int_a^b f(x) dx = \lim_{h\to 0} \sum_{x:a\to b} f(x),
$$
then 
$$
\int_0^1 \chi_\mathbb{Q} dx = 0.
$$
But using the "right" definition of the Riemann integral via the Riemann sum and all possible tagged partition will show that this function is not Riemann integral (see wikipedia) for Details why that is.
