Units of a discrete Fourier transform Normally a Fourier transform (FT) of a function of one variable is defined as
$$f_k=\int^\infty_{-\infty}f(x)\exp\left(-2\pi i k x\right) dx.$$ This means that $f_k$ gets the units of $f$ times the units of $x$: $[f_k]=[f]\times[x]$. For an array of inputs $\{f_n\equiv f(x_n)\}$ of length $N$ the discrete Fourier transform (DFT) is normally defined as $$f_k=\sum_{n=0}^{N-1}f_n \exp\left(-2\pi i k n/N\right). $$
This means that $f_k$ has the same units as $f$: $[f_k]=[f]$. What is missing here? For the DFT to be an approximation of the FT, surely the units must be the same. In other words, if I am to compare the theoretical value of some Fourier transformed quantity to one obtained numerically from a discrete Fourier transform, how do I consistently do this? The obvious norm would be $\frac{x_0-x_{N-1}}{N}$, but is it really so? 
 A: The formula for $N$-DFT should be:
$$\tilde{F}[k] = \sum_{n=0}^{N-1} \tilde{f}[n]\exp(-2\pi ikn/N),$$
where $\tilde{f}[n]$ is the discrete input and $\tilde{F}[k]$ is the discrete frequency output. One can optionally scale by $N^{-1/2}$.
The indexes $n$ and $k$ are dimensionless.  Usually $\tilde{f}[n]$ is obtained from $f(x)$ by sampling, which involves choosing sample period $X$ (same unit as $x$), and extracting $\tilde{f}[n] = f(nX)$.  Since $nX$ has the unit of $X$, so $n$ has no unit.
Comparing continuous FT $\mathscr{F}\{f\}$ and the "DFT approximation" is thorny because one must consider discretization and preprocessing.  If we sample by $\tilde{f}[n] = f(nX)$, then the following data are discarded:


*

*$f(x)$ for $x < 0$ and $x > (N-1)X$,

*$f(x)$ between sample points $x = 0$, $x = X$, ..., $x = (N-1)X$.


Working entirely in continuous domain, this corresponds to multiplying $f(x)$ by a $c(x)$: a "comb" consisting of shifted delta functions:
$$c(x) = \sum_{n=0}^{N-1} \delta(x-nX).$$
Therefore (ignoring scaling), DFT on sampled data involves:


*

*Computing $\mathscr{F}\{f \times c\}$,

*Sampling the result at discrete frequencies. To find these, $\tilde{F}[k]$ corresponds to period $\frac{NX}{k}$, so the angular frequency is $\omega = \frac{2\pi k}{NX}$.  Note that here $k$ is dimensionless, and the factor $\frac{2\pi}{NX}$ has dimension of $x^{-1}$ (Edit: this assumes $k < N/2$; if $k > N/2$ we should consider negative frequencies $k - N$).


From convolution theorem, $\mathscr{F}\{f(x) \times c(x)\}$ is proportional to the convolution of $\mathscr{F}\{f\}$ and $\mathscr{F}\{c\}$.  Therefore $\mathscr{F}\{c\}$ acts to distort the desired $\mathscr{F}\{f\}$.
By inspection, $\mathscr{F}\{c\}$ consists of sums of complex exponentials with frequencies occurring in regular intervals. Plotting on Wolfram Alpha (e.g., $\mathtt{abs(1+exp(ix)+exp(2ix)+exp(3ix))}$) we see that this has a periodic peaks starting from $\omega = 0$. We want this to be as close to $\delta(\omega)$ as possible, and you can do the following to improve results (ASSUMING $f(x)$ has finite bandwidth, i.e., $\mathcal{F}(f)$ drops off at high frequency).


*

*Increase the number of samples $N$: this "sharpens" the peaks of $\mathscr{F}\{c\}$. Indeed, in the limit the FT of an infinite delta-train is another infinite delta-train.

*Decrease the sampling interval $X$: this increases the distance between peaks. With the finite-bandwidth assumption, after convolution sampling frequencies would still yield pretty good results.

*Additional signal processing on $f(x)$ (beyond my scope).


(Edit: fixed grammar and redundant phrasing).
A: To obtain a limit $$f_k=\int^\infty_{-\infty}f(x)\exp\left(-2\pi i k x\right) dx$$ from a discrete Fourier transform, you must actually look what you mean by the formula
$$f_k=\sum_{i=-N/2}^{N/2}f_i \exp\left(-2\pi i k x_i\right).\;\;  (*)$$
This formula is taking samples of undetermined step in the $x$ direction and it will not converge to a Fourier transform just for $N\to\infty$. To converge, you must also soften your step and not forget to go into $-\infty$, that is ($i$ is a confusing index with the imaginary $i$ also occuring, let's use $j$)
$$f_k=\Delta x\sum_{j=-\infty}^{\infty}f_j \exp\left(-2\pi i k (j \Delta x)\right). $$
And send $\Delta x$ to zero. The factor in front is needed because otherwise your $f_k$ would diverge. Or you can understand it in the way that the samples must be weighed in respect to the original sampling when getting finer. You can easily see that for $[k]=1$, $[\Delta x]=[x]$ and this is exactly the factor corresponding to $dx$ in the continuous FT.
In the formula $(*)$, if you take any other step than the unit-dimension step, it would not give a good numerical approximation to the continuous FT. E.g. for a half-unit step, the numerical value of $f_k$ would be roughly two times that of $f(k)$. I think this illustrates perhaps most strongly why the step $\Delta x$ has to be in front of the DFT to give an actual approximation.

(This post has been revised quite a lot.)
A: You're mistakingly assuming that a sequence must have exactly the same unit to be considered the approximation of a signal. In general, a sequence $f_k$ can be considered as an approximation of $f(x)$ if $f_k=\int^{x_{k+1}}_{x_k} f(x) dx $. Clearly the units differ here.
Note that by this rule, $f_k=0$ is an approximation of $f(x)=0$, $c*f_k$  is an approximation of $c*f(x)$, and $f_k+g_k$ is an approximation of $f(x)+g(x)$. 
