Improper integral - equivalent definition? Intuitively, it is rather obvious that
$$\lim_{l\to\infty}\sum_{n=-\infty}^{\infty}f(n\Delta x)\Delta x = \int_{-\infty}^{\infty}f(x)dx \tag{1}$$
where $\Delta x = \frac{1}{l}$, assuming $f$ is integrable and the limit exists.
The fact that this equality is true is the core part of deriving Fourier transform from Fourier series, see page 4, eq. 4.7 in this document. Or maybe we cannot consider this derivation as formal, as it was never intended to be formal, but I thought n mathematics there's no place for informal thinking.
My question is how can we prove it's true from the definitions and properties of improper integral, definite integral and limits?
I've listed the important definitions below in case you would like to refer to some of these in your answers.
Oh, and please ignore mrf's answer - it doesn't refer to my question anymore, I've reformulated it.

If function $f$ is integrable on $[a,b]$, then:
$$\int_{a}^{b}f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\Delta x \tag{2}$$
where $\Delta x = \frac{b-a}{n}$ and $x_i = a+i\Delta x$.
Improper integral definitions
$$\int_{a}^{\infty}f(x)dx=\lim_{t\to\infty}\int_{a}^{t}f(x)dx \tag{3}$$
$$\int_{-\infty}^{b}f(x)dx=\lim_{t\to-\infty}\int_{t}^{b}f(x)dx \tag{4}$$
$$\int_{-\infty}^{\infty}f(x)dx=\int_{a}^{\infty}f(x)dx + \int_{-\infty}^{a}f(x)dx \tag{5}$$
 A: I think this is a counterexample.
Let $g$ be any non-negative function on $\mathbb{R}$ such that the improper Riemann integral $\int_{-\infty}^\infty g$ exists. Let $S = \{ q + (p/q) : q, p = 1, 2, \ldots \}$, and let $f(x)$ be equal to $g(x)$ except on $S$, where $f(x) = 1$.
Since $S$ has only finitely many points in any finite interval, $f$ is continuous and equal to $g$ except on the countable set $S$, therefore $\int_{-\infty}^\infty f$ exists, and is equal to $\int_{-\infty}^\infty g$.
But for $q = 1, 2, \ldots$, the sum $\sum_{n = -\infty}^\infty f(n/q)$ diverges to $+\infty$, therefore the limit on the left hand side of (1) does not exist.
Update: it gets worse, I'm afraid.
One might still reasonably hope that, if $f$ is continuous (which rules out this counterexample as it stands), and $f$ is non-negative, and $\int_{-\infty}^\infty f$ exists,  then all the infinite sums on the left hand side of (1) exist, so there's a fair chance that (1) holds under these (arguably not too restrictive) conditions.
Define the countable set $S \subset \mathbb{R}$, in the same way as before.  Because $S$ has only finitely many points in any finite interval, it can be arranged as a strictly increasing sequence, $s_1 < s_2 < \ldots$.
Choose a convergent series $\sum_{k=1}^\infty t_k$ such that $t_k > 0$ and $s_k + t_k \leqslant s_{k+1} - t_{k+1}$ ($k = 1, 2, \ldots$).
Let $h: (-1, 1) \to \mathbb{R}$ be a "bump function", such as:
$$
h(y) = e^{y^2/(y^2 - 1)} \qquad (-1 < y < 1).
$$
Define:
$$
f(s_k + yt_k) = \frac{h(y)}{s_k} \qquad (k = 1, 2, \ldots; \ -1 < y < 1),
$$
and let $f$ have the value $0$ everywhere outside the pairwise disjoint open intervals $(s_k - t_k, s_k + t_k)$.
Observe that for $l = 1, 2, \ldots$, we have $n/l \in S$ and $f(n/l) = l/n$ for all $n > l^2$, and therefore:
$$
\sum_{n = -\infty}^\infty f\left(\frac{n}{l}\right) \geqslant \sum_{n = l^2 + 1}^\infty f\left(\frac{n}{l}\right) = l\sum_{n = l^2 + 1}^\infty \frac{1}{n} = +\infty.
$$
Thus: $f$ is smooth everywhere on $\mathbb{R}$; $f(x) \geqslant 0$ for all $x \in \mathbb{R}$; $f(x) \to 0$ as $x \to \pm \infty$; the improper Riemann integral $\int_{-\infty}^\infty f$ exists (it is bounded above by $2\sum_{k=1}^\infty t_k/s_k$, and therefore by $2\sum_{k=1}^\infty t_k$); yet, the inner series on the left hand side of (1) diverges to $+\infty$ for all positive integral values of $l$.
So, even though the parameter $\Delta x$ on the left hand side of (1) may assume any strictly positive value, the series expression under the outer limit sign becomes undefined for arbitrarily small values of $\Delta x$, so the limit itself is not well defined, even for this quite "reasonable" function $f$.
Further update: essentially the same construction, and same argument, with these minor changes:
$$
f(s_k + yt_k) = \frac{h(y)}{s_k\log s_k}, \\
\sum_{n = -\infty}^\infty f\left(\frac{n}{l}\right)
\geqslant l\sum_{n = l^2 + 1}^\infty \frac{1}{n\log n} = +\infty,
$$
shows that (1) fails even for smooth $f$ such that $\int_{-\infty}^\infty f$ exists and $f(x) = O\left(\frac{1}{\left\lvert{x}\right\rvert\log\left\lvert{x}\right\rvert}\right)$ for large $\left\lvert{x}\right\rvert$.
