Does $ \frac{1}{N} \sum_{n=-N^2}^{N^2} f(\tfrac{n}{N}) \to \int_{-\infty}^\infty f(x) \, dx $ work for integrable functions?

Question

I was reading about the Fourier transform over $\mathbb{R}$, in that case, the Fourier coefficients read: $$ \hat{g}(\xi) = \int_{-\infty}^\infty g(x) e^{2\pi i x \xi} \, dx $$ Is it possible to approximate this with a Riemann sum? What if I discretize the real line with increments of length $\frac{1}{N}$, then we need $N^2$ such intervals to cover from $[-N, N]$: $$ \frac{1}{N} \sum_{n=-N^2}^{N^2} f(\tfrac{n}{N}) \to \int_{-\infty}^\infty f(x) \, dx $$ In particular we could try to do the same thing for $f(x) = e^{2\pi i x \xi} g(x)$. $$ \frac{1}{N} \sum_{n = - N^2}^{N^2} g(\tfrac{n}{N}) e^{2\pi i \frac{n}{N}\xi} \stackrel{?}{\to} \int_{\mathbb{R}} g(x) e^{2\pi i x \xi} \, dx = \hat{g}(\xi) $$ I would have to look up properties of the Riemann integral. Possibly these notes: $$ \liminf_{|\Delta x| \to 0} \sum_n f(t_n) (x_{n+1} - x_n) < \int f(x) \, dx < \limsup_{|\Delta x| \to 0} \sum_n f(t_n)(x_{n+1} - x_n) $$ where $x_n < t_n < x_{n+1}$ for each interval $[x_n, x_{n+1}]$.

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what does it mean for a function to be riemann integrable