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I have to evaluate the following distributional limit:

$$ \lim_{n \to \infty} T_n = \lim_{n \to \infty} \frac 1n \sum_{k=-2n}^{5n} \delta_{\frac kn}$$ We have that $$\lim_{n \to \infty}\langle T_n, \phi\rangle =\lim_{n \to \infty} \frac 1n \sum_{k=-2n}^{5n} \phi\left({\frac kn}\right)$$ Now my idea is to use an integral sum to evaluate this limit, in particular, I know that: $$ \lim_{n \to \infty}\underbrace{\frac{b-a}{n} \sum_{k=1}^n g\left(a+\frac{b-a}{n}k\right)}_{\text{integral sum}} = \int_a^b g(x) \: \mathrm{d}x$$ so i put $ \frac kn = a+\frac{b-a}{n}h=-2+\frac 7n h \implies h=\frac{k+2n}{7}$, and substituting this into my summation I get: \begin{align} \lim_{n \to \infty} \frac 1n \sum_{k=-2n}^{5n} \phi\left({\frac kn}\right) &= \lim_{n \to \infty} \frac 1n \sum_{h=0}^{n} \phi\left(-2+\frac 7n k\right)\\ & =\lim_{n \to \infty} \frac 17 \frac 7n \sum_{h=0}^{n} \phi\left(-2+\frac 7n k\right)\\ &=\frac 17 \int_{-2}^{5} \phi(x) \: \mathrm{d}x \end{align}

Is that correct? Because my book says that the result should be what I got but without that $\frac 17$ coefficient.

EDIT: I just realized that my substitution $h=\frac{k+2n}{7}$ doesn't seem correct since $h$ is not an integer in general as $k$ varies... so any idea?

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I'm not sure if one can see directly that the sum is the right Riemann sum (I guess it depends a bit on the defintions). But we can do the following: fix $\varepsilon>0$. Since $\phi$ is continuous on $[a,b]$, it is uniformly continuous. So there exists $\delta>0$ with $|\phi(x)-\phi(y)|<\varepsilon$ whenever $|x-y|<\delta$. Then, if $n>1/\delta$, \begin{align} \left|\frac1n\,\phi\left(\frac kn\right)-\int_{k/n}^{(k+1)/n}\phi(t)\,dt\right| &=\left|\int_{k/n}^{(k+1)/n}(\phi(k/n)-\phi(t))\,dt \right|\\ \ \\ &\leq\int_{k/n}^{(k+1)/n}|\phi(k/h)-\phi(t)|\,dt\\ \ \\ &\leq\frac{\varepsilon}n. \end{align} Thus \begin{align} \left| \frac 1n \sum_{k=-2n}^{5n} \phi\left(\frac kn\right) -\int_{-2}^5\phi(t)\,dt \right| &=\left|\sum_{k=-2n}^{5n}\int_{k/n} ^{(k+1)/n}\left[ \phi\left(\frac kn\right) -\phi(t)\right]\,dt\right|\\ \ \\ &\leq\sum_{k=-2n}^{5n}\frac{\varepsilon}{n}=7\varepsilon. \end{align} As $\varepsilon$ was arbitrary, we get that $$ \lim_{n \to \infty} \frac 1n \sum_{k=-2n}^{5n} \phi\left({\frac kn}\right) =\int_{-2}^5\phi(t)\,dt. $$

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Your spacing between point evaluations is $1/n$, not $7/n$, because you have asymptotically $7n$ points instead of $n$ points.

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