Distributional limit of a sequence of Dirac delta

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 $$ \lim_{n \to \infty} \frac 1n \sum_{k=-2n}^{5n} \phi\left({\frac kn}\right) =\int_{-2}^5\phi(t)\,dt. $$