Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function.
Suppose $\langle h_n \space \space | \space \space n \in \mathbb{N} \rangle$ is a sequence of real functions given by,
$\forall x \in \mathbb{R}: \space h_n(x) = \sum_{j=0}^{n-1} \space \frac{1}{n}f(x+\frac{j}{n})$
Show that if $\alpha, \beta \in \mathbb{R}$ and $\alpha < \beta$$\space$ then $\langle h_n \space \space | \space \space n \in \mathbb{N} \rangle$ uniformly converges at $[\alpha, \beta]$.
I observed that given some $\xi \in \mathbb{R}$, we can define a partitions sequence $(\mathscr{P_n})_{n\in\mathbb{N}}$ of the interval $[\xi, \xi + 1]$ as follows:
$\forall \space n\in\mathbb{N}$ : $\mathscr{P}_n = ${$\xi, \xi + \frac{1}{n}, \xi + \frac{2}{n}, \dots, \xi + \frac{n}{n}$}
Then we can denote $\mathscr{P}^{*}_{n} = ${$\xi, \xi + \frac{1}{n}, \xi + \frac{2}{n}, \dots, \xi + \frac{n-1}{n}$} and consider the Riemann sum,
$S(f, \mathscr{P}_n, \mathscr{P}^{*}_n)$ as $\mathscr{P}^{*}_n$ represents the points we choose, which is exactly $h_n(\xi)$ for each $n$. Since $f$ is continuous over, it is integrable over $[\xi, \xi +1]$ and it tells us that $lim_{n \rightarrow\infty}S(f, \mathscr{P}_n, \mathscr{P}^{*}_n) \space$ exists which is what we wanted (obviously $\space (\frac{1}{n})_{n\in\mathbb{N}}$ approaches zero at infinity).
Now it's left to show the convergence is uniform over any interval $[\alpha, \beta]$. Although it seems intuitively true, I find it challenging to reason it. I know that $ F(x) = \int_{x}^{x+1} f(t) \, dt$ is the function the sequence converges to (by the FTC). I would like if someone could help me to figure what I miss out.
Let $\varepsilon > 0$. According to Heine's Theorem $f$ is uniformly continuous on each compact subset of $\mathbb{R}$. Application to $[\alpha,\beta+1]$ yields some $\delta > 0$ such that $x,y \in [\alpha,\beta+1]$, $|x-y| < \delta$ implies $|f(x)-f(y)|< \varepsilon$. Let $n_0 > 1/\delta$. Then for $n \ge n_0$ and $x \in [\alpha,\beta]$ $$ \left|h_n(x)-\int_x^{x+1}f(t)dt\right| =\left| \sum_{j=0}^{n-1} \frac{1}{n}f(x+j/n)-\sum_{j=0}^{n-1}\int_{x+j/n}^{x+(j+1)/n}f(t)dt\right| $$ $$ =\left| \sum_{j=0}^{n-1} \int_{x+j/n}^{x+(j+1)/n}f(x+j/n)dt -\sum_{j=0}^{n-1}\int_{x+j/n}^{x+(j+1)/n}f(t)dt\right| $$ $$ \le \sum_{j=0}^{n-1} \int_{x+j/n}^{x+(j+1)/n}|f(x+j/n)-f(t)|dt<\sum_{j=0}^{n-1}(\frac{1}{n}\varepsilon)= \varepsilon. $$ Note that the last inequality follows from $|(x+j/n)-t| \le 1/n < \delta$ for $t \in [x+j/n,x+(j+1)/n]$.