Can someone tell me if this proof is correct (and in case how to improve it) or it is wrong?
Let $f:[a,b]\subset\mathbb{R} \to \mathbb{R}$ Riemann integrable and let $f(x) \geq 0$ for all $x\in [a,b]$. Then the integral function $F$ of $f$ is increasing.
Let $s,t\in[a,b]$ such that $s\leq t$.
Lets decompose $[a,s]\subset[a,b]$ in a decomposition of $n$ subintervals of width $n$ such that $a=x_1<...<x_n=s \leq t$ and build the Riemann sum for $f$: we have $$\int_a^s f(x) dx=\lim_{n \to \infty} \frac{s-a}{n} \sum_{k=1}^n f(x_k)$$
Since by hypothesis $f(x) \geq 0$ for all $x\in[a,b]$ it is $f(x_k) \geq 0$ for all $k\in\{1,...,n\}$, and since $t \geq s$ it is $$\lim_{n\to\infty}\frac{s-a}{n} \sum_{k=1}^n f(x_k) \leq \lim_{n\to\infty} \frac{t-a}{n} \sum_{k=1}^n f(x_k)=\int_a^t f(x) dx$$ That is $$\int_a^s f(x) dx \leq \int_a^t f(x) dx$$ So we have concluded that $$s \leq t \implies \int_a^s f(x) dx \leq \int_a^t f(x) dx$$ Which is the definition of increasing function for the integral function $F$ of $f$.
I'm a bit unsure about this because of the decomposition of the intervals: I'm not so sure that since $[a,s]$ is decomposed in $a=x_1<...<x_n=s$ then $a=x_1<...<x_n=s \leq t$ is a valid decomposition for the integral of $f$ in $[a,t]$ too.
Thanks.
This $$ \lim_{n\to\infty} \frac{t-a}{n} \sum_{k=1}^n f(x_k)=\int_a^t f(x) dx $$ is wrong because $x_1, \ldots, x_n$ is a partition of $[a, s]$ but not of $[a, t]$. You should also make clear that all $x_k$ depend on $n$.
For a simpler approach note that for $a \le s < t \le b$ $$ F(t) - F(s) = \int_s^t f(x)\, dx $$ and that is $\ge 0$ because any Riemann sum for $f$ on the interval $[s, t]$ is $\ge 0$.