There are two kind of ways to extend the Riemann integral to unbounded sets.
First is the one that you would probably guess first
$$
\int_0^\infty f(x) dx = \lim_{b \to \infty}\int_0^b f(x)dx
$$
as long as the integrals on the right exist for all $b$ and the limit on the right hand side exists. The other way is to say a sequence of unions of disjoint compact intervals $E_i = \cup_{k=1}^{n_i}[a_k,b_k]$ exhausts a set $E$ if $E_{i+1} \supseteq E_i$ and $\cup_{i=1}^\infty E_i = E$. (I will stick to unions of disjoint unions of intervals in order to avoid discussion about Peano-Jordan measurability, which you may look up if you wish.) Then you can define
$$
\int_0^\infty f(x)dx = \lim_{n\to \infty} \int_{E_i} f(x)dx
$$
whenever the limit on the right exists and is independent of the exhausting sequence $E_i$.
At first these definitions may look similar, but consider $$\int_0^\infty \frac{\sin(x)}{x} dx$$
If you take the limit as $b \to \infty$ as in the first definition, you get $\frac{\pi}{2}$. However, if you take the second definition, you will find that the integral does not exist because the integral of the positive portion and the negative portion are both $\infty$.
One thing that is easy to see is that if $f \geq 0$ is continuous, then the two kinds of integrals with both exist and equal eachother. Thus, of $f^+,f^-\geq 0$ are the positive and negative parts of $f$ and $$\int_0^\infty f^+(x)dx,\int_0^\infty f^-(x)dx < \infty$$
or equivalently $\int_0^\infty |f(x)| dx < \infty$, then the two kinds of integrals with both exist, be finite, and equal eachother.
Now to answer your question. Let's suppose that $f$ is continuous and $\int_0^\infty |f(x)|dx < \infty$. Let's investigate when $$\lim_{n\to\infty}\frac{1}{n}\sum_{k \geq 0} f\left(\frac{k}{n}\right) = \int_0^\infty f(x)dx.$$
Write
$$
\frac{1}{n}\sum_{k \geq 0} f\left(\frac{k}{n}\right) = \sum_{m=0}^\infty \frac{1}{n}\sum_{k = mn}^{(m+1)n-1}f\left(\frac{k}{n}\right) = \sum_{m=0}^\infty S(f;1/n;[m,m+1])
$$
Where $S(f;\Delta x,[a,b])$ denotes the left Riemann sum of $f$ of width $\Delta x$ over the interval $[a,b]$. Since $f$ is continuous on $[m,m+1]$ for every $m$, we have $S(f;1/n;[m,m+1]) \to \int_{m}^{m+1} f(x)dx$ as $n \to \infty$. Then what we would like to say is that
$$
\lim_{n\to \infty}\sum_{m=0}^\infty S(f;1/n;[m,m+1]) =
\sum_{m=0}^\infty\lim_{n\to \infty} S(f;1/n;[m,m+1])
$$
$$
=\sum_{m=0}^\infty \int_{m}^{m+1} f(x)dx =
\int_{0}^\infty f(x)dx
$$
so what let's us justify these steps? The second equality is just definition, and since we assumed $\int_0^\infty |f(x)|dx < \infty$ we don't have to worry about the last equality since we know that $\{\cup_{m=0}^k[m,m+1]\}_k$ is an exhausting sequence of $[0,\infty]$ (single point overlap is okay, don't worry), so all that we have to justify is the first one, exchanging the limit and sum.
Let $S^*$ and $S_*$ represent the upper and lower Riemann sums respectively. Then of course
$$
S_*(f;1/n;[m,m+1]) \leq S(f;1/n;[m,m+1]) \leq S^*(f;1/n;[m,m+1])
$$
and of more importance for this problem
$$
|S_*(f;1/n;[m,m+1])|, |S^*(f;1/n;[m,m+1]) | \leq S^*(|f|;1/n;[m,m+1])
$$
Then one sufficient case to be able to exchange the sum and the limit is that for some $n$ $$\sum_{m=0}^\infty S^*(|f|;1/n;[m,m+1]) < \infty.$$
The justification for this is that you are bounding the partial sums by an integrable function and thus may apply Lebesgue's dominated convergence theorem.
As a corollary, we get the friendly
Corollary. If $|f(x)| \leq g(x)$ where $g(x)$ is decreasing and $\int_0^\infty g(x)dx < \infty$, then $$\lim_{n\to\infty}\frac{1}{n}\sum_{k \geq 0} f\left(\frac{k}{n}\right) = \int_0^\infty f(x)dx. $$
Proof. $$\sum_{m=0}^N S^*(|f|;1;[m,m+1]) \leq S^*(g;1;[0,N+1]) \leq g(0) + \int_{0}^{N+1} g(x)dx \leq g(0) + \int_0^\infty g(x)dx$$ for all $N$. Thus $$\sum_{m=0}^\infty S^*(|f|;1;[m,m+1]) \leq g(0) + \int_0^\infty g(x)dx < \infty. $$
Also note, that you only really need $|f(x)| \leq g(x)$ for all $x>x_0$ for some fixed $x_0$.
The statement is false.
Take $f(x) = x$ then we have $$\int_0^1 f(u) du = \frac{1}{2}$$ and \begin{align*}\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right) &= \frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}+\frac{1}{2n}\right) \\\ &= \frac{1}{n}\left(\sum_{k=0}^{n-1}\frac{k}{n}+\sum_{k=0}^{n-1}\frac{1}{2n}\right) \\\\ &= \frac{1}{n}\left(\frac{n-1}{2} + \frac{1}{2}\right) \\\\ &= \frac{1}{n}\cdot\frac{n}{2} = \frac{1}{2}\end{align*}
hence:
$$\left|\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}+\frac{1}{2n}\right)-\int_{0}^{1}f\left(u\right)du\right| = \left|\frac{1}{2} - \frac{1}{2}\right| = 0\leq\frac{\varepsilon}{n}.\tag{1}$$ for all $\varepsilon > 0, n\in\mathbb{N}$
Although $f(x) = x$ is not positive on $[0,1]$ consider that a shift upwards doesn't change anything (it's added as a constant to the LHS and to the integral as well so it will cancel out).
So each function $f(x) = x + c$ with $c>0$ contradicts your claim.
Best Answer
$$f(x)-g(x)=\int_0^1\frac{(\lceil nt\rceil/n)^x-t^x}{x}\,dt.$$ For each fixed $t\in(0,1)$, the integrand is either identically zero or decreasing (in $x$).
[For $0<b<a\leqslant 1$ and $x>0$, $x\mapsto(a^x-b^x)/x$ is decreasing, as $$\frac{a^x-b^x}{x}=ca^xh(cx),\quad c=\ln\frac{a}{b},\quad h(x)=\frac{1-e^{-x}}{x},$$ and $h(x)$ is decreasing because $h'(x)=-(e^x-1-x)/(x^2 e^x)$ and $e^x>1+x$.]