Given a sequence of partitions which size doesn't tend to 0, show that a Riemann integrable function does not converge to its value I found the next problem and I really stucked in the proof:
Let $a < b$. Suppose $P_i$ where $i = 1, 2, 3...$ is a sequence of partitions of $[a,b]$ whose size/mesh does not tend to 0. Show that there is a Riemann integrable function $f$ on [a,b] and a sequence of Riemann sums such that $S(f, P_i) \nrightarrow \smallint_a^bf$
My advances:
The main problem I found is that I'm unable to define a sequence of partitions between $[a, b]$ whose size doesn't tend to 0. I'm just unable to find a suitable sequence. ¿Any help here?
EDIT: my sub-question here is maybe more like, is there an expression for a sequence of partitions whose size doesn't tend to 0? Away from the trivial ones like the constant partition, or keeping some interval fixed?
From that point I found that the idea is intuitively easy: if some of the intervals of the partition does not tend to 0, then that part of the Riemman sum will be dominated by the value $f(\xi_i)$ (for some $\xi$ chosen in the interval) which is not reflecting the real variation of the function on that interval, and obviously the sum will not converge to the real integral.
 A: Without loss of generality, consider the interval $[a,b] = [0,1]$.
Let $(P_n)$ be a sequence of partitions of $[0,1]$ such that the mesh $\|P_n\| \not \to 0$. The partition points of $P_n$ are denoted as $0 = x_0^{(n)} < x_1^{(n)} < \ldots < x_{m_n}^{(n)} = 1$.
Take the Riemann integrable function $f: x \mapsto x$. If the sequence of Riemann sums converges as  $S(P_n,f) \to \int_0^1f(x) \, dx = 1/2$, then for any $\epsilon > 0$, there exists $N(\epsilon) \in \mathbb{N}$ such that for all $n \geqslant N(\epsilon)$ we have $|S(P_n,f) - 1/2| < \epsilon$.  In particular, for the sequence of Riemann sums using the right endponts of subintervals as tags we would have for all $n \geqslant N(\epsilon)$,
$$\tag{1}S(P_n,f) = \sum_{j=1}^{m_n}x_j^{(n)}(x_j^{(n)} -x_{j-1}^{(n)}) < \frac{1}{2} + \epsilon$$
However, since $\|P_n\| \not\to 0$ it follows that there exists $\epsilon_0 > 0$ such that for any $N \in \mathbb{N}$ there exists $k_N \geqslant N$ such that
$$\|P_{k_N}\| = \max_{1 \leqslant j \leqslant m_{k_N}}(x_j^{(k_N)} - x_{j-1}^{(k_N)}) > \epsilon_0,$$
and this implies that there exists $j_{k_N}$ such that $x_{j_{k_N}}^{(k_N)} - x_{j_{k_N}-1}^{(k_N)} > \epsilon_0$.
We have
$$\tag{2}S(P_{k_N},f) = x_{j_{k_N}}^{(k_N)}(x_{j_{k_N}}^{(k_N)} - x_{j_{k_N}-1}^{(k_N)}) + \sum_{j \neq j_{k_N}}x_j^{(k_N)}(x_j^{(k_N)} -x_{j-1}^{(k_N)})$$
Since $f(x) = x$ is increasing, it follows that the sum on the RHS of (2) overestimates the sum of the integrals over the intervals $[0, x_{j_{k_N}-1}^{(k_N)}]$ and $[x_{j_{k_N}}^{(k_N)},1]$. Thus, for any $N \in \mathbb{N}$, there exists $k_N \geqslant N$ such that
$$S(P_{k_N},f) > x_{j_{k_N}}^{(k_N)}(x_{j_{k_N}}^{(k_N)} - x_{j_{k_N}-1}^{(k_N)})+ \int_0^{x_{j_{k_N}-1}^{(k_N)}}x \, dx+ \int_{x_{j_{k_N}}^{(k_N)}}^1x \, dx\\ =x_{j_{k_N}}^{(k_N)}(x_{j_{k_N}}^{(k_N)} - x_{j_{k_N}-1}^{(k_N)})+\frac{1}{2}[x_{j_{k_N}-1}^{(k_N)}]^2 + \frac{1}{2} - \frac{1}{2}[x_{j_{k_N}
}^{(k_N)}]^2 \\ = \frac{1}{2}+ \frac{1}{2}(x_{j_{k_N}}^{(k_N)}- x_{j_{k_N}-1}^{(k_N)})^2 > \frac{1}{2} + \frac{\epsilon_0^2}{2}$$
This contradicts (1) when $\epsilon = \epsilon_0^2/2$ is chosen.  Therefore, there is a Riemann integrable function $f$ and a sequence of Riemann sums such that
$$S(P_n,f) \not\to \int_0^1 f(x) \, dx$$
