Upper and Lower Sums Close In

On the uniform partition $P_n$, $x^2$ is increasing, so the lower sum uses left endpoints and the upper sum right endpoints: $L_n = \frac{1}{n^3}\sum_{k=0}^{n-1}k^2$, $U_n = \frac{1}{n^3}\sum_{k=1}^{n}k^2$.

Substituting the sum-of-squares formula gives $L_n = \frac{(n-1)n(2n-1)}{6n^3}$ and $U_n = \frac{n(n+1)(2n+1)}{6n^3}$.

Both tend to $\frac{2n^3}{6n^3} = \frac13$ as $n \to \infty$.

The uniform partitions suffice because the lower and upper integrals are the supremum/infimum over all partitions: for each $n$, $L_n = L(P_n, f) \le \underline{\int_0^1} f \le \overline{\int_0^1} f \le U(P_n, f) = U_n$.

Both ends tend to $\frac13$, so by squeezing the lower and upper integrals both equal $\frac13$, and the integral is $\frac13$.

Every interval contains both rationals and irrationals, so on every subinterval $\sup D = 1$ and $\inf D = 0$.

Hence for every partition $P$, $U(D,P) = \sum 1\cdot\Delta x_k = 1$ and $L(D,P) = \sum 0\cdot\Delta x_k = 0$.

Since $U - L = 1 \not\to 0$, the necessary and sufficient condition for Riemann integrability fails.

Splitting one subinterval in two: each piece's supremum is at most the original's, so the upper sum decreases (or stays).

Likewise each piece's infimum is at least the original's, so the lower sum increases (or stays).

So $U - L$ decreases monotonically under refinement — the sums squeeze toward each other.