If $f:[a,b]\to\mathbb{R}$ is bounded then $L(f,[a, b])=\lim _{n\to\infty} L(f, P_n,[a, b])$ and $U(f,[a, b])=\lim _{n\to\infty} U(f, P_n,[a, b])$

Solution 1:

The lower Darboux integral $L(f)$ is the supremum of lower Darboux sums formed with every partition. Hence, for any $\epsilon > 0$ there exists a partition$P_\epsilon$ such that

$$L(f) - \frac{\epsilon}{2} < L(f,P_\epsilon) \leqslant L(f)$$

Let $N_\epsilon$ denote the number of subintervals in the partition $P_\epsilon$. With $M = \sup_{x \in [a,b]}|f(x)|$ it follows that the oscillation of $f$ over any subinterval $I$ satisfies $$\sup_{x \in I}f(x) - \inf_{x \in I} f(x) \leqslant 2M$$

Let $P_n$ be a dyadic partition where the length of each of the $2^N$ subintervals is bounded as

$$\|P_N\|= \frac{b-a}{2^N}\leqslant \frac{\epsilon}{4MN_\epsilon}$$

If we can show that

$$L(f) - \epsilon < L(f,P_N) \leqslant L(f)$$

then we have proved that $\lim_{n \to \infty}L(f,P_n) = L(f)$ since the lower sums $L(f,P_n)$ form an increasing sequence and, consequently, $L(f) - \epsilon < L(f,P_n) \leqslant L(f)$ for all $n \geqslant N$. Note that the inequality $L(f,P_N) \leqslant L(f)$ on the RHS is always satisfied regardless of how $P_N$ is chosen since the lower integral is an upper bound for all lower sums.

The correct approach is to introduce the common refinement $Q = P_N \cup P_\epsilon$. Since $Q \supset P_\epsilon$ we immediately have $L(f,P_\epsilon) \leqslant L(f,Q)$ and, thus,

$$\tag{*}L(f) - \frac{\epsilon}{2} < L(f,Q) \leqslant L(f)$$

It is now easy to compare the magnitude of lower sums $L(f,P_N)$ and $L(f,Q)$ because every subinterval of $P_N$ is a union of subintervals of $Q$. (This is not the case in your approach where you tried to compare -- incorrectly -- $L(f,P_N)$ and $L(f,P_\epsilon)$ since the subintervals may overlap.)

The partition $Q$ contains at most $N_\epsilon - 1$ more subintervals than $P_N$. If a subinterval $I$ of $P_N$ contains $q$ points from $P_\epsilon$ in its interior, then $I = \cup_{j=1}^qI_j$ where the $I_j$ are subintervals of $Q$. We then have

$$\inf_{x \in I}f(x)l(I) \leqslant \sum_{j=1}^q\inf_{x \in I_j}f(x)l(I_j),$$

where $l(I)$ denotes the length of the subinterval. Since $l(I) = \sum_{j=1}^q l(I_j)$, we can bound the difference in contributions to lower sums as

$$\left|\inf_{x \in I}f(x)l(I)- \sum_{j=1}^q\inf_{x \in I_j}f(x)l(I_j)\right| = \left|\sum_{j=1}^q(\inf_{x \in I}f(x)- \inf_{x \in I_j}f(x))l(I_j)\right|\\ \leqslant q\cdot 2M \cdot \|P_N\| $$

Summing over all subintervals of $P_N$, we get

$$|L(f,P_N) - L(f,Q)| \leqslant (N_\epsilon-1)\cdot 2M \cdot \|P_N\| \leqslant (N_\epsilon-1)\cdot 2M \cdot \frac{\epsilon}{4MN_\epsilon}\leqslant \frac{\epsilon}{2}$$

Thus,

$$L(f,P_N) \geqslant L(f,Q) - \frac{\epsilon}{2},$$

and from (*) we get

$$L(f) \geqslant L(f,P_N) \geqslant L(f,Q) - \frac{\epsilon}{2} \geqslant L(f) - \frac{\epsilon}{2} - \frac{\epsilon}{2} = L(f) - \epsilon,$$

completing the proof.