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 $U(f,[a, b])=\lim _{n\to\infty} U(f, P_n,[a, b])$ I am trying to prove the following statement from Axler's MIRA book:
"Suppose $f:[a, b]\to\mathbb{R}$ is a bounded function. For $n \in \mathbf{Z}^{+}$, let $P_n$ denote the partition that divides $[a, b]$ into $2^n$ intervals of equal size. Prove that
$L(f,[a, b])=\lim _{n\to\infty} L(f, P_{n},[a, b])$ and $U(f,[a, b])=\lim _{n\to\infty} U(f, P_n,[a, b])$."
I have proved that $\lim_{\to\infty} (,_,[,])\leq (,[,])$ (Since $(,[,])=\sup_ (,,[,])$ it follows that $(,_.[,])\leq (,[,])$ for every $n\in\mathbb{Z}^+$ and result follows) but I have found it difficult to prove the other inequality, namely that $\lim_{\to\infty} (,_,[,])\leq (,[,])$.

EDIT: Proof that $\lim_{\to\infty} (,_,[,])\leq (,[,])$.
Since $L(f,[a,b])=\sup_P (f,P,[a,b])$ there exists a partition $P_{\varepsilon}=\{x_{0_{\varepsilon}},x_{1_{\varepsilon}},x_{2_{\varepsilon}},\dots,x_{N_{\varepsilon}}\}$ such that $L(f,[a,b])-\varepsilon\leq L(f,P_{\varepsilon},[a,b])$.
Let $\varepsilon>0$, $M:=\sup_{[a,b]}f+1$,
$L:=\inf_{P_\varepsilon}|x_{j_{\varepsilon}}-x_{j_{\varepsilon-1}}|$ and $P_N$ be the partition of $[a,b]$ into $2^n$ subintervals of length $\frac{b-a}{2^N}<\min\{L,\frac{\varepsilon}{2MN_{\varepsilon}}\}$.
Then we can see $L(f,P_N,[a,b])$ as composed of two pieces: one where no subinterval of $P_N$ contains any point of $P_{\varepsilon}$ and where the summands of $L(f,P_N,[a,b])$ are bigger than those of $L(f,P_{\varepsilon},[a,b])$ (since $\inf(A)\geq\inf(B)$ if $A\subset B$), and another, where the subintervals of $P_N$ contain points of $P_{\varepsilon}$, and which can subtract at most $N_{\varepsilon}\sup_{[a,b]}f\frac{\varepsilon}{2MN_{\varepsilon}}<\frac{\varepsilon}{2}$. Now, since $L(f,P_N,[a,b])\geq L(f,P_{\varepsilon},[a,b])-\frac{\varepsilon}{2}$ for arbitrary $\varepsilon>0$, we have that $L(f,P_N,[a,b])\geq L(f,P_{\varepsilon},[a,b])$, as desired.
Is the above reasoning correct? Can it be improved? Thanks
 A: 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.
