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Suppose $f:[a, b] \rightarrow \mathbf{R}$ 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 $L(f,[a, b])=\lim _{n \rightarrow \infty} L\left(f, P_{n},[a, b]\right)$ and $U(f,[a, b])=\lim _{n \rightarrow \infty} U\left(f, P_{n},[a, b]\right)$

I am trying to prove the first part. The sequence $x_n= L\left(f, P_{n},[a, b]\right)$ is non-decreasing and is bounded above. I can use the Weierstrass theorem to show that it converges to some point $c$. However, I can't find a way to prove that $c=L(f,[a, b])$.

Here, $L(f,[a, b])$ and $U(f,[a, b])$ are the lower and upper Riemann integrals, respectively.

frysauce
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    Focus on proving one of these equalities first, say the one for lower sums, the one for upper sums will be very similar. Now a very common technique of proving an equality is to show the two inequalities. From the definition of lower sums, one direction of the inequality should be very trivial (which one)? Now for the other direction, first write down the definition of a lower sum in terms of supremum. – peek-a-boo Apr 17 '21 at 00:18
  • See this answer which proves a more general result. – Paramanand Singh Apr 17 '21 at 01:20
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    @peek-a-boo I am trying to prove the same statement; since $L(f,[a,b])\overset{def}{=}\sup_P L(f,P,[a,b]), L(f,P_n.[a,b])\leq L(f,[a,b])$ for every $n\in\mathbb{Z}^+$ so $\lim_{n\to\infty}L(f,P_n,[a,b])\leq L(f,[a,b])$. Now, given $\varepsilon>0$ there must be a partition $P_{\varepsilon}$ such that $L(f,[a,b])-\varepsilon)\leq L(f,P_{\varepsilon},[a,b])$ so now I should find a partition $P_n$ such that $L(f,P_n,[a,b])\geq L(f,P_{\varepsilon},[a,b])$ to prove the other inequality but I am hgaving difficulties with this step. – lorenzo Nov 26 '21 at 11:59
  • @peek-a-boo It's clear that if I choose a partition $P_n$ such that $\frac{b-a}{2^n}$ then in every interval where $[x_{j-1},x_j]$ does not contain a point of $P_{\varepsilon}$ we have that the terms from the lower sum P_n are "higher" than the corresponding terms from $P_{\varepsilon}$ since the inf of a subset is bigger than the inf of the superset. But, I have difficulties writing down this bounding precisely, and estimating how things go in the places where the intervals $[x_{j-1},x_j]$ contain a point of $P_{\varepsilon}$. Could you help me see how to go ahead with the proof? Thanks – lorenzo Nov 26 '21 at 12:05

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