# Proof that Darboux upper/lower sums always converge to the Darboux upper/lower integral as the partition gets thinner [duplicate]

I'm going through my calculus lecture notes, and at some point the following claim is made: Let $f:\left[a,b\right]\to\mathbb{R}$ be a bound function. For any $\epsilon>0$ there exists some $\delta>0$ s.t. for any partition $P$ of $\left[a,b\right]$ for which $\Delta P<\delta$, it holds that

$$\left|U\left(f,P\right)-\bar{\int}f\left(x\right)dx\right|<\epsilon$$

Where $\Delta P$ is the size of the biggest segment in the partition (The text, which is not in English, refers to $\Delta P$ as the “partition parameter”, I don't think this is the correct translation into English since I did not find any reference to it). $U\left(f,P\right)$ is the upper Darboux sum for said partition and $\bar{\int}f\left(x\right)dx$ is the infimum for the upper Darboux sums of all possible partitions.

I have failed in proving this, and it seems important, could someone supply a source where this is proved?

## marked as duplicate by Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 30 '18 at 17:36

• See for example Spivak's book "An ntroduction to real analysis" – dmtri Jun 30 '18 at 9:09
• The "partition parameter" is more popularly known as the "norm" or "mesh" of the partition. – Paramanand Singh Jun 30 '18 at 17:41

The upper integral you mentioned is the infimum of all these upper sums, so by definition these sums have to be as close as possible to that integral....

• True. But what guarantees that I can find a partition whose upper sum is arbitrarily close to the upper integral by only controlling the size of the biggest segment? – D.M. Jun 30 '18 at 9:19
• Please note that partition more thin is closer to upper integral – dmtri Jun 30 '18 at 9:23
• And it can be shown in this way that any given sequence of partitions that get thinner is decreasing. But it is not obvious a-priori [to me] that any such sequence will converge to the upper integral. Maybe different ways of making the partition thinner result in different lower bounds? (I realize this is false, but am looking for the formal argument that shows why) – D.M. Jun 30 '18 at 10:57
• Infimum does not necessarily mean that it is a limit under particular circumstances. The proof of the result in question is not trivial but at the same time not too difficult. See math.stackexchange.com/a/2047959/72031 – Paramanand Singh Jun 30 '18 at 17:39

For a continuous function on a closed interval $[a, b]$, it follows from uniform continuity that there is a union of partitions $\bigcup_i h_i= [a, b]$ such that $\sum_i 1_{h_i} |max${f}-$min${f}$|$ $<\epsilon$. Where $\epsilon > 0$ The inequalities remain true for any partition $d$ such that $length (d) \le length (h_i)$ for all $i$ .

$L(f, p) \le \int f dx \le U (f, p)$

But $|L(f, p) - U (f, p) | <( b-a )\epsilon$

• But $f$ need not be continuous. – D.M. Jun 30 '18 at 12:17
• @D.M. then follow dmtri's advice and read the book. – ibnAbu Jun 30 '18 at 12:19