0
$\begingroup$

In our lecture we have defined integrability as follows

Let $f:[a, b] \to \mathbb{R}$. We say that $f$ is integrable if

$sup\{L(f, P), P$ is a partitition of $[a, b]\}= inf\{U(f, P), P$ is a partitition of $[a, b]\}$,

where $L(f, P)$ and $U(f, P)$ are Darboux sums.

Based on this definition the professor showed that from integrability one can deduce:

If $f:[a, b] \to \mathbb{R}$ is integrable $\Rightarrow$ for every $\epsilon >0$ there exists a partitition $P$ of $[a, b]$ such that $U(f, P)-L(f, P)<\epsilon$.

However, I am not sure if I have understood his proof. So, I will try to reproduce it on my own:

Let $M:=sup\{L(f, P), P$ is a partitition of $[a, b]\}$ and $m:= inf\{U(f, P), P$ is a partitition of $[a, b]\}$. By definition of the $sup$ we know that there must be a $P'$ such that $M<L(f, P')+2\epsilon$. Likewise there must exist a $P''$ with $m>U(f, P'')-\epsilon$. We already know from lecture that both inequalities hold for any finer partitition $P$, because this means that $U(f, P)$ decreases and $L(f, P)$ increases.

It follows that $-M>-L(f, P)-2\epsilon$ and if we add the other inequality, $m>U(f, P)+\epsilon$, we get the desired result: $U(f, P)-L(f, P)<\epsilon$.

Is this correct? I am not sure if I have correctly applied the definitions of sup and inf.

$\endgroup$
3
  • 1
    $\begingroup$ Your proof looks fine. You could avoid any complications if you used $\int_{[a,b]}f$ instead of $M$ & $m$ because $M=m$ since $f$ is integrable. $\endgroup$
    – SL_MathGuy
    Commented Jan 21, 2020 at 23:09
  • $\begingroup$ @SL_MathGuy, are the Darboux sums sequences? $\endgroup$
    – Philipp
    Commented Jan 22, 2020 at 14:02
  • 1
    $\begingroup$ @Philipp: in general they are not sequences but suppose for every $n\in\mathbb {N} $ you form a partition $P_n$ of $[a, b] $. Then $U(f, P_n), L(f, P_n) $ are sequences and you can discuss their limiting behavior. Also note the definition of sequence as a function of the form $f:\mathbb{N} \to X$ where $X$ is any set. $\endgroup$
    – Paramanand Singh
    Commented Jan 26, 2020 at 5:19

0

You must log in to answer this question.