2
$\begingroup$

From Spivak's Calculus,

For the theorem:

If $f$ is bounded on $[a,b]$, then $f$ is integrable on $[a,b]$ if and only if for every $\varepsilon > 0$ there is a partition $\mathcal{P}$ of $[a,b]$ such that $U( f, \mathcal{P}) - L( f, \mathcal{P}) < \varepsilon.$

Part of the proof is:

If $f$ is integrable sup${L(f, \mathcal{P})}$ $=$ inf $\{U(f, \mathcal{P})\}$.

This means that for each $\varepsilon > 0$ there are partitions \mathcal{P}', \mathcal{P}" with U(f, \mathcal{P}") - L(f, \mathcal{P}') < \varepsilon .

I am not getting this part. Could someone explain a little more about how "this means that"?

  • 0
    I'm sorry, I thought these were universal.$U(f,P)$denotes the upper sum: http://mathworld.wolfram.com/UpperSum.html. And L, the lower sum2011-11-21

1 Answers 1

8

Here is the explanation: By definition, $f$ is integrable if and only if $\sup\{L(f, P)\}= \inf\{U(f, P)\}$, and now we want to prove the only if part of your above statement. Note that the $\sup$ and $\inf$ are taken among all the partitions of the intervals on $[a,b]$. Now by definition of $\sup$, given any any $\epsilon>0$, there exists a partition P' such that L(f, P')>\sup\{L(f, P)\}-\frac{\epsilon}{2}. Similarly, be definition of $\inf$, there exists a partition P'' such that U(f, P'')<\inf\{U(f, P)\}+\frac{\epsilon}{2}. Combining these two inequalities, we have U(f, P')-L(f, P'')<\inf\{U(f, P)\}-\sup\{L(f, P)\}+\epsilon=\epsilon since $\sup\{L(f, P)\}= \inf\{U(f, P)\}.$

Moreover, it's easy to check that U(f, P'\cup P'')\leq U(f, P') and L(f, P'\cup P'')\geq L(f, P''). Therefore, we can conclude that U(f, P'\cup P'')-L(f, P'\cup P'')<\epsilon. This proves the only if part of your above statement.