# Proof of the following fact: $f$ is integrable, $U(f,\mathcal{P})-L(f,\mathcal{P})<\varepsilon$ for any $\varepsilon>0$

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"?

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Can you define your notations? What is $U(f,P)$ and $L(f,P)$? –  Paul Nov 21 '11 at 1:35
I'm sorry, I thought these were universal. U(f,P) denotes the upper sum: mathworld.wolfram.com/UpperSum.html. And L, the lower sum –  MathMathCookie Nov 21 '11 at 1:38

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.