Question regarding mesh and Cauchy Criterion for integration I'm trying to prove the following theorem using a different proof than what is provided in the text, and I'm not sure if it's correct.  I feel like I'm overlooking something.  The theorem is a biconditional, but I'm only interested in this direction:
If $f$ is integrable on $[a,b]$, then for each $\epsilon>0$ exists $\delta>0$ such that $mesh(P)<\delta$ implies $U(f,P)-L(f,P)<\epsilon$ for all partitions $P$ of $[a,b]$
My attempted proof goes as follows:
Suppose $f$ integrable on $[a,b]$.  Fix $\epsilon>0$.  Then, by the Cauchy Criterion of Integrability, exists a partition $P$ such that $$U(f,P)-L(f,P)<\epsilon$$  Let $Q$ be a refinement of $P$.  Then $mesh(Q)\leq mesh(P)$, and (from a lemma we previously proved), $$U(f,Q)-L(f,Q)\leq U(f,P)-L(f,P)<\epsilon$$ for all refinements $Q$ of $P$.  
Let $M$={$mesh(P)$: $P$ partition $[a,b]$ and $U(f,P)-L(f,P)<\epsilon$}.
We have shown that $M$ is nonempty.  Furthermore, $M$ is bounded above by $b-a$.  Hence $sup$ $M$ exists.  Let $\delta=sup$ $M$.
Then for all partitions $P$ such that $mesh(P)$<$\delta$, $U(f,P)-L(f,P)<\epsilon.$  Since our selection of $\epsilon$ was arbitrary, we conclude that for any $\epsilon$ such a $\delta$ exists.
The proof in the text is a bit more involved, which makes me think that this is incorrect.  Any help would be appreciated...
 A: As you suspect, your proof is not valid.  
By your construction of the set $M$, if $P$ is a partition such that $U(f,P)−L(f,P)<\epsilon$ then $mesh(P) < \delta$. The converse is not immediately true.   (If $A \Rightarrow B$, then it does not follow automatically that $B \Rightarrow A$.) 
By asserting that $mesh(P) < \delta$ now implies $U(f,P)−L(f,P)<\epsilon$ you are making a circular argument. A correct proof requires a bit more work. 
If $f$ is integrable on $[a,b]$ with integral $I$, then there is a partition $P_\epsilon$ such that $U(f,P_\epsilon) < I + \epsilon/4$. (The integral is the greatest lower bound of upper sums).
Let $D = sup\{|f(x)-f(y)|:x,y \in  [a,b] \}$ denote the maximum oscillation of $f$ and let $\delta = \epsilon/4mD$ where $m$ is the number of points in the partition $P_\epsilon$. 
Now let $P$  be any partition with $mesh(P)<\delta$. Form the common refinement $Q = P \cup P_\epsilon$.
You will see that the upper sums $U(f,P)$ and $U(f,Q)$ differ at at most $m$ sub-intervals and at each the deviation is bounded by $\delta D$. 
It follows that $U(f,P)<U(f,Q) + \epsilon/4 < U(f,P_\epsilon)+\epsilon/4<I + \epsilon/2$.
By a similar argument, you can show $L(f,P)> I - \epsilon/2$. Hence $U(f,P)-L(f,P) < \epsilon$.
