TITLE: A puzzle about integrability
QUESTION [7 upvotes]: I know there is a Proposition:

for $f(x)$ is bounded on $[a,b]$,then $f(x)$ is integrable if and only if given $\epsilon>0$, there exists a partition such that $U(f,P)-L(f,P)<\epsilon$

But my question is that:
Assume $f(x)$ is bounded and integrable on $[a,b]$, then given  $\epsilon>0$   does it must exist a $\delta>0$ for any partition $|P|<\delta$,$U(f,P)-L(f,P)<\epsilon$.
If it is true, how to prove it? Thanks in advance.

REPLY [3 votes]: I will try to answer with slightly different approach. It almost follows paramanand's approach.

Let $f$ be a bounded function on $[a,b]$. The set $P=\{a=x_0<x_1...<x_i<....<x_n=b\}$ is defined to be a partition of $[a,b]$. If we additionally specify a set $T=\{t_i\}_1^n, t_i \epsilon[x_{i-1},x_i]$, $(P,T)$ will be referred as a tagged partition with tags in T.
Let $U(f,P)=\sum_i^nM_i(x_i-x_{i-1})$ where $M_i=sup\{f(x)\}_{[x_{i-1},x_i]}$
Let $K=sup\{|f(x)|\}_{[a,b]}$
    $L(f,P)=\sum_i^nm_i(x_i-x_{i-1})$ where $m_i=inf\{f(x)\}_{[a,b]}$  
As $f$ is bounded, $inf\{U(f,P)\}=\int^-f$ and $sup\{L(f,P)\}=\int_-f$ both exist.  
$\forall \epsilon>0 \exists P, U(f,P)-\int^-f<\epsilon$
$P \subset P^{'} \implies U(f,P^{'})\le U(f,P)$  
Let $Q$ be any partition and let $P^{'}=P \bigcup Q\implies P\subset P^{'}\implies \int^-f \le U(f,P^{'})\le U(f,P) < \int^-f+\epsilon$
Let $Q=\{a=x_0<x_1...<x_j<....<x_m=b\}$
$U(f,Q)-U(f,P^{'})=\sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[M_i(x_i-x_{i-1})-M_j^{'}(x_j-x_{i-1})-M_j^{"}(x_i-x_j)]$  
$|U(f,Q)-U(f,P^{'})|=U(f,Q)-U(f,P^{'})$  
$=|\sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[M_i(x_i-x_{i-1})-M_j^{'}(x_j-x_{i-1})-M_j^{"}(x_i-x_j)]|$  
$\le \sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[|M_i|(x_i-x_{i-1})+|M_j^{'}|(x_j-x_{i-1})+|M_j^{"}|(x_i-x_j)]$  
$\le \sum_{\{i| \exists x_i \in P, x_i \in (x_{j-1},x_j)\}}[2K(x_i-x_{i-1})]$  
Choose $Q$ with $||Q||< \epsilon / (2nK)$ we get  
$U(f,Q)-U(f,P^{'}) < \epsilon$ we already have  
$U(f,P^{'})\le U(f,P) < \int^-f+\epsilon$
so $U(f,Q)-\int^-f<2\epsilon$
i.e $\forall \epsilon > 0 \exists \delta_1 > 0,||Q|| < \delta_1 \implies U(f,Q)-\int^-f<\epsilon$  
similarly we obtain  
$\forall \epsilon > 0 \exists \delta_2 > 0, ||Q|| < \delta_2 \implies\int_-f - L(f,Q)<\epsilon$  
In case $f$ is darboux integrable, $\int_-f = \int^-f =I$
Taking $\delta < min\{\delta_1, \delta_2 \}$ and combining last two inequalities  
$\forall \epsilon > 0 \exists \delta > 0, ||Q|| < \delta \implies U(f,Q)-L(f,Q) < 2\epsilon$