I was wondering, given a partition P and a refinement of P, namely Q, and a bounded function f:[a,b] --> R,

how can I prove that U(Q,f)<=U(P,f)?

Here is what I have, Im hoping someone can confirm that its right or let me know where I went wrong??...

Proof:

Let $\displaystyle P\subseteq Q$.

Suppose P={$\displaystyle x_0,x_1,...,x_n$} and let Q be a refinement of P such that Q contains one additional point of [a,b] say c where $\displaystyle x_{i-1}\leq c \leq x_{i}$.

let $\displaystyle r_1$=sup{f(x)|$\displaystyle x\in[x_{i-1},c$]} and

let $\displaystyle r_2$=sup{f(x)|$\displaystyle x\in[c,x_{i}$]}.

let $\displaystyle M_{i}$=max{$\displaystyle r_1,r_2$}. [should this be min??]

Then U(P,f)=$\displaystyle \sum_{k=1}^{n}M_k\Delta x_k$

=$\displaystyle \sum_{k=1}^{i-1}M_k(x_k-x_{k-1})+M_i(x_i-x_{i-1})+\sum_{k=i+1}^{n}M_k(x_k-x_{k-1})$

$\displaystyle \geq \sum_{k=1}^{i-1}M_k(x_k-x_{k-1}) + r_1(c-x_{i-1})+r_2(x_i-c)+\sum_{k=i+1}^{n}M_k(x_k-x_{k-1})$=U(Q,f).