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I am proving an integral property. Is the following manipulation valid in sums?

$\sum\limits_{i=1}^n x_i = \sum\limits_{i=1}^n y_i$

Then $\sum\limits_{i=1}^n x_i\cdot p_i = \sum\limits_{i=1}^n y_i\cdot p_i$

I am given $f$ and $g$ are integrable functions (I am using Darboux integration). So we know: $\sum\limits_{k=1}^n m(f,[t_{k-1}, t_k])\cdot(t_k - t_{k-1}) = \sum\limits_{k=1}^n M(f,[t_{k-1}, t_k])\cdot(t_k - t_{k-1})$

$\sum\limits_{k=1}^n m(g,[t_{k-1}, t_k])\cdot(t_k - t_{k-1}) = \sum\limits_{k=1}^n M(g,[t_{k-1}, t_k])\cdot(t_k - t_{k-1})$

So: $\sum\limits_{k=1}^n m(f,[t_{k-1}, t_k])\cdot m(g,[t_{k-1}, t_k])\cdot(t_k - t_{k-1})$

$= \sum\limits_{k=1}^n M(f,[t_{k-1}, t_k])\cdot m(g,[t_{k-1}, t_k])\cdot(t_k - t_{k-1})$

$= \sum\limits_{k=1}^n M(f,[t_{k-1}, t_k])\cdot M(g,[t_{k-1}, t_k])\cdot(t_k - t_{k-1})$

I just want to make sure the above sum manipulations are correct? Pretty much I am just substituting.

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For a general $p_i$ the second formula does not follow from first. –  Maesumi Nov 27 '12 at 2:24
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1 Answer 1

up vote 2 down vote accepted

The property you want for sums does not hold. For instance, $1+1=0+2,$ but $\frac23\times1+\frac13\times1\ne\frac23\times0+\frac13\times2$.

I don't know what property you are trying to prove, but even if $f$ is integrable you cannot assume that an upper sum and a lower sum are equal.

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The definition in my book says that a function is integrable provided $U(f) = L(f)$ –  CodeKingPlusPlus Nov 27 '12 at 2:37
    
Of course. But read what $U(f)$ is; it is the supremum of the $U(f;t_1,\ldots,t_n)$ (which is the sum you have) running over all partitions. Same with $L(f)$. –  Martin Argerami Nov 27 '12 at 2:39
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