Theorem 3: Let `U` be the set of all upper sums of `f`, over all nets
`N`. Then `U` is bounded below.
Proof: Let `U = { U(N) : N` is a net `}`.
By definition: `U(N)=sum_(i=1)^n(M_i\Deltax_i)`.
We know that `f ` is bounded, moreover, we know that `f` is bounded
below.
This means that `EE m in RR` such that: `f(x) >= m` for all `x in
[a, b]`
If we can show that `U(N) >= m(b  a)` then our proof will be done.
( we need to show this for all nets `N` )
Take the inequality: `M_i\Deltax_i>=\Deltax_im`. It obviously holds
since m is a lower bound,
then `m <= M_i` for all `i`.
Let's take the finite sum of both sides, we get:
`sum_(i=1)^nM_i\Deltax_i>=sum_(i=1)^n(m\Deltax_i)`
The LHS of the inequality is `U(N)` by definition, and on the right
hand side, we can take the constant
`m` outisde to get:
`U(N)>=msum_(i=1)^n(\Deltax_i)`
Since the sum of all `\Deltax_i` is the interval `[a,b]` itself, we
finally conclude:
`U(N)>=m(ba)`
And the result follows.
This proof was obviously inspired by the one E... did for exercice 4.
Thanks E...
Good luck all!
...
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