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(b-a)`

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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