Proving integration for a bounded increasing function

[STEMucator]

Homework Statement

Suppose that f is a bounded, increasing function on [a,b]. If p is the partition of [a,b] into n equal sub intervals, compute S_p - s_p and hence show f is integrable on [a,b]. What can you say about a decreasing function?

Homework Equations

We partition [a,b] into sub-intervals.

For each i, we let : [itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

Now, we define [itex]s_p = \sum_{i=1}^{n} m_i Δx_i[/itex] as the lower sum and [itex]S_p = \sum_{i=1}^{n} M_i Δx_i[/itex] as the upper sum.

Some more info in my notes :

Let [itex]M = sup \left\{{f(x)|x \in [a,b]}\right\}[/itex] and [itex]m = inf \left\{{f(x)|x \in [a,b]}\right\}[/itex]. Then we get s_p ≤ M(b-a) so the set of all possible s_p is bounded above.

Let I = sup{s_p} and J = inf{S_p}

Definition : if I = J, then f(x) is integrable.

Now another theorem I could use : For a bounded function f on [a,b]. f is integrable if and only if :

[itex]\forall ε>0[/itex] there is a partition p of [a,b] such that S_p < s_p + ε.

The Attempt at a Solution

So we are given that f is a bounded increasing function. This means that m ≤ f ≤ M for some lower bound m and upper bound M. For any partition p of [a,b] into n sub-intervals, we also have :

[itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

Now before I jump any further I want to confirm the direction I'm going in. I have two theorems I provided and I'm wondering which one is more appropriate to use here.

 

[STEMucator]

I think I have an idea about how the information given relates, so continuing from my first post :

So we are given that f is a bounded increasing function. This means that m ≤ f ≤ M for some lower bound m and upper bound M. For any partition p of [a,b] into n sub-intervals, we also have :

[itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

Now, writing out the upper sum we get : [itex]S_p = \sum_{i=1}^{n} M_iΔx_i[/itex] and the lower sum : [itex]s_p = \sum_{i=1}^{n} m_iΔx_i[/itex]

So we get :

[itex]S_p - s_p = \sum_{i=1}^{n} (M_i - m_i)Δx_i = \sum_{i=1}^{n} (f(x_i) - f(x_{i-1}))(\frac{b-a}{n})[/itex]

I think this the path I'm trying to take.

 

[STEMucator]

Sorry for all the posting, but since no one has responded I believe I have pieced this puzzle together. So I'll write it out cleanly here.

So we are given that f is a bounded increasing function on [a,b]. This means that f(a) ≤ f(x) ≤ f(b) for any x in [a,b]. Hence f(a) is a lower bound for f and f(b) is an upper bound so that f is bounded. We need to show that for any ε>0, we can choose any partition p of [a,b] into n sub-intervals such that S_p - s_p < ε. Suppose p is the partition [itex]\left\{{x_0, x_1, ..., x_n}\right\}[/itex].

We let : [itex]L = \displaystyle\max_{1≤i≤n}(x_i - x_{i-1}) < \frac{ε}{f(b) - f(a)} = Q[/itex]

Now, we define :

[itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

Now, writing out the upper sum we get : [itex]S_p = \sum_{i=1}^{n} M_iΔx_i[/itex] and the lower sum : [itex]s_p = \sum_{i=1}^{n} m_iΔx_i[/itex]

So we get :

[itex]S_p - s_p = \sum_{i=1}^{n} (M_i - m_i)Δx_i ≤ \sum_{i=1}^{n} (f(x_i) - f(x_{i-1}))L = (f(b) - f(a))L < (f(b) - f(a))Q = ε[/itex]

Hence f is integrable on [a,b] since S_p - s_p < ε.

Is this good? It's my first time trying one of these.