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

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

Some more info in my notes :

Let M = sup \left\{{f(x)|x \in [a,b]}\right\} and m = inf \left\{{f(x)|x \in [a,b]}\right\}. 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 :

\forall ε&gt;0 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 :

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

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 :

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

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

So we get :

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})

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 \left\{{x_0, x_1, ..., x_n}\right\}.

We let : L = \displaystyle\max_{1≤i≤n}(x_i - x_{i-1}) &lt; \frac{ε}{f(b) - f(a)} = Q

Now, we define :

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

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

So we get :

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 &lt; (f(b) - f(a))Q = ε

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.