Proof of integration via Riemann

Question

Riemann's condition

Let $f$ be defined and bounded on $[a,b]$. Then $f$ is Riemann integrable on $[a,b]$ if and only if for every $\varepsilon> 0$ there exists a Partition $P$ of $[a,b]$ such that $U(P) - L(P) <\varepsilon$

Use the Riemann condition to prove that if a function $f$ is integrable on $[a, b]$ and $a\leq c < d \leq b$, then $f$ is integrable on $[c, d]$.

I don't even know where to begin with the proof :( Any answer will be very much appreciated

Answer 1

To prove this, you first need some basic properties of upper and lower sums. By hypothesis, since $f$ is Riemann integrable on $[a,b]$, for any $\epsilon > 0$ there is a partition $P = (x_0,x_1,\ldots, x_n)$ of $[a,b]$ such that $U(P) - L(P) < \epsilon.$

Note that upper and lower sums are defined by

$$U(P) = \sum_{k=1}^n M_k(x_k-x_{k-1}), \\ L(P) = \sum_{k=1}^n m_k(x_k-x_{k-1}),$$

where $M_k$ and $m_k$ are the least upper and greatest lower bounds, respectively, of the values of $f(x)$ for $x \in [x_{k-1},x_k]$:

$$M_k = \sup_{x \in [x_{k-1},x_k]} f(x), \\ m_k = \inf_{x \in [x_{k-1},x_k]} f(x). $$

Since $M_k \geqslant m_k$ for all $k$ we have the Property I that

$$U(P) - L(P) = \sum_{k=1}^n (M_k - m_k)(x_k - x_{k-1}) \geqslant 0.$$

Now the partition $P$ may or may not include the points $c$ and $d$. If both are included we can proceed. This includes the case where $c=a$ and $d =b.$ Otherwise, if at least one of $c$ and $d$ is not included, we construct a new partition $P'$ by inserting the necessary point or points into $P$. I will proceed assuming without loss of generality that both points are inserted. Suppose for some indexes $i$ and $j$ we have $x_{i-1} < c < x_i$ and $x_{j-1} < d < x_j.$ The refined partition is

$$P' = (x_0,x_1, \ldots, x_{i-1},c,x_i, \ldots, x_{j-1},d,x_j, \ldots , x_n).$$

we can relabel the partition points as

$$P' = (x_0',x_1', \ldots, x_{i-1}',x_i',x_{i+1}' \ldots, x_{j}',x_{j+1}',x_{j+2}' \ldots , x_{n+2}'),$$

where $x_{i}' = c$ and $x_{j+1}'=d.$

The second property we need is that adding points to a partition results in a smaller upper sum and a larger lower sum. Hence, $U(P') \leqslant U(P),$ and $L(P') \geqslant L(P).$ I will leave it up to you to prove this by considering the effect of adding one point at a time.

It then follows that

$$U(P') - L(P') \leqslant U(P) - L(P) < \epsilon.$$

However, we can write this as

$$\sum_{k=1}^i(M_k-m_k)(x_k'-x_{k-1}') + \sum_{k=i+1}^j(M_k-m_k)(x_k'-x_{k-1}') + \sum_{k=j+1}^{n+2}(M_k-m_k)(x_k'-x_{k-1}')= U(P') - L(P') < \epsilon.$$

Each sum is positive (by Property I). Hence,

$$\sum_{k=i+1}^j(M_k-m_k)(x_k'-x_{k-1}') < U(P') - L(P') < \epsilon.$$

Notice that the points in the second sum form a partition $P'' = (c=x_i', x_{i+1}', \ldots, x_{j}',x_{j+1}'=d)$ of $[c,d].$ Hence the second sum is the difference of upper and lower sums of $f$ with respect to the partition $P''$, and we have

$$U(P'')-L(P'') < \epsilon.$$

Thus, $f$ is Riemann integrable on $[c,d].$

Answer 2

So you want to show that the function f is integrable on [c,d].

1. By the definition, this means you want to find a suitable partition on [c,d].
2. What do you have? Well, f is integrable on [a,b] so you have a suitable partition on [a,b].
3. So now the question is, given a partition P on [a,b], how do you produce one P' on [c,d], and can you say anything about U(P')-L(P') in terms of U(P)-L(P)?

Answer 3

To elaborate a little. You allready know that $f$ is integrable on $[a, b]$, so Riemann condition is satisfied there: for any $\epsilon > 0$ there is some partition $P$ of the domain of integration $[a, b]$ such that $U_f (P) - L_f (P) < \epsilon$. On the other hand, to show that $f$ is also integrable on $[c, d]$ you need a new partition $P'$ of that interval, and $P'$ has to include $c$ and $d$. You already have $P$, so you should just restrict in on $[c, d]$ and add points $c$ and $d$ in it. Thus you will have $||P'|| \leq ||P||$; the rest is up to you...