2
$\begingroup$

Let $f$ be integrable on $[a,b]$, and let Let $P = \{x_0,x_1,x_2,...,x_{n−1},x_n\}$ be any partition of $[a,b]$. Show that $f$ is integrable in each subinterval, $[x_{i−1}, x_i]$, and further, $\int_a^bfdx=\sum\limits_{x_{k-1}}^{x_k}\int\limits_{x_{k-1}}^{x_k}fdx$.

I'm not sure where to start but, my idea is, using this definition to prove the integrability of $f$.

Let f be defined on $[a,b]$ . We say that $f$ is Riemann integrable on $[a,b]$ if there is a number $L$ with the following property: For every $\epsilon> 0$, there is a $\delta> 0$ such that $|\sigma - L|<\epsilon$ if $\sigma$ is any Riemann sum of $f$ over a partition $P$ of $[a,b]$ such that $||P||<\delta$.In this case,we say that $L$ is the Riemann integral of $f$ over $[a,b]$ and write $\int_a^bf(x)dx=L$

$\endgroup$
  • 2
    $\begingroup$ It seem to me, you should have writen $\sum_{k=1}^{n}$. The idea is simple. I would use the limit definition of riemann integral, rather than Darboux sum's. Take partition to have $x_{i}$ in itself. (I mean the partition for which you will define $\int_{a}^{b}$). $\endgroup$ – kolobokish Nov 6 '16 at 14:33
-1
$\begingroup$

I'm assuming Q2 is asking to how prove

$$\sum_{k=1}^n \int_{x_{i-1}}^{x_i} f dx = \int_{a}^{b} f dx$$


Q1 Using indicator functions (like here or here):

  1. $1_{[x_{i-1}, x_i]}$ is integrable on $[a,b]$ (I guess because $1$ is integrable on $[x_{i-1}, x_i]$).

  2. Finite products of functions that are integrable on $[a,b]$ are integrable on $[a,b]$ (Not sure if we are allowed to use this).

Hence, $f1_{[x_{i-1}, x_i]}$ is integrable on $[a,b]$

$\to f1_{[x_{i-1}, x_i]}$ is integrable on $[x_{i-1}, x_i] (*)$

Now, on $[x_{i-1}, x_i], f1_{[x_{i-1}, x_i]} = f$ so $f$ is integrable on $[x_{i-1}, x_i]$.


Q2 Denote such integral $\int_{x_{i-1}}^{x_i} f dx$

Now observe that

$$\sum_{i=1}^n \int_{x_{i-1}}^{x_i} f dx \stackrel{(**)}{=} \int_{x_0}^{x_n} f dx = \int_{a}^{b} f dx$$


Q1 Alternatively, we could directly use Cauchy Criterion (see p.8 of UCDavis - The Riemann Integral):

$f$ is integrable on $[a,b]$ if

$$\forall \varepsilon > 0, \exists P \ \text{s.t.} \ U(f;P) - L(f;P) < \varepsilon$$

where $P = \{x_0, \cdots, x_n\}$ is a partition of $[x_0,x_n] = [a,b]$

We are given that

$f$ is integrable on $[a,b]$

$\to f$ is bounded on $[a,b]$

$\to f$ is bounded on $[x_{i-1},x_i]$ (obvious but knowing your prof, I'd probably justify this)

Now since $f$ is bounded, by CC, $f$ is integrable on $[x_{i-1},x_i]$ if

$$\forall \varepsilon > 0, \exists Q \ \text{s.t.} \ U(f;Q) - L(f;Q) < \varepsilon$$

where $Q = \{y_0, \cdots, y_m\}$ is a partition of $[y_0,y_m] = [x_{i-1},x_i]$

Now we can do one of two things:

  1. Prove that if $f$ is integrable on a closed interval $[a,b]$, then $f$ is integrable on any closed subinterval $[c,d]$ (see p.10 in TAMU Lecture 19 and then choose the closed subinterval to be $[x_{i-1},x_i]$.

Pf: Let $P' = P \cup \{c,d\}$

Then $$U(f,P') - L(f,P') \le U(f,P) - L(f,P) < \varepsilon$$ since $P'$ is finer than $P$ (further justification needed: apparently finer partitions yield smaller upper-lower sums?)

Now let $Q = P \cap [c,d]$

Then $$U(f,Q) - L(f,Q) \le U(f,P') - L(f,P') < \varepsilon$$

QED

So if you've already proven such a fact in class, just apply to it $[x_{i-1},x_i]$. Otherwise, prove it as given remembering to prove the 'further justification needed' part.

  1. Follow the proof of 1 to prove that if $f$ is integrable on a closed interval $[a,b]$, then $f$ is integrable on any closed subinterval whose endpoints are adjacent elements in some (ordered?) partition $P$ of $[a,b]$.

Pf:

Here

  1. $$P' = P \cup \{c,d\} = P \cup \{x_{i-1},x_i\} = P$$

so

$$U(f,P') - L(f,P') \color{red}{=} U(f,P) - L(f,P) < \varepsilon$$

  1. $Q$ consists of only two elements:

$$Q = P' \cap [c,d] = P \cap [x_{i-1}, x_i] = \{x_{i-1}, x_i\}$$

so $U(f,Q)$ and $L(f,Q)$ each consist of only one term:

$$U(f,Q) - L(f,Q) := M_i(x_i - x_{i-1}) - m_i(x_i - x_{i-1}) = (M_i - m_i)(x_i - x_{i-1})$$

$$ \le \sum_{k=0}^{n-1} (M_k - m_k)(x_k - x_{k-1}) := U(f,P') - L(f,P')$$

where the last inequality follows because $M_k \ge m_k$ and $x_k \ge x_{k-1}$.

QED


$(*),(**)$ I'm not sure if we're allowed to do this. I think I made use of p. 11 in TAMU Lecture 19 whose first line of proof relies on p.10 which we are trying to prove. I guess it depends on the textbook. Does the text in your book prove p.11 in TAMU Lecture 19 without making use of what we're trying to prove?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.