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 $\int_a^bf(x)dx=L$
Best Answer
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):
$1_{[x_{i-1}, x_i]}$ is integrable on $[a,b]$ (I guess because $1$ is integrable on $[x_{i-1}, x_i]$).
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):
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:
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.
Pf:
Here
so
$$U(f,P') - L(f,P') \color{red}{=} U(f,P) - L(f,P) < \varepsilon$$
$$Q = P' \cap [c,d] = P \cap [x_{i-1}, x_i] = \{x_{i-1}, x_i\}$$
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?