# Show $f$ is integrable in each subinterval, $[x_{i−1}, x_i]$, and further, $\int_a^bfdx=\sum\limits_{x_{k-1}}^{x_k}\int\limits_{x_{k-1}}^{x_k}fdx$.

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 $\int_a^bf(x)dx=L$

• 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 $\int_{a}^{b}$). – kolobokish Nov 6 '16 at 14:33

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?