That the first condition implies the first is immediate, since (using your notation) you always have $m_i \le f(\xi_i) \le M_i$, so the sums in the second definition are caught between the $L$ und $U$ sums.
Edit in response to a comment an additional explanation is necessary here. For this direction
it suffices to show that $I^* = lim_{||P||\rightarrow 0} L(f,P)$ and $I_* = lim_{||P||\rightarrow 0} U(f,P)$ Since both parts are similar it suffices to show, e.g., the first equality.
First it is easy to see that for partitions $P\subset P^\prime$ we have $L(f,P)\le L(f,P^\prime)$. A remaining hurdle is that for two partitions we do not necessarily know that one is a subset of the other one. This is resolved by looking at common refinements:
Assume $P$ satisfies $L(f,P) > I^* - \varepsilon$ and $Q$ is an arbitrary partition. We need to show that then there is a refinement $Q^\prime$ of $Q$ such that $L(f,Q^\prime)\ge L(f,P)$ (and, consequently, $L(f,Q^\prime)>I^*-\varepsilon$).
For $Q^\prime$ one can choose the common refinement $R$: if $P=\{x_1,\ldots x_n \}$ and $Q=\{y_1,\ldots y_m \}$ then we just let $R = P\cup Q$. Since this is a refinement of both $P$ and $Q$ we have both $L(f,R)\ge L(f,P)$ as well as $L(f,R)\ge L(f,Q)$
Second edit: the original version was not correct:
For the other direction it suffices to show that if the function is integrable in the sense of the second definition then both $I_*$ and $I^*$ agree with the of the sums from the second definition. Since the reasoning is the same in both cases I'll just look at $I_*$.
So fix $\varepsilon >0$ and a given partition $P$ such that
$$|L - \sum_{i=1}^n f(\xi_i)\Delta x_i |< \varepsilon$$
if only the partition is fine enough.
Choose such a partition $P=\{x_0,\dots x_n\}$ and to $[x_{i-1},x_{i}]$ choose $\eta_i\in[x_{i-i},x_{i}]$ such that for
$m_i:=\inf \{ f(x):x\in [x_{i-1},x_i]\} $
we have $$0\le f(\eta_i)-m_i\le \frac{\varepsilon}{2n}$$
Then
\begin{eqnarray}
| L -\sum_{i=1}^n m_i \Delta x_i|
& = & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i + \sum_{i=1}^n f(\eta_i)\Delta x_i
-\sum_{i=1}^n m_i\Delta x_i| \\
&\le & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i| + \sum_{i=1}^n | f(\eta_i)
- m_i|\Delta x_i \\
& < & \frac{\varepsilon}{2} + \sum_{i=1}^n \frac{\varepsilon}{2n}=\varepsilon
\end{eqnarray}
If you 'see' that $0 <L -I_*< L -\sum_{i}m_i \Delta x_i$ then you are done here, otherwise it follows easily from the last estimate that the $\sum_i m_i \Delta x_i$ are, for any partition which is fine enough, $\varepsilon $ close to the fixed real number $L$, which of course implies that the $\sup$ over these sums exists and equals $L$ (here you need to use again the fact that you will approach the $\sup$, if it exists, if the width of the partitions goes to $0$).
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):
$f$ 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:
- 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.
- 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
- $$P' = P \cup \{c,d\} = P \cup \{x_{i-1},x_i\} = P$$
so
$$U(f,P') - L(f,P') \color{red}{=} U(f,P) - L(f,P) < \varepsilon$$
- $Q$ 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?
Best Answer
If you can establish that the Riemann-Stieltjes integral $\int_a^b\alpha \, d\alpha$ exists by hypothesis or otherwise, then it is straightforward to show that
$$I= \int_a^b \alpha \, d\alpha = \frac{1}{2}[\alpha^2(b) - \alpha^2(a)]$$
Note that for any partition $a = x_0 < x_1 < \ldots < x_n = b$ we have
$$\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]=\frac{1}{2}\sum_{j=1}^n[\alpha^2(x_j) - \alpha^2(x_{j-1})] =\frac{1}{2}\sum_{j=1}^n[\alpha(x_j) + \alpha(x_{j-1})][\alpha(x_j) - \alpha(x_{j-1})] \\ = \frac{1}{2}\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})] +\frac{1}{2}\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})] $$
Thus,
$$\left|\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]-I\right|\leqslant \\ \frac{1}{2}\left|\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})]-I\right| +\frac{1}{2}\left|\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})]-I\right|$$
Note that the sums on the RHS are both Riemann-Stieltjes sums. Since the Riemann-Stieltjes integral, $I$, exists, for any $\epsilon > 0$ there exists a partition such that
$$\left|\sum_{j=1}^n\alpha(x_j)[\alpha(x_j) - \alpha(x_{j-1})]-I\right|< \epsilon, \\\left|\sum_{j=1}^n \alpha(x_{j-1})[\alpha(x_j) - \alpha(x_{j-1})]-I\right| < \epsilon$$
Hence, for any $\epsilon > 0$ we have
$$\left|\frac{1}{2}[\alpha^2(b) - \alpha^2(a)]-I\right|< \epsilon,$$
and it follows that
$$I = \frac{1}{2}[\alpha^2(b) - \alpha^2(a)]$$
If $\alpha$ is increasing then it might have jump discontinuities in which case the RS integral of $\alpha $ with respect to $\alpha$ cannot exist. To carry on we must assume that $\alpha$ is continuous. Then it is not difficult to show that $\int_a^b\alpha \, d\alpha$ exists using both continuity and the fact that $\alpha$ is increasing (or more generally because it is of bounded variation).