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 $0$).
$\Rightarrow$:
Suppose $A=A(\epsilon)$ is a finite union of open* intervals with disjoint closures* and total measure less than $\epsilon$ that covers all points of discontinuity of $f$. Incorporate the endpoints of the intervals in $A$ into a partition. Control the error coming from $A$ by brute force using the fact that $f$ is bounded. Meanwhile $A^c$ is a finite union of compact intervals on which $f$ is continuous; partition that as fine as needed. Sum up the errors on $A$ and $A^c$ to finish.
* Here "open" is relative to the topology of $[a,b]$ itself, not that of $\mathbb{R}$.
$\Leftarrow$:
Introduce the concept of oscillation at a point, $\omega_f(x)$. Let $D_i=\{ x : \omega_f(x) \geq a_i \}$ where $a_i$ is a sequence to be determined that decreases to zero. Then $x$ is a point of discontinuity of $f$ if and only if there exists $i$ such that $x \in D_i$. We want to show $\lambda \left ( \bigcup_{n=1}^\infty D_i \right )<\epsilon$. We'll do that by choosing our favorite $b_i>0$ with $\sum_{i=1}^\infty b_i=\epsilon$ followed by choosing $a_i$ to ensure that $\lambda(D_i)<b_i$.
Now take a partition $P_i$ with $U(f,P_i)-L(f,P_i)<c_i$ with $c_i>0$ to be determined. The subintervals of $P_i$ which have nonempty intersection with $D_i$ have some measure $d_i \geq \lambda(D_i)$. Thus it suffices to show $d_i<b_i$. These subintervals contribute at least $d_i a_i$ to $U(f,P_i)-L(f,P_i)$. Therefore $d_i a_i<c_i$ and so $d_i<\frac{c_i}{a_i}$. Thus it suffices to choose $c_i,a_i$ both going to zero such that $\frac{c_i}{a_i}<b_i$.
A way to prove that the Lebesgue integral of a (properly) Riemann integrable function is its Riemann integral starts by noting that the Riemann and Lebesgue integrals of step functions are equal (this may or may not be immediate from the definitions, it depends which definitions you chose to use). Then use the fact that finite unions of intervals have arbitrarily close measure to an arbitrary measurable set. From here you can argue that the Lebesgue integral of a simple function is arbitrarily close to the Lebesgue integral of a step function. Then you just use density of simple functions as usual.
Best Answer
Since $1/f$ is bounded (above), there is $Μ>0$ such that for all $x\in [a,b]$ we have $f(x)\geqslant M.$ Let $\varepsilon>0$. Since $f$ is integrable, there is a partition $P=\left\{ a=x_0<x_1<...<x_{n-1}<x_n=b \right\}$ of $[a,b]$ such that: $$U(f,P)-L(f,P)<\varepsilon M^2.$$ Since for $k=0,1,\ldots,n-1$: $$Μ_k\left(\frac{1}{f}\right)= \frac{1}{m_k(f)},~~~m_k\left(\frac{1}{f}\right)= \frac{1}{M_k(f)},$$ we get: $$Μ_k\left(\frac{1}{f}\right)-m_k\left(\frac{1}{f}\right)=\frac{M_k(f)-m_k(f)}{M_k(f)m_k(f)}\leqslant\frac{M_k(f)-m_k(f)}{M^2},$$since $M_k(f)\geqslant m_k(f)\geqslant M$. So: $$U\left(\frac{1}{f},P\right)-L\left(\frac{1}{f},P\right)\leqslant\frac{U(f,P)-L(f,P)}{M^2}<\varepsilon$$ and $1/f$ is integrable.